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This video is about the exponent rules, rules
that govern expressions like two to the fifth,

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or x to the n.

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two to the fifth is just shorthand for two
times two times two times two times two, written

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five times. And similarly x to the n is just
x multiplied by itself. And times

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when we write these expressions, the number
on the bottom that's being multiplied by itself

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is called the base. And the number at the
top telling us how many times we're multiplying

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the base by itself is called the exponent.

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Sometimes the exponent is also called the
power.

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The product rule says that a 5x to the power
of n times x to the power of m, that's the

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same thing as x to the n plus m power. In
other words, when I multiply two expressions

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with the same base,

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then I can add their exponents.

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For example, if I have two cubed times two
to the fourth, that's equal to two to the

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seventh. And that makes sense, because two
cubed times two to the fourth, means I multiply

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two by itself three times. And then I multiply
that by two multiplied by itself four times.

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And in the end, I have to multiplied by itself
seven times,

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which is two to the seventh,

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I'm just adding up the number of times as
multiplied in each piece, to get the number

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of times as multiplied total.

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The quotient rule says that if I have x to
the n power divided by x to the m power, that's

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equal to x to the n minus m power. In other
words, if I divide two expressions with the

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same base,

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then I can subtract their exponents.

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For example, three to the six divided by three
squared is going to be three to the six minus

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two, or three to the fourth. And this makes
sense, because three to the six means I multiply

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three by itself six times, and then I divide
that by three multiplied by itself twice.

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So when I cancel out threes, I have four threes
left,

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notice that I have to subtract the number
of threes on the bottom, from the number of

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threes at the top to get my number of threes
remaining. That's why subtract my exponents.

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The power rule tells us if I have x to the
n power raised to the m power, that's the

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same thing as x to the n times M power. In
other words, when I raise a power to a power,

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I get to multiply the exponents.

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For example, five to the fourth cubed is equal
to five to the four times three, or five to

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the 12th. And this makes sense, because five
to the fourth cubed can be thought of as five

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to the fourth times five to the fourth, times
five to the fourth, expanding this out some

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more, that's five times five times five times
five times the same thing times the same thing

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again.

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So I have three groups of four, or five, which
is a total of three times four, or 12. fives.

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The next rule involves what happens when I
raise a number, or a variable to the zeroeth

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power, it turns out that anything to the zeroeth
power is equal to one.

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Usually, this is just taken as a definition.
But here's why it makes sense to me.

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If you have something like two cubed divided
by two cubed, Well, certainly that has to

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equal one, anything divided by itself is just
one. But using the quotient rule,

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we know that this is the same thing as two
to the three minus three, because when we

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divide two things with the same base, we get
to subtract their exponents. Therefore, this

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is the same thing as two to the zero. So two
to the zero has to equal one in order to make

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it work with the quotient rule.

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And the same argument shows that anything
to the zero power

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has to be equal to one.

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What happens when we take something to a negative
power,

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x to the n is equal to one over x to the n.
Again, most people just take this as a definition

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of a negative exponent. But here's why it
makes sense.

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If I take something like five to seven times
five to the negative seven, then buy the product

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roll the SAS to equal five to the seven plus
negative seven, which is five to the zero,

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and we just said that that is equal to one.

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Now I have the equation of five to the seventh
times five to the negative seventh equals

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one, if I divide both sides by five, the seventh,
I get that five to the negative seventh has

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to equal one over five to the seventh. So
that's where this rule about negative exponents

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comes from. That has to be true in order to
be consistent with the product rule.

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Finally, let's look at a fractional exponents.
What does an expression like x to the one

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over N really mean? Well, it means the nth
root of x, for example, 64 to the 1/3 power

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means the cube root of 64, which happens to
be four, and nine to the one half means the

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square root of nine, which is usually written
without that little superscript up there.

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Now, the square root of nine is just three.

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fractional exponents also makes sense. For
example, if I have five to the 1/3, and I

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cube that, then by the power role, that's
equal to five to the 1/3, times three,

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which is just five to the one or five. So
in other words, five to the 1/3, is the number

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that when you cube it, you get five. And that's
exactly what's meant by the cube root of five,

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the cube root of five is also a number that
when you cube it, you get five.

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The next rule tells us we can distribute an
exponent over a product.

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In other words, if we have a product, x times
y, all raised to the nth power, that's equal

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to x to the n times y to the N.

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For example, five times seven,

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all raised to the third power is equal to
five cubed times seven cubed. And this makes

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sense, because five times seven, all raised
to the cube power can be expanded as five

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times seven, times five times seven, times
five times seven. But if I rearrange the order

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of multiplication, this is the same thing
as five times five times five, times seven

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times seven times seven, or five cubed times
seven cubed.

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Similarly, we could distribute an exponent
over a quotient, if we have the quotient X

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over Y, all raised to the n power, that's
the same as x to the n over y to the N. For

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example, two sevenths raised to the fifth
power is the same thing as two to the fifth

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over seven to the fifth. This makes sense,
because two sevens to the fifth can be expanded

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as two sevens multiply by itself five times,
which can be written rewritten as two multiplied

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by itself five times divided by seven multiplied
by itself five times, and that's two to the

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fifth, over seven to the fifth, as wanted.

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We've seen that we can distribute an exponent

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over multiplication, and division.

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But be careful, because we cannot distribute
next bowknot.

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over addition, or subtraction, for example,
a plus b to the n is not generally equal to

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a to the n plus b to the n, a minus b to the
n is not generally equal to a to the n minus

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b to the n. And if you're not sure, just try
an example with numbers.

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For example, two plus three squared is not
the same thing as two squared plus three squared,

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and two minus three squared is definitely
not equal to two squared minus three squared.

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In this video, I gave eight exponent rules,
which I'll list again here. There's the product

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rule,

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the quotient rule, the power rule,

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the zero exponent, the negative exponent,
the fractional exponent,

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and the two rules involving distributing exponents.

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Across multiplication,

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and division.

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In another video, I'll use these exponent
rules to rewrite and simplify expressions

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involving exponents.

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In this video, I'll work out some examples
of simplifying expressions using exponent

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rules.

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I'll start by reviewing the exponent rules.

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The product rule says that when you multiply
two expressions with the same base, you add

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the exponents. The quotient rule says that
when you divide two expressions with the same

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base, you subtract the exponents. The power
rule says that when you take a power to a

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power, you multiply the exponents. The power
of zero rule says that anything to the zero

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power is one, as long as the base is not zero.

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Since zero to the zero is undefined, it doesn't
make sense.

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negative exponents to evaluate x to the minus
n, we take the reciprocal one over x to the

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n. To evaluate a fractional exponent, like
x to the one over n, we take the nth root

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of x,

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we can distribute an exponent over a product,
a times b to the n is equal to A to the N

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times b to the n. And we can distribute an
exponent over a quotient a over b to the n

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is a to the n over b to the n.

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In the rest of this video, we'll use these
exponent rules to simplify expressions.

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For our first example, we want to simplify
three times x to the minus two divided by

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x to the fourth, there's several possible
ways to proceed.

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For example, we could use the negative exponent
rule dr x to the minus two as one over x squared,

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all that gets divided by x to the fourth,
still,

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notice that we only take the reciprocal of
the x squared, the three stays where it is.

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And that's because the exponent of negative
two only applies to the x not to the three.

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Now if we think of three as three over one,
we have a product of two fractions and our

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numerator. And so we evaluate that by taking
the product of the numerators times the product

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of the denominators, which is three over x
squared, all divided by x to the fourth,

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I can think of x to the fourth as x to the
fourth over one. So now I have a fraction

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of a fraction, which I can evaluate by multiplying
by the reciprocal, that simplifies to three

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times one divided by x squared times x to
the fourth, which is three over x to the six,

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using the product rule. Since x squared times
x to the fourth is equal to x to the two plus

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four, or x to the six.

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An alternate way of solving this problem

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is to start by using the quotient rule,

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I can rewrite this as three times x to the
minus two over x to the fourth, and by the

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quotient rule, that's three times x to the
minus two minus four, or three times x to

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the minus six.

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Now using the negative exponent roll, x to
the minus six is one over x to the six. And

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this product of fractions simplifies to three
over x to the six, the same answer I got before.

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The second problem can be solved in similar
ways. Please pause the video and try it before

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going on.

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One way to simplify would be to use the negative
exponent rule first, and rewrite y to the

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minus five as one over wide the fifth.

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thinking of this as a fraction, divided by
a fraction, I can multiply by the reciprocal

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and get four y cubed y to the fifth over one
by the product rule. The numerator here is

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four y to the eight. And so my final answer
is just for one of the eight.

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Alternatively, I could decide to use the quotient
rule first.

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As in the previous problem, I can write this
as for y cubed minus negative five by the

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quotient rule. And so that's for y to the
eighth as before.

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I'd like to show you one more method to solve
these two problems, kind of a shortcut method

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before we

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To go on, that shortcut relies on the principle
that a negative exponent in the numerator

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corresponds to a positive exponent in the
denominator. For example, the x to the negative

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two in the numerator here, after some manipulations
became an X to the positive two in the denominator.

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Furthermore, a negative exponent in the denominator

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is equivalent to a positive exponent in the
numerator.

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That's what happened when we had the y to
the negative five in the denominator, and

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translated into a y to the positive five in
the numerator.

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Sometimes people like to talk about this principle,
by saying that you can pass a factor

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across the fraction bar

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by switching the sine of the exponent that
is making a positive exponent negative, or

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a negative exponent positive.

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Let's see how this principle gives us a shortcut
for solving these two problems.

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In the first problem, 3x to the minus two
over x to the four, we can move the negative

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exponent in the numerator and make it a positive
exponent the denominator, so we get three

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over x to the four plus two or x to the six.

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In the second example, for y cubed over y
to the minus five, we can change the Y to

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the minus five in the denominator into a y
to the five in the numerator and get our final

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answer of four y to the three plus five or
eight.

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We'll use this principle again in the next
problems.

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In this example, notice that I have I have
y's in the numerator and the denominator,

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and also Z's in the numerator and the denominator.
In order to simplify, I'm going to try to

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get all my y's either in the numerator or
the denominator. And similarly for the Z's.

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Since I have more y's in the denominator,
let me move this y to the three downstairs

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and make it a y to the negative three. I'm
using the principle here that a positive exponent

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in the numerator corresponds to a negative
exponent in the denominator.

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Now since I have a positive exponent, z in
the numerator and a negative exponent denominator,

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and I want to get rid of negative exponents,
I'm going to pass the Z's to the numerator

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as e to the minus two in the denominator,
it comes as e to the plus two in the numerator.

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Notice that my number seven doesn't move.
And when I do any of these manipulations,

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because it doesn't have an exponent, and the
exponent of negative two, for example, only

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applies to the Z not to the seven.

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Now that I've got all my Z's in the numerator
and all my y's in the denominator, it's easy

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to clean this up using the product rule.

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And I have my simplified expression.

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In this last example, we have a complicated
expression raised to a fractional power.

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I'm going to start by simplifying the expression
inside the parentheses.

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I can bring all my y's downstairs and all
my x's upstairs and get rid of negative exponents

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at the same time.

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In other words, I can rewrite this as 25x
to the fourth,

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I'll bring the Y to the minus five downstairs
and make it y to the fifth on the denominator,

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bring the x to the minus six upstairs and
make it x to the sixth the numerator, and

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then I still have the Y cubed on the denominator,
all that's raised to the three halves power.

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Using the product rule, I can rewrite the
expression on the inside of the parentheses

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as 25x to the 10th over y to the eighth.

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Recall that we're allowed to distribute an
exponent across a product or across a quotient.

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When I distribute my three halves power,

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I get 25 to the three halves times x to the
10th to the three halves divided by y to the

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eighth to the three halves.

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Now the power rule tells me when I have a
power to a power, I get to multiply the exponents.

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So I can rewrite this as 25 to the three halves
times x to the 10 times three halves of our

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y to the eight times three halves. In other
words, 25 to the three halves times x to the

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15th over y to the 12th. Finally, I need to
rewrite 25 to the three halves. Since three

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halves can be thought of as three times one
half, or as one half times three, I can write

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25 to the three halves as 25 to the three
times one half, or as 25 to the one half times

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00:20:27,019 --> 00:20:28,019
three.

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00:20:28,019 --> 00:20:33,299
Well, using the power rule in reverse, I can
think of this as 25 cubed to the one half,

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or as 25 to the one half cubed. Since when
I take a power to a power, I multiply the

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00:20:39,710 --> 00:20:40,710
exponents

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00:20:40,710 --> 00:20:47,819
25 cubed to the one half might be hard to
evaluate, since 25 cubed is a huge number,

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00:20:47,819 --> 00:20:58,439
but 25 to the one half is just the square
root of 25. So I have the square root of 25

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00:20:58,440 --> 00:21:02,100
cubed, or five cubed, which is 125.

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00:21:02,099 --> 00:21:09,139
Therefore, my original expression is going
to be 120 5x to the 15 over y to the 12th.

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In this video, we use the exponent rules to
simplify complicated expressions.

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This video goes through a few tricks for simplifying
expressions with radicals in them.

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00:21:23,799 --> 00:21:30,669
Recall that this notation means the nth root
of x, so this notation here means the cube

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00:21:30,670 --> 00:21:38,360
root of eight, the number that when you cube
it, you get eight, that number would be two,

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00:21:38,359 --> 00:21:44,079
when we write the root sign without a little
number, that just means the square root so

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00:21:44,079 --> 00:21:47,230
the two is implied.

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00:21:47,230 --> 00:21:54,599
In this case, the square root of 25 is five
since five squared is 25.

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00:21:54,599 --> 00:22:00,149
Let's start by reviewing some rules for radical
expressions. First, if we have the radical

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00:22:00,150 --> 00:22:06,940
of a product, we can rewrite that as the product
of two radicals.

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00:22:06,940 --> 00:22:14,830
For example, the square root of nine times
16 is the same thing as the square root of

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00:22:14,829 --> 00:22:21,179
nine times the square root of 16, you can
check that both of these evaluate to 12.

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00:22:21,180 --> 00:22:29,150
Similarly, it's possible to distribute a radical
sine across division, the radical of a divided

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00:22:29,150 --> 00:22:37,350
by b is the same thing as the radical of A
divided by the radical of B. For example,

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00:22:37,349 --> 00:22:45,299
the cube root of 64 over eight is the same
thing as the cube root of 64 over the cube

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00:22:45,299 --> 00:22:48,980
root of eight, and you can check that both
of these evaluated to

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00:22:48,980 --> 00:22:57,039
you have to be a little bit careful though,
because it's not okay to distribute a radical

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00:22:57,039 --> 00:23:03,819
sign across addition. In general, the nth
root of a plus b is not equal to the nth root

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00:23:03,819 --> 00:23:11,179
of A plus the nth root of b. And similarly,
it's not okay to distribute a radical across

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00:23:11,180 --> 00:23:14,130
subtraction.

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00:23:14,130 --> 00:23:19,450
If you're ever in doubt, you can always check
with simple examples. For example, the square

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00:23:19,450 --> 00:23:24,490
root of one plus one is not the same thing
as the square root of one plus the square

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00:23:24,490 --> 00:23:26,450
root of one,

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00:23:26,450 --> 00:23:32,269
the right side evaluates to one plus one or
two, and the left side is square root of two

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00:23:32,269 --> 00:23:34,650
and the irrational number.

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00:23:34,650 --> 00:23:40,019
The second expression to show that that fails,
I don't think it'll work to use the square

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00:23:40,019 --> 00:23:44,889
root of one minus one it'll actually hold
in that case, but I can show it's false by

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00:23:44,890 --> 00:23:49,890
using say the square root of two minus one,
which does not equal the square root of two

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00:23:49,890 --> 00:23:53,050
minus the square root of one.

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00:23:53,049 --> 00:23:57,240
You might notice that these Rules for Radicals,
the ones that hold and the ones that don't

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00:23:57,240 --> 00:24:04,420
hold, remind you of rules for exponents. And
that's no coincidence. Because radicals can

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00:24:04,420 --> 00:24:09,920
be written in terms of exponents. For example,
if we look at the first rule, we can rewrite

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00:24:09,920 --> 00:24:19,000
this, the nth root of A times B is the same
thing as the one of our nth power. And by

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00:24:19,000 --> 00:24:26,099
exponent rules, I can distribute an exponent
across multiplication. And so this radical

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00:24:26,099 --> 00:24:34,819
rule can be restated completely in terms of
an exponent rule. Similarly, the second rule

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00:24:34,819 --> 00:24:41,119
can be restated in terms of exponents as a
or b to the one over n is equal to A to the

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00:24:41,119 --> 00:24:46,779
one over n divided by b to the one over n.
We can use the relationship between radicals

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00:24:46,779 --> 00:24:54,700
and the exponents to rewrite a to the m over
n. a to the m over N is the same thing as

251
00:24:54,700 --> 00:25:00,250
a to the m with the N through taken. That's
also the same thing

252
00:25:00,250 --> 00:25:06,640
As the nth root of A, all taken to the nth
power to see whether that's true, think about

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00:25:06,640 --> 00:25:13,011
exponent rules. So a to the m over N is the
same thing as a to the m, taken to the one

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00:25:13,010 --> 00:25:19,029
over nth power. That's because when we take
a power to a power, we multiply exponents,

255
00:25:19,029 --> 00:25:25,680
and M times one over n is equal to M over
n.

256
00:25:25,680 --> 00:25:31,660
But a one over nth power is the same thing
as an nth root. And therefore, this expression

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00:25:31,660 --> 00:25:39,070
is the same thing as this expression. And
that proves the first equivalence. The second

258
00:25:39,069 --> 00:25:48,169
equivalence can we prove similarly, by writing
a to the m over n as a to the one over n times

259
00:25:48,170 --> 00:25:55,380
M. Again, this is works because when I take
the power to the power, I'm multiply exponents,

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00:25:55,380 --> 00:26:02,330
one over n times M is the same thing as M
over n. But now, these two expressions are

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00:26:02,329 --> 00:26:08,119
the same, because the one over nth power is
the same as the nth root.

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00:26:08,119 --> 00:26:15,729
One mnemonic for remembering these relationships
is flower over root. So flour is like power,

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00:26:15,730 --> 00:26:20,779
and root is like root, so that tells us we
can write a fractional exponent, the M becomes

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00:26:20,779 --> 00:26:26,259
the power, and the n becomes the root in either
of these two orders.

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00:26:26,259 --> 00:26:33,109
Now let's use these rules in some examples.
If we want to compute 25 to the negative three

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00:26:33,109 --> 00:26:39,159
halves power, well, first I'll use my exponent
rule to rewrite that negative exponent as

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00:26:39,160 --> 00:26:43,029
one over 25 to the three halves power.

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00:26:43,029 --> 00:26:52,839
Next, I'll use the power of a root mnemonic
to rewrite this as 25 to the third power square

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00:26:52,839 --> 00:26:59,500
rooted, or, as 25 square rooted to the third
power.

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00:26:59,500 --> 00:27:04,650
I wrote the two's there for the square root
for emphasis, but most of the time, people

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00:27:04,650 --> 00:27:09,900
will omit this and just write the square root
without a little number there.

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00:27:09,900 --> 00:27:15,750
Now, I could use either of these two equivalent
expressions to continue, but I'd rather use

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00:27:15,750 --> 00:27:20,569
this one because it's easier to compute without
a calculator. The square root of 25 is just

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00:27:20,569 --> 00:27:28,589
five, five cubed is 125. So my answer is one
over 125.

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00:27:28,589 --> 00:27:34,750
If I tried to compute the cube of 25, first,
I'd get a huge number. In general, it's usually

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00:27:34,750 --> 00:27:41,859
easier to compute the route before the power
when you're working without a calculator.

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00:27:41,859 --> 00:27:46,209
Now let's do an example simplifying a more
complicated expression with with exponent

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00:27:46,210 --> 00:27:52,450
cynet. I want to take the square root of all
this stuff. And since I don't really like

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00:27:52,450 --> 00:27:58,471
negative exponents, I'm first going to rewrite
this as the square root of 60x squared y to

280
00:27:58,471 --> 00:28:04,319
the sixth over z to the 11th. So I'll change
that negative exponent to a positive exponent

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00:28:04,319 --> 00:28:07,879
by moving this, this factor to the denominator.

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00:28:07,880 --> 00:28:13,450
Now, when you're asked to simplify radical
expression, that generally means to pull as

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00:28:13,450 --> 00:28:17,950
much as possible, out of the radical side.

284
00:28:17,950 --> 00:28:23,000
To pull things out of the square root side,
I'm going to factor my numbers and try to

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00:28:23,000 --> 00:28:28,410
rewrite everything in terms of squares as
much as possible. Since the square root of

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00:28:28,410 --> 00:28:34,890
a square, those two operations undo each other.
So I'll show you what I mean first, our factor

287
00:28:34,890 --> 00:28:43,210
60. So 60 is going to be two squared times
three times five. And I'll just copy everything

288
00:28:43,210 --> 00:28:47,259
over for now.

289
00:28:47,259 --> 00:28:52,690
Now I'll break things up into squares as much
as possible. So I've got a two squared, and

290
00:28:52,690 --> 00:28:57,360
three times five, I've already got an x squared,
I write the wider the six as y squared times

291
00:28:57,359 --> 00:29:03,579
y squared times y squared. And I'll write
the Z to the 11th as z squared times z squared,

292
00:29:03,579 --> 00:29:12,048
I guess five times 12345 times when extra
z, that should add up to z to the 11th. I

293
00:29:12,048 --> 00:29:15,410
add all those exponents together.

294
00:29:15,410 --> 00:29:20,430
Now I know that I can distribute my radical
sign across multiplication and division. So

295
00:29:20,430 --> 00:29:25,860
I'll write this with a zillion different radicals
here.

296
00:29:25,859 --> 00:29:32,479
And every time I see the square root of something
squared, I can just cancel those square roots

297
00:29:32,480 --> 00:29:37,410
and the squares out and get what's what's
left here. So So after doing that cancellation,

298
00:29:37,410 --> 00:29:46,080
I get two times the square root of three times
five times x times y times y times y over

299
00:29:46,079 --> 00:29:53,569
z times itself, I guess five times times the
square root of z. And now I can clean that

300
00:29:53,569 --> 00:29:59,069
up with exponents. I'll write that as the
square root of 15 I guess two times a squared

301
00:29:59,069 --> 00:30:00,189
of 15

302
00:30:00,190 --> 00:30:07,950
times x times y cubed over z to the fifth
the square root of z.

303
00:30:07,950 --> 00:30:15,480
I'm gonna leave this example as is. But sometimes
people prefer to rewrite radical expressions

304
00:30:15,480 --> 00:30:20,769
without radical signs in the denominator.
That's called rationalizing the denominator.

305
00:30:20,769 --> 00:30:24,789
I won't do it here, but I'll show you how
to do it in the next example.

306
00:30:24,789 --> 00:30:31,859
This example asks us to rationalize the denominator,
that means to rewrite as an equivalent expression

307
00:30:31,859 --> 00:30:36,549
without radical signs in the denominator.

308
00:30:36,549 --> 00:30:44,069
To get rid of the radical sine and the denominator,
I want to multiply my denominator by square

309
00:30:44,069 --> 00:30:49,389
root of x. But I can't just multiply the denominator
willy nilly by something unless I multiply

310
00:30:49,390 --> 00:30:55,240
the numerator by the same thing. So then just
multiply my expression by one in a fancy forum

311
00:30:55,240 --> 00:30:57,529
and I don't change the value of my expression.

312
00:30:57,529 --> 00:31:06,319
Now, if I just multiply together numerators
3x squared of x and multiply denominators

313
00:31:06,319 --> 00:31:12,019
squared of x 10 squared of x is the square
root of x squared squared of x squared is

314
00:31:12,019 --> 00:31:14,210
just x.

315
00:31:14,210 --> 00:31:19,620
Now I can cancel my access from the numerator
denominator, and my final answer is three

316
00:31:19,619 --> 00:31:24,750
times the square root of x. I rationalize
my denominator, and the process got a nicer

317
00:31:24,750 --> 00:31:26,400
looking expression.

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00:31:26,400 --> 00:31:34,110
In this video, we went over the Rules for
Radicals. And we simplified some radical expressions

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00:31:34,109 --> 00:31:37,889
by working with fractional exponents,

320
00:31:37,890 --> 00:31:41,360
pulling things out of the radical sign

321
00:31:41,359 --> 00:31:46,159
and rationalizing the denominator.

322
00:31:46,160 --> 00:31:51,170
This video goes over some common methods of
factoring. Recall that factoring an expression

323
00:31:51,170 --> 00:31:56,890
means to write it as a product. So we could
factor the number 30, by writing it as six

324
00:31:56,890 --> 00:32:04,670
times five, we could factor it more completely
by writing it as two times three times five.

325
00:32:04,670 --> 00:32:12,670
As another example, we could factor the expression
x squared plus 5x plus six by writing it as

326
00:32:12,670 --> 00:32:16,500
x plus two times x plus three.

327
00:32:16,500 --> 00:32:23,700
In this video, I'll go over how I get from
here to here, how I know how to do the factoring.

328
00:32:23,700 --> 00:32:29,340
But for right now, I just want to review how
I can go backwards how I can check that the

329
00:32:29,339 --> 00:32:35,259
factoring is correct. And that's just by multiplying
out or distributing. If I distribute x plus

330
00:32:35,259 --> 00:32:40,799
two times x plus three, then I multiply x
by x, that gives me x squared, x times three

331
00:32:40,799 --> 00:32:48,119
gives me 3x. Two times x gives me 2x. And
two times three gives me six. So that simplifies

332
00:32:48,119 --> 00:32:54,569
to x squared plus 5x plus six, which checks
out with what I started with. So you can think

333
00:32:54,569 --> 00:33:00,240
of factoring as the opposite of distributing
out. And you can always check your factoring

334
00:33:00,240 --> 00:33:01,730
by distributing or multiplying out

335
00:33:01,730 --> 00:33:09,140
a bit of terminology, when I think of an expression
as a sum of a bunch of things, then the things

336
00:33:09,140 --> 00:33:17,610
I sum up are called the terms. But if I think
of the same expression as a product of things,

337
00:33:17,609 --> 00:33:22,929
then the things that I multiply together are
called factors.

338
00:33:22,930 --> 00:33:28,350
Now let's get started on techniques of factoring.
When I have to factor something, I always

339
00:33:28,349 --> 00:33:33,339
like to start by pulling out the greatest
common factor, the greatest common factor

340
00:33:33,339 --> 00:33:38,269
means the largest thing that divides each
of the terms.

341
00:33:38,269 --> 00:33:46,069
In this first example, the largest thing that
divides both 15 and 25x is five.

342
00:33:46,069 --> 00:33:53,990
So the GCF is five. So I pull the five out,
and then I divide each of the terms by that

343
00:33:53,990 --> 00:33:58,769
number. And so I get three plus 5x.

344
00:33:58,769 --> 00:34:04,990
Pause the video for a moment and see if you
can find the greatest common factor of x squared

345
00:34:04,990 --> 00:34:08,690
y and y squared x cubed.

346
00:34:08,690 --> 00:34:15,648
The biggest thing that divides both x squared
y and y squared x cubed is going to be x squared

347
00:34:15,648 --> 00:34:18,098
times y.

348
00:34:18,099 --> 00:34:24,030
One way to find this is to look for the power
of x that smallest in each of these terms.

349
00:34:24,030 --> 00:34:29,149
So that's x squared. And the power of y that
smallest in each of these terms is just y

350
00:34:29,148 --> 00:34:31,710
to the one or y.

351
00:34:31,710 --> 00:34:38,599
Now if I factor out the x squared y from each
of the terms, that's like dividing each term

352
00:34:38,599 --> 00:34:44,129
by x squared y, if I divide the first term
by x squared y, I just get one. If I divide

353
00:34:44,128 --> 00:34:50,878
the second term by x squared y, I'm going
to be left with an X and a Y. I'll write this

354
00:34:50,878 --> 00:34:57,199
out on the side just to make it more clear,
y squared x cubed over x squared y. That's

355
00:34:57,199 --> 00:34:59,980
like two y's on the end.

356
00:34:59,980 --> 00:35:07,838
Three x's on the top and two x's and a y on
the bottom. So I'm left with just an X and

357
00:35:07,838 --> 00:35:16,130
a Y. So I'll write the x, y here, and I factored
my expression. As always, I can check my answer

358
00:35:16,130 --> 00:35:23,329
by multiplying out. So if I multiply out my
factored expression, I get x squared y is

359
00:35:23,329 --> 00:35:28,849
the first term and the second term, I get
x there. Now let's see three X's multiplied

360
00:35:28,849 --> 00:35:35,710
together and two y's multiplied together.
And that checks out with what I started with.

361
00:35:35,710 --> 00:35:40,880
The next technique of factoring, I'd like
to go over his factoring by grouping. In this

362
00:35:40,880 --> 00:35:46,940
example, notice that we have four terms, factoring
by grouping is a handy method to look at.

363
00:35:46,940 --> 00:35:50,200
If you have four terms in your expression,
you need to factor

364
00:35:50,199 --> 00:35:56,368
in order to factor by grouping, I'm first
going to factor out the greatest common factor

365
00:35:56,369 --> 00:36:00,820
of the first two terms, and then separately,
factor out the greatest common factor of the

366
00:36:00,820 --> 00:36:08,318
last two terms. The greatest common factor
of x cubed, and 3x squared is x squared. So

367
00:36:08,318 --> 00:36:14,518
I factor out the x squared, and I get x plus
three. And now the greatest common factor

368
00:36:14,518 --> 00:36:23,939
of forex and 12 is just four. So I factor
out the four from those two terms.

369
00:36:23,940 --> 00:36:31,009
Notice that the factor of x plus three now
appears in both pieces. So I can factor out

370
00:36:31,009 --> 00:36:36,039
the greatest common factor of x plus three,
and I'll factor it out on the left side instead

371
00:36:36,039 --> 00:36:45,549
of the right. And now I have an x squared
from this first piece, and I have a four from

372
00:36:45,548 --> 00:36:52,170
this second piece. And that completes my factoring
by grouping. You might wonder if we could

373
00:36:52,170 --> 00:36:57,320
factor further by factoring the expression
x squared plus four. But in fact, as we'll

374
00:36:57,320 --> 00:37:03,269
see later, this expression, which is a sum
of two squares, x squared plus two squared

375
00:37:03,268 --> 00:37:06,078
does not factor any further over the integers.

376
00:37:06,079 --> 00:37:13,220
Next, we'll do some factoring of quadratics.
a quadratic is an expression with a squared

377
00:37:13,219 --> 00:37:15,209
term,

378
00:37:15,210 --> 00:37:20,480
just a term with x in it, and a constant term
with no x's in it.

379
00:37:20,480 --> 00:37:29,500
I'd like to factor this expression as a product
of x plus or minus some number times x plus

380
00:37:29,500 --> 00:37:32,289
or minus some other number.

381
00:37:32,289 --> 00:37:37,329
The key idea is that if I can find those two
numbers, then if I were to distribute out

382
00:37:37,329 --> 00:37:41,440
this expression, those two numbers would have
to multiply to give me my constant term of

383
00:37:41,440 --> 00:37:48,889
eight. And these two numbers would end up
having to add to give me my negative six,

384
00:37:48,889 --> 00:37:53,319
because when I multiply out, this number will
be a coefficient of x, and this number will

385
00:37:53,320 --> 00:37:59,509
be also another coefficient of x, they'll
add together to the negative six.

386
00:37:59,509 --> 00:38:04,820
So if I look at all the pairs of numbers that
multiply together to give me eight, so that

387
00:38:04,820 --> 00:38:11,519
could be one and eight, two, and four, four
and two, but that's really the same thing

388
00:38:11,519 --> 00:38:16,340
as I had before. And that's sort of the same
thing I had before. I shouldn't forget the

389
00:38:16,340 --> 00:38:20,470
negatives, I could have negative one, negative
eight, or I could have negative two, negative

390
00:38:20,469 --> 00:38:25,399
four, those alternate multiply together to
give me eight. Now I just have to find, see

391
00:38:25,400 --> 00:38:31,579
if there's a pair of these numbers that add
to negative six, and it's not hard to see

392
00:38:31,579 --> 00:38:39,190
that these ones will work. So now I can write
out my factoring as that would be x minus

393
00:38:39,190 --> 00:38:45,720
two times x minus four. And it's always a
good idea to check by multiplying out, I'm

394
00:38:45,719 --> 00:38:53,558
going to get x squared minus 4x minus 2x,
plus eight. And that works out to just what

395
00:38:53,559 --> 00:39:00,730
I want. Now this second examples a bit more
complicated, because now my leading coefficient,

396
00:39:00,730 --> 00:39:04,750
my coefficient of x squared is not just one,
it's the number 10.

397
00:39:04,750 --> 00:39:08,789
Now, there are lots of different methods for
approaching a problem like this. And I'm just

398
00:39:08,789 --> 00:39:13,889
going to show you one method, my favorite
method that uses factoring by grouping, but

399
00:39:13,889 --> 00:39:18,139
but to start out, I'm going to multiply my
coefficient of x squared by my constant term,

400
00:39:18,139 --> 00:39:25,548
so I'm multiplying 10 by negative six, that
gives me negative 60. And I'll also take my

401
00:39:25,548 --> 00:39:31,690
coefficient of x the number 11. And write
that down here. Now I'm going to look for

402
00:39:31,690 --> 00:39:38,280
two numbers that multiply to give me negative
60. And add to give me 11. You might notice

403
00:39:38,280 --> 00:39:42,410
that this is exactly what we were doing in
the previous problem. It's just here, we didn't

404
00:39:42,409 --> 00:39:47,210
have to multiply the coefficient of x squared
by eight, because the coefficient of x squared

405
00:39:47,210 --> 00:39:53,769
was just one. So to find the two numbers that
multiply to negative 60 and add to 11, you

406
00:39:53,769 --> 00:39:58,528
might just be able to come up with them in
your head thinking about it, but if not, you

407
00:39:58,528 --> 00:40:00,550
can figure it out. Pretty simple.

408
00:40:00,550 --> 00:40:05,528
thematically by writing out all the factors,
pairs of factors that multiply to negative

409
00:40:05,528 --> 00:40:12,579
60. So I can start with negative one and 60,
negative two and 30, negative three and 20.

410
00:40:12,579 --> 00:40:19,960
And keep going like this until I have found
factors that actually add together to give

411
00:40:19,960 --> 00:40:28,318
me the number 11. And, and now that I look
at it, I've already found them. 15 minus four,

412
00:40:28,318 --> 00:40:34,159
gives me 11. So I don't have to continue with
my chart of factors. Now, once I found those

413
00:40:34,159 --> 00:40:40,858
factors, I write out my expression 10x squared,
but instead of writing 11x, I write negative

414
00:40:40,858 --> 00:40:49,210
4x plus 15x. Now I copy down the negative
six, notice that negative 4x plus 15x equals

415
00:40:49,210 --> 00:40:55,679
11x. That's how I chose those numbers. And
so this expression is evaluates is the same

416
00:40:55,679 --> 00:41:01,018
as, as this expression, I haven't changed
my expression. But I have turned it into something

417
00:41:01,018 --> 00:41:05,250
that I can apply factoring by grouping on
Look, I've got four terms here. And so if

418
00:41:05,250 --> 00:41:11,139
I factor out my greatest common factor of
my first two terms, that's let's see, I think

419
00:41:11,139 --> 00:41:18,009
it's 2x. So I factor out the 2x, I get 5x
minus two, and then I factor out the greatest

420
00:41:18,010 --> 00:41:25,160
common factor of 15x and negative six, that
would be three, and I get a 5x minus two,

421
00:41:25,159 --> 00:41:31,409
again, this is working beautifully. So I have
a 5x minus two in each part. And so I put

422
00:41:31,409 --> 00:41:38,920
the 5x minus two on the right, and I put what's
left from these terms in here. So that's 2x

423
00:41:38,920 --> 00:41:43,710
plus three. And I have factored my expression.

424
00:41:43,710 --> 00:41:50,929
There are a couple special kinds of expressions
that appear frequently, that it's handy to

425
00:41:50,929 --> 00:41:55,149
just memorize the formula for. So the first
one is the difference of squares. If you see

426
00:41:55,150 --> 00:42:02,460
something of the form a squared minus b squared,
then you can factor that as a plus b times

427
00:42:02,460 --> 00:42:08,230
a minus b. And let's just check that that
works. If I do a plus b times a minus b and

428
00:42:08,230 --> 00:42:17,358
multiply that out, I get a squared minus a
b plus b A minus B squared, and those middle

429
00:42:17,358 --> 00:42:21,710
two terms cancel out. So it gives me back
the difference of squares just like I want

430
00:42:21,710 --> 00:42:30,590
it. So for this first example, I if I think
of x squared minus 16 as x squared minus four

431
00:42:30,590 --> 00:42:34,170
squared, then I can see that's a difference
of squares. And I can immediately write it

432
00:42:34,170 --> 00:42:41,329
as x plus four times x minus four. And the
second example, nine p squared minus one,

433
00:42:41,329 --> 00:42:50,140
that's the same thing as three p squared minus
one squared. So that's three p plus one times

434
00:42:50,139 --> 00:42:54,139
three p minus one.

435
00:42:54,139 --> 00:42:57,949
Notice that if I have a sum of squares,

436
00:42:57,949 --> 00:43:01,169
for example,

437
00:43:01,170 --> 00:43:09,389
x squared plus four, which is x squared plus
two squared, then that does not factor.

438
00:43:09,389 --> 00:43:13,239
The difference of squares formula doesn't
apply. And there is no formula that applies

439
00:43:13,239 --> 00:43:19,989
for a sum of squares. There is, however, a
formula for both a difference of cubes and

440
00:43:19,989 --> 00:43:27,449
a sum of cubes. The difference of cubes formula,
a cubed minus b cubed is a minus b times a

441
00:43:27,449 --> 00:43:34,879
squared plus a b plus b squared. The formula
for the sum of cubes is pretty much the same,

442
00:43:34,880 --> 00:43:41,318
you just switch the negative and positive
sign here in here. So that gives us a plus

443
00:43:41,318 --> 00:43:49,018
b times a squared minus a b plus b squared.
As usual, you can check these formulas by

444
00:43:49,018 --> 00:43:55,608
multiplying out. Let's look at one example
of using these formulas. Y cubed plus 27 is

445
00:43:55,608 --> 00:44:03,298
actually a sum of two cubes because it's y
cubed plus three cubed. So I can factor it

446
00:44:03,298 --> 00:44:10,969
using the sum of cubes formula by plugging
in y for a and three for B. That gives me

447
00:44:10,969 --> 00:44:18,028
y plus three times y squared minus y times
three plus three squared. And I can clean

448
00:44:18,028 --> 00:44:23,849
that up a little bit to read y plus three
times y squared minus three y plus nine.

449
00:44:23,849 --> 00:44:31,190
So in this video, we went over several methods
of factoring. We did factoring out the greatest

450
00:44:31,190 --> 00:44:40,259
common factor. We did factoring by grouping.
We did factoring quadratics. And we did a

451
00:44:40,259 --> 00:44:47,789
difference of squares. And we did a difference
and a psalm of cubes

452
00:44:47,789 --> 00:44:54,210
and more complicated problems, you may need
to apply several these techniques in order

453
00:44:54,210 --> 00:44:58,470
to get through a single problem. For example,
you might need to start by pulling out a greatest

454
00:44:58,469 --> 00:45:00,328
common factor and then

455
00:45:00,329 --> 00:45:03,920
Add to a factoring of quadratics, or something
similar.

456
00:45:03,920 --> 00:45:07,700
Now go forth and factor.

457
00:45:07,699 --> 00:45:13,848
This video gives some additional examples
of factoring. Please pause the video and decide

458
00:45:13,849 --> 00:45:20,009
which of these first five expressions factor
and which one does not.

459
00:45:20,009 --> 00:45:27,469
The first expression can be factored by pulling
out a common factor of x from each term.

460
00:45:27,469 --> 00:45:34,199
So that becomes x times x plus one.

461
00:45:34,199 --> 00:45:41,358
The second example can be factors as a difference
of two squares, since x squared minus 25 is

462
00:45:41,358 --> 00:45:44,630
something squared, minus something else squared.
And we know that anytime we have something

463
00:45:44,630 --> 00:45:55,099
like a squared minus b squared, that's a plus
b times a minus b. So we can factor this as

464
00:45:55,099 --> 00:45:58,509
X plus five times x minus five.

465
00:45:58,509 --> 00:46:05,139
The third one is a sum of two squares, there's
no way to factor a sum of two squares over

466
00:46:05,139 --> 00:46:09,690
real numbers. So this is the one that does
not factor.

467
00:46:09,690 --> 00:46:16,608
just for completeness, let's look at the next
to this next one does factor by grouping.

468
00:46:16,608 --> 00:46:23,259
When we factor by grouping, we pull up the
biggest common factor out of the first two

469
00:46:23,260 --> 00:46:29,720
terms, that would be an x squared, that becomes
x squared times x plus two. And then we factor

470
00:46:29,719 --> 00:46:36,019
as much as we can add the next two terms,
that would be a three times x plus two. Notice

471
00:46:36,019 --> 00:46:42,420
that the x plus two factor now occurs in both
of the resulting terms. So we can pull that

472
00:46:42,420 --> 00:46:51,528
x plus two out and get x plus two times x
squared plus three,

473
00:46:51,528 --> 00:46:57,050
we can't factor any further because x squared
plus three doesn't factor.

474
00:46:57,050 --> 00:47:04,810
Finally, we have a quadratic, this also factors.
And I like to factor these also using a factoring

475
00:47:04,809 --> 00:47:12,849
by grouping trick. So first, what I do is
I multiply the coefficient of x squared and

476
00:47:12,849 --> 00:47:19,019
the constant term, five times eight is 40.
I'll write that on the top of my x. Now I

477
00:47:19,019 --> 00:47:24,318
take the coefficient of the x term, that's
negative 14, and I write that on the bottom

478
00:47:24,318 --> 00:47:32,480
part of the x. Now I'm looking for two numbers
that multiply to 40 and add to negative 14.

479
00:47:32,480 --> 00:47:37,659
Sometimes I can just guess numbers like this,
but if not, I start writing out factors of

480
00:47:37,659 --> 00:47:46,868
40. So factors of 40, I could do one times
40. Well, now I just noticed, I'm trying to

481
00:47:46,869 --> 00:47:51,568
add to a negative number. So if I use two
positive factors, there's no way there's going

482
00:47:51,568 --> 00:47:56,929
to add to a negative number. It's better for
me to use negative numbers, that factor 40,

483
00:47:56,929 --> 00:48:02,278
a negative times a negative still multiplies
to 40. But they have a chance of adding to

484
00:48:02,278 --> 00:48:06,460
a negative number. So but negative one and
negative 40, of course, don't work, they don't

485
00:48:06,460 --> 00:48:11,220
add to negative 14, they add to negative 41.
So let me try some other factors. The next

486
00:48:11,219 --> 00:48:15,838
biggest number that divides 40, besides one
is two, so I'll try negative two and negative

487
00:48:15,838 --> 00:48:22,518
20. Those add to negative 22. That doesn't
work. Next one that divides 40 would be four,

488
00:48:22,518 --> 00:48:29,818
so I'll try negative four and negative 10.
Aha, we have a winner. So negative four plus

489
00:48:29,818 --> 00:48:35,460
negative 10 is negative 14, negative four
times negative 10 is positive 40. We've got

490
00:48:35,460 --> 00:48:41,869
it. Alright, so the next step is to use factoring
by grouping, we're going to first split up

491
00:48:41,869 --> 00:48:50,809
this negative 14x as negative 4x minus 10x.
And carry down the eight and the 5x squared.

492
00:48:50,809 --> 00:48:57,300
Notice that this works, because I picked negative
four and negative 10 to add up to negative

493
00:48:57,300 --> 00:49:04,680
14, so so negative 4x minus 10x, will add
up to negative 14x. So I've got the same expression,

494
00:49:04,679 --> 00:49:09,528
just just expand it out a little bit. Now
I have four terms, I can do factoring by grouping,

495
00:49:09,528 --> 00:49:13,989
so I can group the first two terms and factor
out the biggest thing I can that will be n

496
00:49:13,989 --> 00:49:19,348
x times 5x minus four. And now I'll factor
the biggest thing I can out of these two numbers,

497
00:49:19,349 --> 00:49:26,499
including the the negative. So that becomes,
let's see, I can factor out a negative two

498
00:49:26,498 --> 00:49:33,588
and that becomes 5x minus four since negative
two times minus four is eight. All right,

499
00:49:33,588 --> 00:49:40,068
I've got the same 5x plus four in both my
terms, so factoring by grouping is going swimmingly,

500
00:49:40,068 --> 00:49:45,858
I can factor out the 5x minus four from both
those terms and I get the x minus two, and

501
00:49:45,858 --> 00:49:51,440
I factored this quadratic. If I want to, of
course, I can always check my work by distributing

502
00:49:51,440 --> 00:49:59,179
out by multiplying out. So a check here would
be multiplying 5x times x is 5x.

503
00:49:59,179 --> 00:50:08,288
squared 5x minus 10 to minus two is minus
10x minus four times x is minus 4x. And minus

504
00:50:08,289 --> 00:50:13,460
four times minus two is plus eight. So let's
see, this does check out to exactly what it

505
00:50:13,460 --> 00:50:22,119
should be. So that was the method of factoring
a quadratic.

506
00:50:22,119 --> 00:50:27,358
And all of these factors except for the sum
of squares.

507
00:50:27,358 --> 00:50:35,009
So we saw that factoring by grouping is handy
for factoring this expression here. It was

508
00:50:35,009 --> 00:50:39,929
also handy for factoring the quadratic indirectly,
after splitting up

509
00:50:39,929 --> 00:50:44,389
the middle term

510
00:50:44,389 --> 00:50:52,279
into two terms. So how can you tell when a
an expression is is appropriate to factor

511
00:50:52,280 --> 00:50:57,690
by grouping, there's, there's an easy way
to tell that it might be a candidate, and

512
00:50:57,690 --> 00:51:00,099
that's that it has four terms.

513
00:51:00,099 --> 00:51:05,539
So if you see four terms, or in the case of
quadratic, you can split it up into four terms,

514
00:51:05,539 --> 00:51:09,030
then that's a good candidate for factoring
by grouping because you can group the first

515
00:51:09,030 --> 00:51:16,210
two terms group the second two terms, factoring
by grouping all always work on, on expressions

516
00:51:16,210 --> 00:51:21,559
with four terms. But, but that's like the
first thing to look for. So let's just review

517
00:51:21,559 --> 00:51:25,829
what are the same main techniques of factoring.
We saw these on the previous page, we saw

518
00:51:25,829 --> 00:51:32,230
there was pull out common factors. There's
difference of squares.

519
00:51:32,230 --> 00:51:36,230
There's factoring by grouping.

520
00:51:36,230 --> 00:51:41,849
There's factoring quadratics.

521
00:51:41,849 --> 00:51:49,970
And one more that I didn't mention is factoring
sums and differences of cubes.

522
00:51:49,969 --> 00:51:57,328
that uses the formulas, aq minus b cubed is
a minus b times a squared plus a b plus b

523
00:51:57,329 --> 00:52:07,930
squared. And a cubed plus b cubed is a plus
b times a squared minus a b plus b squared.

524
00:52:07,929 --> 00:52:11,068
One important tip when factoring,

525
00:52:11,068 --> 00:52:17,150
I always recommend doing this first, pull
out the common factors first.

526
00:52:17,150 --> 00:52:21,798
That'll simplify things and making the rest
of factoring easier. One more tip is that

527
00:52:21,798 --> 00:52:28,130
you might need to do several these factoring
techniques in one problem, for example,

528
00:52:28,130 --> 00:52:31,869
you might have to first pull out a common
factor, then factor a difference of squares.

529
00:52:31,869 --> 00:52:35,338
And then you might notice that one of your
factors is itself a difference of squares,

530
00:52:35,338 --> 00:52:41,038
and you have to apply a difference of squares
again, so don't stop when you factor a little

531
00:52:41,039 --> 00:52:43,910
bit, keep factoring as far as you can go.

532
00:52:43,909 --> 00:52:49,411
Here are some extra examples of factoring
quadratics. For you to practice, please pause

533
00:52:49,411 --> 00:52:52,619
the video and give these a try.

534
00:52:52,619 --> 00:53:01,130
For the first one, let's multiply two times
negative 14. That gives us negative 28. And

535
00:53:01,130 --> 00:53:06,338
then we'll bring the three down in the bottom
of the x. Now we're looking for two numbers

536
00:53:06,338 --> 00:53:14,269
that multiply to negative 28 and add to three.
Well, to multiply two numbers to get a negative

537
00:53:14,269 --> 00:53:23,518
28, we'll need one of them to be negative
and one of them to be positive. So to be one,

538
00:53:23,518 --> 00:53:26,808
negative 128, or one, negative 28, those don't
work. Let's see negative 214 or two, negative

539
00:53:26,809 --> 00:53:30,309
14, those don't work. Hey, I just noticed
the positive number had better be bigger than

540
00:53:30,309 --> 00:53:37,400
the negative number, so they add to a positive
number. Let's see what comes next. How about

541
00:53:37,400 --> 00:53:43,460
negative four times seven, four times negative
seven, I think four, negative four times seven

542
00:53:43,460 --> 00:53:49,199
will work. So I'll write those here at negative
four, seven,

543
00:53:49,199 --> 00:53:58,009
copy down the two z squared. And I'll split
up the three z into negative four z plus seven

544
00:53:58,009 --> 00:54:05,170
z and then minus 14. Now factoring by grouping,
pull out a to z, that becomes z minus two,

545
00:54:05,170 --> 00:54:12,650
pull out a seven and that becomes z minus
two again, looking good. I've got two z plus

546
00:54:12,650 --> 00:54:18,259
seven times z minus two as my factored expression.

547
00:54:18,259 --> 00:54:23,048
My second expression, I could work at the
same way, drawing my axe and factoring by

548
00:54:23,048 --> 00:54:28,619
grouping that kind of thing. But it's actually
going to be easier if I notice first that

549
00:54:28,619 --> 00:54:32,170
I can pull out a common factor from all of
my terms, though, that'll make things a lot

550
00:54:32,170 --> 00:54:36,130
simpler to deal with. So notice that a five
divides each of these terms, and in fact,

551
00:54:36,130 --> 00:54:39,660
I'm going to go ahead and pull out the negative
five because I don't like having negatives

552
00:54:39,659 --> 00:54:44,690
in front of my squared term. So I'm going
to pull out a common factor of negative five.

553
00:54:44,690 --> 00:54:49,470
Again, it would work if I forgot to do this,
but it would be a lot more complicated. So

554
00:54:49,469 --> 00:54:56,699
negative five v squared, this becomes minus,
this becomes plus nine V, since nine times

555
00:54:56,699 --> 00:54:59,868
negative five is negative 45. And this becomes
minus

556
00:54:59,869 --> 00:55:06,650
10 cents native 10 times negative five is
positive 50. Now I can start my x and my factoring

557
00:55:06,650 --> 00:55:10,849
by grouping, or I can use kind of a shortcut
method, which you may have seen before. So

558
00:55:10,849 --> 00:55:15,588
I can just put these here, and then I know
that whatever numbers go here, they're gonna

559
00:55:15,588 --> 00:55:21,150
have to multiply to the negative 10. And they're
gonna have to add to the nine. So that would

560
00:55:21,150 --> 00:55:26,289
be plus 10, and a minus one will do the trick.

561
00:55:26,289 --> 00:55:32,230
Those are all my factoring examples for today.
I hope you enjoy your snow morning and have

562
00:55:32,230 --> 00:55:37,929
a chance to spend some time working in ALEKS.
Bye.

563
00:55:37,929 --> 00:55:43,028
This video is about working with rational
expressions. A rational expression is a fraction

564
00:55:43,028 --> 00:55:49,460
usually with variables in it, something like
x plus two over x squared minus three is a

565
00:55:49,460 --> 00:55:55,360
rational expression. In this video, we'll
practice adding, subtracting, multiplying

566
00:55:55,360 --> 00:56:01,660
and dividing rational expressions and simplifying
them to lowest terms.

567
00:56:01,659 --> 00:56:07,808
We'll start with simplifying to lowest terms.
Recall that if you have a fraction with just

568
00:56:07,809 --> 00:56:14,599
numbers in it, something like 21 over 45,
we can reduce it to lowest terms by factoring

569
00:56:14,599 --> 00:56:17,430
the numerator

570
00:56:17,429 --> 00:56:22,199
and factoring the denominator

571
00:56:22,199 --> 00:56:26,019
and then canceling common factors.

572
00:56:26,019 --> 00:56:35,478
So in this example, the three is cancel, and
our fraction reduces to seven over 15.

573
00:56:35,478 --> 00:56:40,259
If we want to reduce a rational expression
with the variables and add to lowest terms,

574
00:56:40,259 --> 00:56:47,309
we proceed the same way. First, we'll factor
the numerator, that's three times x plus two,

575
00:56:47,309 --> 00:56:54,680
and then factor the denominator. In this case
of factors 2x plus two times x plus two, we

576
00:56:54,679 --> 00:57:00,710
could also write that as x plus two squared.
Now we cancel the common factors. And we're

577
00:57:00,710 --> 00:57:08,579
left with three over x plus two. Definitely
a simpler way of writing that rational expression.

578
00:57:08,579 --> 00:57:15,900
Next, let's practice multiplying and dividing.
Recall that if we multiply two fractions with

579
00:57:15,900 --> 00:57:20,789
just numbers in them, we simply multiply the
numerators and multiply the denominators.

580
00:57:20,789 --> 00:57:28,599
So in this case, we would get four times two
over three times five or 8/15.

581
00:57:28,599 --> 00:57:35,338
If we want to divide two fractions, like in
the second example, then we can rewrite it

582
00:57:35,338 --> 00:57:43,170
as multiplying by the reciprocal of the fraction
on the denominator. So here, we get four fifths

583
00:57:43,170 --> 00:57:50,970
times three halves, and that gives us 12 tenths.
But actually, we could reduce that fraction

584
00:57:50,969 --> 00:57:53,818
to six fifths,

585
00:57:53,818 --> 00:58:00,000
we use the same rules when we compute the
product or quotient of two rational expressions

586
00:58:00,000 --> 00:58:06,548
with the variables. And then here, we're trying
to divide two rational expressions. So instead,

587
00:58:06,548 --> 00:58:12,889
we can multiply by the reciprocal. I call
this flipping and multiplying.

588
00:58:12,889 --> 00:58:18,828
And now we just multiply the numerators.

589
00:58:18,829 --> 00:58:21,859
And multiply the denominators.

590
00:58:21,858 --> 00:58:28,889
It might be tempting at this point to multiply
out to distribute out the numerator and the

591
00:58:28,889 --> 00:58:32,748
denominator. But actually, it's better to
leave it in this factored form and factored

592
00:58:32,748 --> 00:58:38,139
even more completely. That way, we'll be able
to reduce the rational expression to cancel

593
00:58:38,139 --> 00:58:46,048
the common factors. So let's factor even more
the x squared plus x factors as x times x

594
00:58:46,048 --> 00:58:52,440
plus one, and x squared minus 16. And that's
a difference of two squares, that's x plus

595
00:58:52,440 --> 00:58:59,510
four times x minus four, the denominator is
already fully factored, so we'll just copy

596
00:58:59,510 --> 00:59:06,569
it over. And now we can cancel common factors
here and here, and we're left with x times

597
00:59:06,568 --> 00:59:12,659
x minus four. This is our final answer.

598
00:59:12,659 --> 00:59:16,170
Adding and subtracting fractions is a little
more complicated because we first have to

599
00:59:16,170 --> 00:59:23,088
find a common denominator. A common denominator
is an expression that both denominators divided

600
00:59:23,088 --> 00:59:29,190
into, it's usually best of the long run to
use the least common denominator, which is

601
00:59:29,190 --> 00:59:34,028
the smallest expression that both denominators
divided into.

602
00:59:34,028 --> 00:59:39,219
In this example, if we just want a common
denominator, we could use six times 15, which

603
00:59:39,219 --> 00:59:46,058
is 90 because both six and 15 divided evenly
into 90. But if we want the least common denominator,

604
00:59:46,059 --> 00:59:55,150
the best way to do that is to factor the two
denominators. So six is two times 315 is three

605
00:59:55,150 --> 00:59:59,900
times five, and then put together only the
factors we need for

606
00:59:59,900 --> 01:00:07,789
Both six and 50 into divider numbers. So if
we just use two times three times five, which

607
01:00:07,789 --> 01:00:14,159
is 30, we know that two times three will divide
it, and three times five will also divide

608
01:00:14,159 --> 01:00:19,440
it. And we won't be able to get a denominator
any smaller, because we need the factors two,

609
01:00:19,440 --> 01:00:24,970
three, and five, in order to ensure both these
numbers divided. Once we have our least common

610
01:00:24,969 --> 01:00:31,598
denominator, we can rewrite each of our fractions
in terms of that denominator. So seven, six,

611
01:00:31,599 --> 01:00:38,940
I need to get a 30 in the denominator, so
I'm going to multiply that by five over five,

612
01:00:38,940 --> 01:00:45,539
and multiply by the factors that are missing
from the current denominator in order to get

613
01:00:45,539 --> 01:00:54,068
my least common denominator of 30. For the
second fraction, for 15th 15 times two is

614
01:00:54,068 --> 01:00:57,829
30. So I'm going to multiply by two over two,

615
01:00:57,829 --> 01:01:08,971
I can rewrite this as 3530 s minus 8/30. And
now that I have a common denominator, I can

616
01:01:08,971 --> 01:01:15,608
just subtract my two numerators. And I get
27/30.

617
01:01:15,608 --> 01:01:20,119
If I factor, I can reduce this

618
01:01:20,119 --> 01:01:28,108
to three squared over two times five, which
is nine tenths. The process for finding the

619
01:01:28,108 --> 01:01:33,759
sum of two rational expressions with variables
in them follows the exact same process. First,

620
01:01:33,759 --> 01:01:39,440
we have to find the least common denominator,
I'll do that by factoring the two denominators.

621
01:01:39,440 --> 01:01:45,659
So 2x plus two factors as two times x plus
1x squared minus one, that's a difference

622
01:01:45,659 --> 01:01:51,858
of two squares. So that's x plus one times
x minus one. Now for the least common denominator,

623
01:01:51,858 --> 01:01:57,558
I'm going to take all the factors, I need
to get an expression that each of these divides

624
01:01:57,559 --> 01:02:02,329
into, so I need the factor two, I need the
factor x plus one, and I need the factor x

625
01:02:02,329 --> 01:02:08,010
minus one, I don't have to repeat the factor
x plus one I just need to have at one time.

626
01:02:08,010 --> 01:02:14,410
And so I will get my least common denominator
two times x plus one times x minus one, I'm

627
01:02:14,409 --> 01:02:17,868
not going to bother multiplying this out,
it's actually better to leave it in factored

628
01:02:17,869 --> 01:02:24,920
form to help me simplify later. Now I can
rewrite each of my two rational expressions

629
01:02:24,920 --> 01:02:29,950
by multiplying by whatever's missing from
the denominator in terms of the least common

630
01:02:29,949 --> 01:02:36,098
denominator. So what I mean is, I can rewrite
three over 2x plus two, I'll write the 2x

631
01:02:36,099 --> 01:02:41,278
plus two is two times x plus one, I'll write
it in factored form. And then I noticed that

632
01:02:41,278 --> 01:02:45,778
compared to the least common denominator,
I'm missing the factor of x minus one. So

633
01:02:45,778 --> 01:02:52,099
I multiply the numerator and the denominator
by x minus one, I just need it in the denominator.

634
01:02:52,099 --> 01:02:56,660
But I can't get away with just multiplying
by the denominator without changing my expression,

635
01:02:56,659 --> 01:03:00,788
I have to multiply by it on the numerator
and the denominator. So I'm just multiplying

636
01:03:00,789 --> 01:03:06,690
by one and a fancy form and not changing the
value here. So now I do the same thing for

637
01:03:06,690 --> 01:03:10,950
the second rational expression, I'll I'll
write the denominator in factored form to

638
01:03:10,949 --> 01:03:16,989
make it easier to see what's missing from
the denominator. What's missing in this denominator,

639
01:03:16,989 --> 01:03:21,588
compared to my least common denominator is
just the factor two, so multiply the numerator

640
01:03:21,588 --> 01:03:27,590
and the denominator by two. Now I can rewrite
everything. So the first rational expression

641
01:03:27,590 --> 01:03:36,059
becomes three times x minus one over two times
x plus 1x minus one, and the second one becomes

642
01:03:36,059 --> 01:03:44,849
five times two over two times x plus 1x minus
one, notice that I now have a common denominator.

643
01:03:44,849 --> 01:03:51,519
So I can just add together my numerators.
So I get three times x minus one plus 10 over

644
01:03:51,518 --> 01:03:58,228
two times x plus 1x minus one. I'd like to
simplify this. And the best way to do that

645
01:03:58,228 --> 01:04:04,218
is to leave the denominator in factored form.
But I do have to multiply out the numerator

646
01:04:04,219 --> 01:04:12,599
so that I can add things together. So I get
3x minus three plus 10 over two times x plus

647
01:04:12,599 --> 01:04:22,130
1x minus one, or 3x plus seven over two times
x plus 1x minus one. Now 3x plus seven doesn't

648
01:04:22,130 --> 01:04:29,160
factor. And there's therefore no factors that
I can cancel out. So this is already reduced.

649
01:04:29,159 --> 01:04:33,129
As much as it can be. This is my final answer.

650
01:04:33,130 --> 01:04:40,778
In this video, we saw how to simplify rational
expressions to lowest terms by factoring and

651
01:04:40,778 --> 01:04:42,980
canceling common factors.

652
01:04:42,980 --> 01:04:47,949
We also saw how to multiply rational expressions
by multiplying the numerator and multiplying

653
01:04:47,949 --> 01:04:53,538
the denominator, how to divide rational expressions
by flipping and multiplying and how to add

654
01:04:53,539 --> 01:05:00,579
and subtract rational expressions by writing
them in terms of the least common denominator

655
01:05:00,579 --> 01:05:07,119
This video is about solving quadratic equations.
a quadratic equation is an equation that contains

656
01:05:07,119 --> 01:05:11,960
the square of the variable, say x squared,
but no higher powers of x.

657
01:05:11,960 --> 01:05:20,449
The standard form for a quadratic equation
is the form a x squared plus b x plus c equals

658
01:05:20,449 --> 01:05:25,498
zero, where A, B and C

659
01:05:25,498 --> 01:05:31,419
represent real numbers. And a is not zero
so that we actually have an x squared term.

660
01:05:31,420 --> 01:05:40,450
Let me give you an example. 3x squared plus
7x minus two equals zero is a quadratic equation

661
01:05:40,449 --> 01:05:49,788
in standard form, here a is three, B is seven,
and C is minus two. The equation 3x squared

662
01:05:49,789 --> 01:05:56,839
equals minus 7x plus two is also a quadratic
equation, it's just not in standard form.

663
01:05:56,838 --> 01:06:02,449
The key steps to solving quadratic equations
are usually to write the equation in standard

664
01:06:02,449 --> 01:06:07,449
form, and then either factor it

665
01:06:07,449 --> 01:06:14,548
or use the quadratic formula, which I'll show
you later in this video.

666
01:06:14,548 --> 01:06:21,119
Let's start with the example y squared equals
18 minus seven y. Here our variable is y,

667
01:06:21,119 --> 01:06:26,650
and we need to rewrite this quadratic equation
in standard form, we can do this by subtracting

668
01:06:26,650 --> 01:06:33,680
18 from both sides and adding seven y to both
sides. That gives us the equation y squared

669
01:06:33,679 --> 01:06:40,848
minus 18 plus seven y equals zero. And I can
rearrange a little bit to get y squared plus

670
01:06:40,849 --> 01:06:48,300
seven y minus 18 equals zero. Now I've got
my equation in standard form. Next, I'm going

671
01:06:48,300 --> 01:06:54,589
to try to factor it. So I need to look for
two numbers that multiply to negative 18 and

672
01:06:54,588 --> 01:06:57,489
add to seven.

673
01:06:57,489 --> 01:07:02,659
two numbers that work are nine and negative
two, so I can factor my expression on the

674
01:07:02,659 --> 01:07:12,039
left as y plus nine times y minus two equals
zero. Now, anytime you have two quantities

675
01:07:12,039 --> 01:07:17,389
that multiply together to give you zero, either
the first quantity has to be zero, or the

676
01:07:17,389 --> 01:07:22,478
second quantity has to be zero, or I suppose
they both could be zero. In some situations,

677
01:07:22,478 --> 01:07:27,879
this is really handy, because that means that
I know that either y plus nine equals zero,

678
01:07:27,880 --> 01:07:36,410
or y minus two equals zero. So I can as my
next step, set my factors equal to zero. So

679
01:07:36,409 --> 01:07:43,278
y plus nine equals zero, or y minus two equals
zero, which means that y equals negative nine,

680
01:07:43,278 --> 01:07:46,619
or y equals two.

681
01:07:46,619 --> 01:07:51,039
It's not a bad idea to check that those answers
actually work by plugging them into the original

682
01:07:51,039 --> 01:07:56,998
equation, negative nine squared, does that
equal 18 minus seven times negative nine,

683
01:07:56,998 --> 01:08:04,459
and you can work out that it does. And similarly,
two squared equals 18 minus seven times two.

684
01:08:04,460 --> 01:08:11,449
In the next example, let's find solutions
of the equation w squared equals 121. This

685
01:08:11,449 --> 01:08:17,729
is a quadratic equation, because it's got
a square of my variable W, I can rewrite it

686
01:08:17,729 --> 01:08:24,278
in standard form by subtracting 121 from both
sides.

687
01:08:24,279 --> 01:08:31,839
Notice that A is equal to one b is equal to
zero because there's no w term, and c is equal

688
01:08:31,838 --> 01:08:39,340
to negative 121 in the standard form, a W
squared plus BW plus c equals zero. Next step,

689
01:08:39,340 --> 01:08:47,980
I'm going to try to factor this expression.
So since 121, is 11 squared, this is a difference

690
01:08:47,979 --> 01:08:56,129
of two squares, and it factors as w plus 11,
times w minus 11 is equal to zero. If I set

691
01:08:56,130 --> 01:08:59,560
the factors equal to zero,

692
01:08:59,560 --> 01:09:07,560
I get w plus 11 equals zero or w minus 11
equals zero. So w equals minus 11, or w equals

693
01:09:07,560 --> 01:09:14,289
11. In this example, I could have solved the
equation more simply, I could have instead

694
01:09:14,289 --> 01:09:20,539
said that if W squared is 121, and then w
has to equal plus or minus the square root

695
01:09:20,539 --> 01:09:26,189
of 121. In other words, W is plus or minus
11.

696
01:09:26,189 --> 01:09:30,939
If you saw the equation this way, it's important
to remember the plus or minus since minus

697
01:09:30,939 --> 01:09:36,019
11 squared equals 121, just like 11 squared
does.

698
01:09:36,020 --> 01:09:41,790
Now let's find the solutions for the equation.
x times x plus two equals seven. Some people

699
01:09:41,789 --> 01:09:46,630
might be tempted to say that, oh, if two numbers
multiply to equal seven, then one of them

700
01:09:46,630 --> 01:09:51,529
better equal one and the other equals seven
or maybe negative one and negative seven.

701
01:09:51,529 --> 01:09:57,590
But that's faulty reasoning in this case,
because x and x plus two don't have to be

702
01:09:57,590 --> 01:10:00,489
whole numbers. They could be crazy.

703
01:10:00,489 --> 01:10:08,729
fractions or even irrational numbers. So instead,
let's rewrite this equation in standard form.

704
01:10:08,729 --> 01:10:16,879
To do that, I'm first going to multiply out.
So x times x is x squared x times two is 2x.

705
01:10:16,880 --> 01:10:23,310
That equals seven, and I'll subtract the seven
from both sides to get x squared plus 2x minus

706
01:10:23,310 --> 01:10:28,770
seven is zero. Now I'm looking to factor it.
So I need two numbers that multiply to negative

707
01:10:28,770 --> 01:10:34,770
seven and add to two, since the only way to
factor negative seven is as negative one times

708
01:10:34,770 --> 01:10:39,580
seven or seven times negative one, it's easy
to see that there are no whole numbers that

709
01:10:39,579 --> 01:10:46,750
will do will work. So there's no way to factor
this expression over the integers. Instead,

710
01:10:46,750 --> 01:10:53,140
let's use the quadratic equation. So we have
our leading coefficient of x squared is one.

711
01:10:53,140 --> 01:11:00,000
So A is one, B is two, and C is minus seven.
And we're going to plug that into the equation

712
01:11:00,000 --> 01:11:06,869
quadratic equation, which goes x equals negative
b plus or minus the square root of b squared

713
01:11:06,869 --> 01:11:10,279
minus four, I see all over two

714
01:11:10,279 --> 01:11:15,059
different people have different ways of remembering
this formula, I'd like to remember it by seeing

715
01:11:15,060 --> 01:11:22,420
it x equals negative b plus or minus the square
root of b squared minus four, I see oh over

716
01:11:22,420 --> 01:11:28,060
to a, but you can use any pneumonic you like.
Anyway, plugging in here, we have x equals

717
01:11:28,060 --> 01:11:34,140
negative two plus or minus the square root
of two squared minus four times one times

718
01:11:34,140 --> 01:11:40,730
negative seven support and remember the negative
seven there to all over two times one.

719
01:11:40,729 --> 01:11:48,429
Now two squared is four, and four times one
times negative seven is negative 28. So this

720
01:11:48,430 --> 01:11:55,360
whole quantity under the square root sign
becomes four minus negative 28, or 32. So

721
01:11:55,359 --> 01:12:00,849
I can rewrite this as x equals negative two
plus or minus the square root of 32, all over

722
01:12:00,850 --> 01:12:01,890
two.

723
01:12:01,890 --> 01:12:11,079
Since 32, is 16 times two and 16 is a perfect
square, I can rewrite this as negative two

724
01:12:11,079 --> 01:12:15,819
plus or minus the square root of 16 times
the square root of two over two, which is

725
01:12:15,819 --> 01:12:19,920
negative two plus or minus four times the
square root of two over two.

726
01:12:19,920 --> 01:12:26,159
Next, I'm going to split out my fraction

727
01:12:26,159 --> 01:12:32,159
as negative two over two plus or minus four
square root of two over two, and then simplify

728
01:12:32,159 --> 01:12:38,729
those fractions. This becomes negative one
plus or minus two square root of two. So my

729
01:12:38,729 --> 01:12:45,039
answers are negative one plus two square root
of two and negative one minus two square root

730
01:12:45,039 --> 01:12:52,039
of two. And if I need a decimal answer for
any reason, I could work this out on my calculator.

731
01:12:52,039 --> 01:12:57,390
As our final example, let's find all real
solutions for the equation, one half y squared

732
01:12:57,390 --> 01:13:04,600
equals 1/3 y minus two. I'll start as usual
by putting it in standard form. So that gives

733
01:13:04,600 --> 01:13:12,810
me one half y squared minus 1/3 y plus two
equals zero, I could go ahead and start trying

734
01:13:12,810 --> 01:13:17,940
to factor or use the quadratic formula right
now. But I find fractional coefficients kind

735
01:13:17,939 --> 01:13:22,649
of annoying. So I'd like to get rid of them.
By doing what I call clearing the denominator,

736
01:13:22,649 --> 01:13:27,879
that means I'm going to multiply the whole
entire equation by the least common denominator.

737
01:13:27,880 --> 01:13:33,020
In this case, the least common denominator
is two times three or six. So I'll multiply

738
01:13:33,020 --> 01:13:37,300
the whole equation by six have to make sure
I multiplied both sides of the equation, but

739
01:13:37,300 --> 01:13:42,220
in this case, six times zero is just zero.
And when I distribute the six, I get three

740
01:13:42,220 --> 01:13:47,140
y squared minus two y plus 12 equals zero.

741
01:13:47,140 --> 01:13:51,760
Now, I could try to factor this, but I think
it's easier probably just to plunge in and

742
01:13:51,760 --> 01:13:54,000
use the quadratic formula.

743
01:13:54,000 --> 01:14:01,909
So I get x equals negative B, that's negative
negative two or two, plus or minus the square

744
01:14:01,909 --> 01:14:12,539
root of b squared, minus four times a times
c, all over to a.

745
01:14:12,539 --> 01:14:17,590
Working out the stuff in a square root sign,
negative two squared is four. And here we

746
01:14:17,590 --> 01:14:21,699
have, let's see 144.

747
01:14:21,699 --> 01:14:28,449
So this simplifies to x equals to plus or
minus the square root of four minus 144. That's

748
01:14:28,449 --> 01:14:35,210
negative 140. All of our sex. Well, if you're
concerned about that negative number under

749
01:14:35,210 --> 01:14:40,399
the square root sign, you should be we can't
take the square root of a negative number

750
01:14:40,399 --> 01:14:45,469
and get an N get a real number is our answer.
There's no real number whose square is a negative

751
01:14:45,470 --> 01:14:54,650
number. And therefore, our conclusion is we
have no real solutions to this quadratic equation.

752
01:14:54,649 --> 01:14:59,689
In this video, we solve some quadratic equations
by first writing them in standard form and

753
01:14:59,689 --> 01:15:00,689
then

754
01:15:00,689 --> 01:15:03,669
Either factoring or using the quadratic formula.

755
01:15:03,670 --> 01:15:08,440
In some examples, factoring doesn't work,
it's not possible to factor the equation.

756
01:15:08,439 --> 01:15:13,969
But in fact, using the quadratic formula will
always work even if it's also possible to

757
01:15:13,970 --> 01:15:19,011
solve it by factoring. So you can't really
lose by using the quadratic formula. It's

758
01:15:19,011 --> 01:15:23,000
just sometimes it'll be faster to factor instead.

759
01:15:23,000 --> 01:15:29,539
This video is about solving rational equations.
A rational equation, like this one is an equation

760
01:15:29,539 --> 01:15:34,189
that has rational expressions in that, in
other words, an equation that has some variables

761
01:15:34,189 --> 01:15:35,359
in the denominator.

762
01:15:35,359 --> 01:15:41,069
There are several different approaches for
solving a rational equation, but they all

763
01:15:41,069 --> 01:15:46,719
start by finding the least common denominator.
In this example, the denominators are x plus

764
01:15:46,720 --> 01:15:52,350
three and x, we can think of one as just having
a denominator of one.

765
01:15:52,350 --> 01:15:57,250
Since the denominators don't have any factors
in common, I can find the least common denominator

766
01:15:57,250 --> 01:16:00,689
just by multiplying them together.

767
01:16:00,689 --> 01:16:04,359
My next step is going to be clearing the denominator.

768
01:16:04,359 --> 01:16:12,460
By this, I mean that I multiply both sides
of my equation by this least common denominator,

769
01:16:12,460 --> 01:16:19,270
x plus three times x, I multiply on the left
side of the equation, and I multiply by the

770
01:16:19,270 --> 01:16:22,310
same thing on the right side of the equation.

771
01:16:22,310 --> 01:16:28,250
Since I'm doing the same thing to both sides
of the equation, I don't change the the value

772
01:16:28,250 --> 01:16:33,020
of the equation. Multiplying the least common
denominator on both sides of the equation

773
01:16:33,020 --> 01:16:38,650
is equivalent to multiplying it by all three
terms in the equation, I can see this when

774
01:16:38,649 --> 01:16:41,189
I multiply out,

775
01:16:41,189 --> 01:16:47,909
I'll rewrite the left side the same as before,
pretty much. And then I'll distribute the

776
01:16:47,909 --> 01:16:56,029
right side to get x plus three times x times
one plus x plus three times x times one over

777
01:16:56,029 --> 01:17:01,920
x. So I've actually multiplied the least common
denominator by all three terms of my equation.

778
01:17:01,920 --> 01:17:07,920
Now I can have a blast canceling things. The
x plus three cancels with the x plus three

779
01:17:07,920 --> 01:17:10,119
on the denominator.

780
01:17:10,119 --> 01:17:15,579
The here are nothing cancels out because there's
no denominator, and here are the x in the

781
01:17:15,579 --> 01:17:18,470
numerator cancels with the x in the denominator.

782
01:17:18,470 --> 01:17:26,310
So I can rewrite my expression as x squared
equals x plus three times x times one plus

783
01:17:26,310 --> 01:17:29,550
x plus three. Now I'm going to simplify.

784
01:17:29,550 --> 01:17:36,690
So I'll leave the x squared alone on this
side, I'll distribute out x squared plus 3x

785
01:17:36,689 --> 01:17:44,039
plus x plus three, hey, look, the x squared
is cancel on both sides. And so I get zero

786
01:17:44,039 --> 01:17:51,199
equals 4x plus three, so 4x is negative three,
and x is negative three fourths. Finally,

787
01:17:51,199 --> 01:17:57,409
I'm going to plug in my answer to check. This
is a good idea for any kind of equation. But

788
01:17:57,409 --> 01:18:01,609
it's especially important for a rational equation
because occasionally for rational equations,

789
01:18:01,609 --> 01:18:05,699
you'll get what's called extraneous solution
solutions that don't actually work in your

790
01:18:05,699 --> 01:18:09,949
original equation because they make the denominator
zero. Now, in this example, I don't think

791
01:18:09,949 --> 01:18:16,029
we're going to get any extraneous equations
because negative three fourths is not going

792
01:18:16,029 --> 01:18:20,670
to make any of these denominators zero, so
it should work out fine when I plug in. If

793
01:18:20,670 --> 01:18:22,590
I plug in,

794
01:18:22,590 --> 01:18:29,489
I get this, I can simplify

795
01:18:29,489 --> 01:18:34,119
the denominator here, negative three fourths
plus three, three is 12 fourths, as becomes

796
01:18:34,119 --> 01:18:42,680
nine fourths. And this is one I'll flip and
multiply to get minus four thirds. So here,

797
01:18:42,680 --> 01:18:49,570
I can simplify my complex fraction, it ends
up being negative three nights, and one minus

798
01:18:49,569 --> 01:18:55,960
four thirds is negative 1/3. So that all seems
to check out.

799
01:18:55,960 --> 01:19:00,890
And so my final answer is x equals negative
three fourths.

800
01:19:00,890 --> 01:19:05,310
This next example looks a little trickier.
And it is, but the same approach will work.

801
01:19:05,310 --> 01:19:13,410
First off, find the least common denominator.
So here, my denominators are c minus five,

802
01:19:13,409 --> 01:19:22,029
c plus one, and C squared minus four c minus
five, I'm going to factor that as C minus

803
01:19:22,029 --> 01:19:29,189
five times c plus one. Now, my least common
denominator needs to have just enough factors

804
01:19:29,189 --> 01:19:35,269
to that each of these denominators divided
into it. So I need the factor c minus five,

805
01:19:35,270 --> 01:19:39,960
I need the factor c plus one. And now I've
already got all the factors I need for this

806
01:19:39,960 --> 01:19:47,480
denominator. So here is my least common denominator.
Next step is to clear the denominators.

807
01:19:47,479 --> 01:19:53,589
So I do this by multiplying both sides of
the equation by my least common denominator.

808
01:19:53,590 --> 01:20:01,560
In fact, I can just multiply each of the three
terms by this least common denominator

809
01:20:01,560 --> 01:20:06,789
I went ahead and wrote my third denominator
in factored form to make it easier to see

810
01:20:06,789 --> 01:20:16,619
what cancels. Now canceling time dies, this
dies. And both of those factors die. cancelling

811
01:20:16,619 --> 01:20:21,229
out the denominator is the whole point of
multiplying by the least common denominator,

812
01:20:21,229 --> 01:20:25,189
you're multiplying by something that's big
enough to kill every single denominator, so

813
01:20:25,189 --> 01:20:28,649
you don't have to deal with denominators anymore.

814
01:20:28,649 --> 01:20:32,069
Now I'm going to simplify by multiplying out.

815
01:20:32,069 --> 01:20:41,569
So I get, let's see, c plus one times four
c, that's four c squared plus four c, now

816
01:20:41,569 --> 01:20:51,130
I get minus just c minus five, and then over
here, I get three c squared plus three,

817
01:20:51,130 --> 01:21:00,440
I can rewrite the minus quantity c minus five
is minus c plus five.

818
01:21:00,439 --> 01:21:06,829
And now I can subtract the three c squared
from both sides to get just a C squared over

819
01:21:06,829 --> 01:21:15,710
here, and the four c minus c that becomes
a three C.

820
01:21:15,710 --> 01:21:21,829
And finally, I can subtract the three from
both sides to get c squared plus three c plus

821
01:21:21,829 --> 01:21:28,979
two equals zero. got myself a quadratic equation
that looks like a nice one that factors. So

822
01:21:28,979 --> 01:21:36,229
this factors to C plus one times c plus two
equals zero. So either c plus one is zero,

823
01:21:36,229 --> 01:21:44,279
or C plus two is zero. So C equals negative
one, or C equals negative two.

824
01:21:44,279 --> 01:21:49,689
Now let's see, we need to still check our
answers.

825
01:21:49,689 --> 01:21:54,519
Without even going to the trouble of calculating
anything, I can see that C equals negative

826
01:21:54,520 --> 01:22:01,160
one is not going to work, because if I plug
it in to this denominator here, I get a denominator

827
01:22:01,159 --> 01:22:09,250
zero, which doesn't make sense. So C equals
minus one is an extraneous solution, it doesn't

828
01:22:09,250 --> 01:22:16,789
actually satisfy my original equation. And
so I can just cross it right out, C equals

829
01:22:16,789 --> 01:22:19,869
negative two.

830
01:22:19,869 --> 01:22:23,809
I can go if I go ahead, and that doesn't make
any of my denominators zero. So if I haven't

831
01:22:23,810 --> 01:22:30,780
made any mistakes, it should satisfy my original
equation, but, but I'll just plug it in to

832
01:22:30,779 --> 01:22:33,619
be sure.

833
01:22:33,619 --> 01:22:36,899
And after some simplifying,

834
01:22:36,899 --> 01:22:39,849
I get a true statement.

835
01:22:39,850 --> 01:22:46,690
So my final answer is C equals negative two.
In this video, we saw the couple of rational

836
01:22:46,689 --> 01:22:53,349
equations using the method of finding the
least common denominator and then clearing

837
01:22:53,350 --> 01:22:55,600
the denominator,

838
01:22:55,600 --> 01:22:59,490
we cleared the denominator by multiplying
both sides of the equation by the least common

839
01:22:59,489 --> 01:23:04,389
denominator or equivalently. multiplying each
of the terms by that denominator.

840
01:23:04,390 --> 01:23:10,250
There's another equivalent method that some
people prefer, it still starts out the same,

841
01:23:10,250 --> 01:23:16,390
we find the least common denominator, but
then we write all the fractions

842
01:23:16,390 --> 01:23:22,100
over that least common denominator. So in
this example, we'd still use the least common

843
01:23:22,100 --> 01:23:27,829
denominator of x plus three times x. But our
next step would be to write each of these

844
01:23:27,829 --> 01:23:32,350
rational expressions over that common denominator
by multiplying the top and the bottom by the

845
01:23:32,350 --> 01:23:38,880
appropriate things. So one, in order to get
the common denominator of x plus 3x, I need

846
01:23:38,880 --> 01:23:44,310
to multiply the top and the bottom by x plus
three times x, one over x, I need to multiply

847
01:23:44,310 --> 01:23:49,640
the top and the bottom just by x plus three
since that's what's missing from the denominator

848
01:23:49,640 --> 01:23:53,380
x. Now, if I simplify a little bit,

849
01:23:53,380 --> 01:24:00,090
let's say this is x squared over that common
denominator, and

850
01:24:00,090 --> 01:24:09,150
here I have just x plus three times x over
that denominator, and here I have x plus three

851
01:24:09,149 --> 01:24:17,139
over that common denominator. Now add together
my fractions on the right side, so they have

852
01:24:17,140 --> 01:24:18,950
a common denominator.

853
01:24:18,949 --> 01:24:27,189
So this is x plus three times x plus x plus
three. And now I have two fractions that have

854
01:24:27,189 --> 01:24:32,939
that are equal that have the same denominator,
therefore their numerators have to be equal

855
01:24:32,939 --> 01:24:37,759
also. So the next step is to set the numerators
equal.

856
01:24:37,760 --> 01:24:46,860
So I get x squared is x plus three times x
plus x plus three. And if you look back at

857
01:24:46,859 --> 01:24:52,079
the previous way, we solve this equation,
you'll recognize this equation. And so from

858
01:24:52,079 --> 01:24:57,809
here on we just continue as before.

859
01:24:57,810 --> 01:25:00,020
When choosing between these two methods, I
personally tend

860
01:25:00,020 --> 01:25:03,600
prefer the clear the denominators method,
because it's a little bit less writing, you

861
01:25:03,600 --> 01:25:07,000
don't have to get rid of those denominators
earlier, you don't have to write them as many

862
01:25:07,000 --> 01:25:12,569
times. But some people find this one a little
bit easier to remember, a little easier to

863
01:25:12,569 --> 01:25:15,359
understand either of these methods is fine.

864
01:25:15,359 --> 01:25:21,529
One last caution, don't forget at the end,
to check your solutions and eliminate any

865
01:25:21,529 --> 01:25:23,460
extraneous solutions.

866
01:25:23,460 --> 01:25:30,670
These will be solutions that make the denominators
of your original equations go to zero.

867
01:25:30,670 --> 01:25:36,250
This video is about solving radical equations,
that is equations like this one that have

868
01:25:36,250 --> 01:25:40,789
square root signs in them, or cube roots or
any other kind of radical.

869
01:25:40,789 --> 01:25:45,621
When I see an equation with a square root
in it, I really want to get rid of the square

870
01:25:45,621 --> 01:25:51,079
root. But it'll be easiest to get rid of the
square root. If I first isolate the square

871
01:25:51,079 --> 01:25:56,750
root. In other words, I want to get the term
with the square root and that on one side

872
01:25:56,750 --> 01:26:01,739
of the equation by itself, and everything
else on the other side of the equation. If

873
01:26:01,739 --> 01:26:07,489
I start with my original equation, x plus
the square root of x equals 12. And I subtract

874
01:26:07,489 --> 01:26:12,929
x from both sides, then that does isolate
the square root term on the left side with

875
01:26:12,930 --> 01:26:19,850
everything else on the right. Once I've isolated
the term with the square root, I want to get

876
01:26:19,850 --> 01:26:27,079
rid of the square root. And I'll do that by
squaring both sides

877
01:26:27,079 --> 01:26:36,479
of my equation. So I'll take the square root
of x equals 12 minus x and square both sides.

878
01:26:36,479 --> 01:26:42,309
Now the square root of x squared is just x,
taking the square root and then squaring those

879
01:26:42,310 --> 01:26:45,370
operations undo each other

880
01:26:45,369 --> 01:26:54,170
to work out 12 minus x squared, write it out
and distribute 12 times 12 is 144 12 times

881
01:26:54,170 --> 01:27:02,539
minus x is minus 12x. I get another minus
12x from here.

882
01:27:02,539 --> 01:27:09,869
And finally minus x times minus x is positive
x squared. So I can combine my minus 12 x's,

883
01:27:09,869 --> 01:27:18,640
that's minus 24x. And now I can subtract x
from both sides to get zero equals 144 minus

884
01:27:18,640 --> 01:27:26,550
25x plus x squared. That's a quadratic equation,
I'll rewrite it in a little bit more standard

885
01:27:26,550 --> 01:27:27,890
form here.

886
01:27:27,890 --> 01:27:34,170
So now I've got a familiar quadratic equation
with no radical signs left in my equation,

887
01:27:34,170 --> 01:27:38,800
I'll just proceed to solve it like I usually
do for a quadratic, I'll try to factor it.

888
01:27:38,800 --> 01:27:46,079
So I'm going to look for two numbers that
multiply to 144 and add to minus 25. I know

889
01:27:46,079 --> 01:27:53,079
I'm going to need negative numbers to get
to negative 25. And in fact, I'll need two

890
01:27:53,079 --> 01:27:57,800
negative numbers. So they still multiply to
a positive number. So I'll start listing some

891
01:27:57,800 --> 01:28:04,730
factors of 144, I could have negative one
and a negative 144, negative two, and negative

892
01:28:04,729 --> 01:28:12,079
72, negative four, negative 36, and so on.
Once I've listed out the possible factors,

893
01:28:12,079 --> 01:28:17,780
it's not hard to find the two that add to
negative 25. So that's negative nine and negative

894
01:28:17,780 --> 01:28:28,029
16. So now I can factor in my quadratic equation
as x minus nine times x minus 16 equals zero,

895
01:28:28,029 --> 01:28:34,269
that means that x minus nine is zero or x
minus 16 is zero. So x equals nine or x equals

896
01:28:34,270 --> 01:28:41,000
16. I'm almost done. But there's one last
very important step. And that's to check the

897
01:28:41,000 --> 01:28:49,170
Solutions so that we can eliminate any extraneous
solutions and extraneous solution as a solution

898
01:28:49,170 --> 01:28:54,149
that we get that does not actually satisfy
our original equation and extraneous solutions

899
01:28:54,149 --> 01:29:00,269
can happen when you're solving equations with
radicals in them. So let's first check x equals

900
01:29:00,270 --> 01:29:06,000
nine. If we plug in to our original equation,
we get nine plus a squared of nine and we

901
01:29:06,000 --> 01:29:11,539
want that to equal 12. Well, the square root
of nine is three, and nine plus three does

902
01:29:11,539 --> 01:29:18,390
indeed equal 12. So that solution checks out.
Now, let's try x equals 16. Plugging in, we

903
01:29:18,390 --> 01:29:25,020
get 16 plus a squared of 16. And that's supposed
to equal 12. Well, that says 16 plus four

904
01:29:25,020 --> 01:29:31,880
is supposed to equal 12. But that most definitely
is not true. And so x equals 16. Turns out

905
01:29:31,880 --> 01:29:47,840
to be extraneous solution, and our only solution
is x equals nine

906
01:29:47,840 --> 01:30:26,100
This next equation might not look like an
equation involving radicals. But in fact,

907
01:30:26,100 --> 01:30:32,210
we can think of a fractional exponent as being
a radical in disguise. Let's start by doing

908
01:30:32,210 --> 01:30:38,829
the same thing we did on the previous problem
by isolating This time, we'll isolate the

909
01:30:38,829 --> 01:30:41,590
part of the equation

910
01:30:41,590 --> 01:30:46,730
that involves the fractional exponent. So
I'll start with the original equation, two

911
01:30:46,729 --> 01:30:53,299
times P to the four fifths equals 1/8. And
I'll divide both sides by two or equivalently,

912
01:30:53,300 --> 01:31:00,000
I can multiply both sides by one half, that
gives me p to the four fifths equals 1/16.

913
01:31:00,000 --> 01:31:05,739
And I've effectively isolated the part of
the equation with the fractional exponent

914
01:31:05,739 --> 01:31:12,920
as much as possible. Now, in the previous
example, the next step was to get rid of the

915
01:31:12,920 --> 01:31:18,980
radical. In this example, we're going to get
rid of the fractional exponent. And I'm going

916
01:31:18,979 --> 01:31:24,799
to actually do this in two stages. First,
I'm going to raise both sides to the fifth

917
01:31:24,800 --> 01:31:26,690
power.

918
01:31:26,689 --> 01:31:34,509
That's because when I take an exponent to
an exponent, I'm multiply my exponents. And

919
01:31:34,510 --> 01:31:41,840
so that becomes just p to the fourth equals
1/16 to the fifth power. Now, I'm going to

920
01:31:41,840 --> 01:31:48,360
get rid of the fourth power by raising both
sides to the 1/4 power, or by taking the fourth

921
01:31:48,359 --> 01:31:53,299
root, there's something that you need to be
careful about though, when taking

922
01:31:53,300 --> 01:32:01,880
an even root, or the one over an even number
power, you always have to consider plus or

923
01:32:01,880 --> 01:32:04,630
minus your answer.

924
01:32:04,630 --> 01:32:10,869
It's kind of like when you write x squared
equals four, and you take the square root

925
01:32:10,869 --> 01:32:16,449
of both sides, x could equal plus or minus
the square root of four, right, x could equal

926
01:32:16,449 --> 01:32:23,659
plus or minus two, since minus two squared
is four, just as well as two squared. So that's

927
01:32:23,659 --> 01:32:32,800
why when you take an even root, or a one over
an even power of both sides, you always need

928
01:32:32,800 --> 01:32:39,440
to include the plus or minus sign, when it's
an odd root or one over an odd power, you

929
01:32:39,439 --> 01:32:46,159
don't need to do that. If you had something
like x cubed equals negative eight, then x

930
01:32:46,159 --> 01:32:51,180
equals the cube root of negative eight, which
is negative two would be your only solution,

931
01:32:51,180 --> 01:32:55,560
you don't need to do the plus or minus because
positive two wouldn't work.

932
01:32:55,560 --> 01:33:02,390
So that aside, explains why we need this plus
or minus power. And now p to the four to the

933
01:33:02,390 --> 01:33:09,590
1/4. When I raise a power to a power, I multiply
by exponents, so that's just p to the one,

934
01:33:09,590 --> 01:33:17,630
which is equal to plus or minus 1/16 to the
fifth power to the 1/4 power.

935
01:33:17,630 --> 01:33:23,010
Now I just need to simplify this expression,
I don't really want to raise 1/16 to the fifth

936
01:33:23,010 --> 01:33:28,760
power, because 16 to the fifth power is like
a really huge number. So I think I'm actually

937
01:33:28,760 --> 01:33:32,420
going to rewrite this first

938
01:33:32,420 --> 01:33:42,649
as P equals plus or minus 1/16. I'll write
it back as the 5/4 power again.

939
01:33:42,649 --> 01:33:49,239
And as I continue to solve using my exponent
rules,

940
01:33:49,239 --> 01:33:58,939
I'm going to prefer to write this as 1/16
to the fourth root to the fifth power, because

941
01:33:58,939 --> 01:34:03,649
it's going to be easier to take the fourth
root, let's see the fourth root of 1/16 is

942
01:34:03,649 --> 01:34:08,149
the same thing as the fourth root of one over
the fourth root of 16. Raise all that to the

943
01:34:08,149 --> 01:34:14,359
fifth power. fourth root of one is just one
and the fourth root of 16 is to raise that

944
01:34:14,359 --> 01:34:19,279
to the fifth power, that's just going to be
one to the fifth over two to the fifth, which

945
01:34:19,279 --> 01:34:24,819
is plus or minus 130 seconds.

946
01:34:24,819 --> 01:34:28,949
The last step is to check answers.

947
01:34:28,949 --> 01:34:35,880
So I have the two answers p equals 130 seconds,
and P equals minus 130 seconds. And if I plug

948
01:34:35,880 --> 01:34:36,880
those both in

949
01:34:36,880 --> 01:34:43,199
1/32 to the fourth fifth power.

950
01:34:43,199 --> 01:34:51,109
That gives me two times one to the fourth
power over 32 to the fourth power, which is

951
01:34:51,109 --> 01:34:59,659
two times one over 32/5 routed to the fourth
power. fifth root of 32 is two

952
01:34:59,659 --> 01:35:07,519
Raisa to the fourth power, I get 16. So this
is two times 1/16, which is 1/8, just as we

953
01:35:07,520 --> 01:35:16,910
wanted in the original equation up here. Similarly,
we can check that the P equals negative 1/32

954
01:35:16,909 --> 01:35:24,039
actually does satisfy the equation, I'll leave
that step to the viewer.

955
01:35:24,039 --> 01:35:30,710
So our two solutions are p equals one over
32, and P equals minus one over 32. I do want

956
01:35:30,710 --> 01:35:35,310
to point out an alternate approach to getting
rid of the fractional exponent, we could have

957
01:35:35,310 --> 01:35:41,270
gotten rid of it all in one fell swoop by
raising both sides of our equation to the

958
01:35:41,270 --> 01:35:45,340
five fourths power.

959
01:35:45,340 --> 01:35:51,199
five fourths is the reciprocal of four fifths.
So when I use my exponent rules, and say that

960
01:35:51,199 --> 01:35:56,859
when I raise the power to the power, I multiply
my exponents, that gives me p to the four

961
01:35:56,859 --> 01:36:03,670
fifths times five fourths is plus or minus
1/16 to the five fourths, in other words,

962
01:36:03,670 --> 01:36:10,569
P to the One Power, which is just P is plus
or minus 1/16, to the five fourths, so that's

963
01:36:10,569 --> 01:36:16,759
an alternate and possibly faster way to get
the solution. Once again, the plus or minus

964
01:36:16,760 --> 01:36:22,440
comes from the fact that when we take the
5/4 power, we're really taking an even root

965
01:36:22,439 --> 01:36:27,859
a fourth root, and so we need to consider
both positive and negative answers.

966
01:36:27,859 --> 01:36:33,449
This video is about solving radical equations,
that is equations like this one that have

967
01:36:33,449 --> 01:36:39,059
square root signs in them, or cube roots or
any other kind of radical.

968
01:36:39,060 --> 01:36:45,680
In this video, we solved radical equations
by first isolating the radical sign, or the

969
01:36:45,680 --> 01:36:48,270
fractional exponent,

970
01:36:48,270 --> 01:36:54,670
and then removing the radical sine or the
fractional exponent

971
01:36:54,670 --> 01:37:01,630
by either squaring both sides or taking the
reciprocal power of both sides.

972
01:37:01,630 --> 01:37:06,550
This video is about solving equations with
absolute values in them.

973
01:37:06,550 --> 01:37:12,130
Recall that the absolute value of a positive
number is just the number, but the absolute

974
01:37:12,130 --> 01:37:17,761
value of a negative number is its opposite.
In general, I think of the absolute value

975
01:37:17,761 --> 01:37:23,170
of a number as representing its distance from
zero on the number line,

976
01:37:23,170 --> 01:37:25,840
the number four,

977
01:37:25,840 --> 01:37:32,360
and the number of negative four are both at
distance for from zero, and so the absolute

978
01:37:32,359 --> 01:37:35,939
value of both of them is four.

979
01:37:35,939 --> 01:37:44,339
Similarly, if I write the equation, the absolute
value of x is three, that means that x has

980
01:37:44,340 --> 01:37:50,319
to be three units away from zero on the number
line.

981
01:37:50,319 --> 01:37:56,319
And so X would have to be either negative
three, or three. Let's start with the equation

982
01:37:56,319 --> 01:38:00,439
three times the absolute value of x plus two
equals four.

983
01:38:00,439 --> 01:38:08,639
I'd like to isolate the absolute value part
of the equation, I can do this by starting

984
01:38:08,640 --> 01:38:14,090
with my original equation,

985
01:38:14,090 --> 01:38:17,670
subtracting two from both sides

986
01:38:17,670 --> 01:38:23,920
and dividing both sides by three.

987
01:38:23,920 --> 01:38:31,100
Now I'll think in terms of distance on a number
line, the absolute value of x is two thirds

988
01:38:31,100 --> 01:38:40,671
means that x is two thirds away from zero.
So x could be here for here at negative two

989
01:38:40,671 --> 01:38:43,289
thirds or two thirds.

990
01:38:43,289 --> 01:38:50,420
And the answer to my equation is x is negative
two thirds, or two thirds,

991
01:38:50,420 --> 01:38:52,711
I can check my answers by plugging in

992
01:38:52,711 --> 01:39:00,090
three times the added value of negative two
thirds plus two, I need to check it that equals

993
01:39:00,090 --> 01:39:04,739
four. Well, the absolute value of negative
two thirds is just two thirds. So this is

994
01:39:04,739 --> 01:39:10,969
three times two thirds plus two, which works
out to four.

995
01:39:10,970 --> 01:39:20,050
Similarly, if I plug in positive two thirds,
it also works out to give me the correct answer.

996
01:39:20,050 --> 01:39:24,239
The second example is a little different,
because the opposite value sign is around

997
01:39:24,239 --> 01:39:29,210
a more complicated expression, not just around
the X.

998
01:39:29,210 --> 01:39:35,930
I would start by isolating the absolute value
part.

999
01:39:35,930 --> 01:39:41,730
But it's already isolated. So I'll just go
ahead and jump to thinking about distance

1000
01:39:41,729 --> 01:39:49,899
on the number line. So on my number line,
the whole expression 3x plus two is supposed

1001
01:39:49,899 --> 01:39:53,979
to be at a distance of four from zero.

1002
01:39:53,979 --> 01:39:59,939
So that means that 3x plus two is here at
four or 3x plus two

1003
01:39:59,939 --> 01:40:03,839
though is it negative four, all right, those
as equations

1004
01:40:03,840 --> 01:40:12,090
3x plus two equals four, or 3x plus two is
minus four. And then I can solve.

1005
01:40:12,090 --> 01:40:20,369
So this becomes 3x equals two, or x equals
two thirds. And over here, I get 3x equals

1006
01:40:20,369 --> 01:40:29,489
minus six, or x equals minus two. Finally,
I'll check my answers.

1007
01:40:29,489 --> 01:40:32,800
I'll leave it to you to verify that they both
work.

1008
01:40:32,800 --> 01:40:38,590
A common mistake on absolute value equations
is to get rid of the absolute value signs

1009
01:40:38,590 --> 01:40:44,569
like we did here, and then just solve for
one answer, instead of solving for both answers.

1010
01:40:44,569 --> 01:40:48,759
Another mistake sometimes people make is,
once they get the first answer, they just

1011
01:40:48,760 --> 01:40:52,640
assume that the negative of that works also.

1012
01:40:52,640 --> 01:40:57,900
But that doesn't always work. In the first
example, our two answers were both the negatives

1013
01:40:57,899 --> 01:41:02,710
of each other. But in our second examples,
or two answers, were not just the opposites

1014
01:41:02,710 --> 01:41:06,619
of each other one was two thirds and the other
was negative two.

1015
01:41:06,619 --> 01:41:12,659
In this third example, let's again, isolate
the absolute value part of the equation.

1016
01:41:12,659 --> 01:41:16,760
So starting with our original equation,

1017
01:41:16,760 --> 01:41:22,550
we can subtract 16 from both sides,

1018
01:41:22,550 --> 01:41:31,320
and divide both sides by five or equivalently,
multiply by 1/5.

1019
01:41:31,319 --> 01:41:35,049
Now let's think about distance on the number
line,

1020
01:41:35,050 --> 01:41:41,770
we have an absolute value needs to equal negative
three. So that means whatever is inside the

1021
01:41:41,770 --> 01:41:48,400
absolute value sign needs to be at distance
negative three from zero, well, you can't

1022
01:41:48,399 --> 01:41:52,849
be at distance negative three from zero. Another
way of thinking about this is you can't have

1023
01:41:52,850 --> 01:41:57,220
the absolute value of something and end up
with a negative number if the value is always

1024
01:41:57,220 --> 01:42:05,530
positive, or zero. So this equation doesn't
actually make sense, and there are no solutions

1025
01:42:05,529 --> 01:42:07,509
to this equation.

1026
01:42:07,510 --> 01:42:15,010
In this video, we solved absolute value equations.
In many cases, an absolute value equation

1027
01:42:15,010 --> 01:42:20,949
will have two solutions. But in some cases,
it'll have no solutions. And occasionally,

1028
01:42:20,949 --> 01:42:25,079
it'll have just one solution.

1029
01:42:25,079 --> 01:42:30,979
This video is about interval notation, and
easy and well known way to record inequalities.

1030
01:42:30,979 --> 01:42:39,159
Before dealing with interval notation, it
is important to know how to deal with inequalities.

1031
01:42:39,159 --> 01:42:45,420
Our first example of an inequality is written
here, all numbers between one and three, not

1032
01:42:45,420 --> 01:42:46,750
including one and three.

1033
01:42:46,750 --> 01:42:53,260
First, we used to write down our variable
x, that it is important to locate the key

1034
01:42:53,260 --> 01:43:01,320
values for this problem that is one and three.
Now here we see that x is between these two

1035
01:43:01,319 --> 01:43:07,529
numbers, meaning we will have one inequality
statement on each side of the variable. Here

1036
01:43:07,529 --> 01:43:12,779
it says not including one and three, which
means that we do not have an or equal to sign

1037
01:43:12,779 --> 01:43:18,670
beneath each inequality. Here we will put
one the lowest key value, and here we will

1038
01:43:18,670 --> 01:43:24,489
put three the highest key value. Next, we're
going to graph this inequality on a number

1039
01:43:24,489 --> 01:43:27,069
line.

1040
01:43:27,069 --> 01:43:34,029
Here we write our key values one, and three.
Because it is not including one, and three,

1041
01:43:34,029 --> 01:43:40,189
we have an empty circle around each number.
Because it isn't the numbers between we have

1042
01:43:40,189 --> 01:43:41,289
a line connecting them.

1043
01:43:41,289 --> 01:43:50,069
The last step of this problem is writing this
inequality in interval notation.

1044
01:43:50,069 --> 01:43:54,949
writing things in interval notation is kind
of like writing an ordered pair. Here we will

1045
01:43:54,949 --> 01:43:57,579
put one and then here we will put three.

1046
01:43:57,579 --> 01:44:03,189
Next we need to put brackets around these
numbers.

1047
01:44:03,189 --> 01:44:09,429
For this problem, it is not including one
in three, which means we will use soft brackets.

1048
01:44:09,430 --> 01:44:15,970
However, if it were including one and three,
we will use hard brackets.

1049
01:44:15,970 --> 01:44:22,960
It is important to note for interval notation,
that the smallest value always goes on the

1050
01:44:22,960 --> 01:44:29,039
left and the biggest value always goes on
the right.

1051
01:44:29,039 --> 01:44:32,680
You also include a comma between your two
key values.

1052
01:44:32,680 --> 01:44:41,770
Now let's work on problem B. Once again we
write our variable x. Again it is between

1053
01:44:41,770 --> 01:44:47,170
negative four and two but this time it is
including a negative four into this requires

1054
01:44:47,170 --> 01:44:53,859
us to have the or equal to assign below the
inequality. Then we put our lowest key value

1055
01:44:53,859 --> 01:44:59,739
here and their highest here. The number line
graph for this problem is slightly different.

1056
01:44:59,739 --> 01:45:00,739
We still

1057
01:45:00,739 --> 01:45:07,399
have our key values negative four and two.
But instead of an open circle, like we have

1058
01:45:07,399 --> 01:45:15,759
right here, we instead use a closed circle,
representing that it is including negative

1059
01:45:15,760 --> 01:45:21,320
four and two, you complete this with a line
in between. Last we write this in interval

1060
01:45:21,319 --> 01:45:27,170
notation. In the last problem, I said that
for numbers, including the outside values,

1061
01:45:27,170 --> 01:45:35,470
we would use these hard brackets. Now we are
actually using this. So we have or hard bracket

1062
01:45:35,470 --> 01:45:40,650
on each side because isn't including foreign
zoo, then we put our smallest value on the

1063
01:45:40,649 --> 01:45:46,799
left, and our highest value on the right,
with the comma in between.

1064
01:45:46,800 --> 01:45:51,029
You now know how to correctly write these
two types of intervals.

1065
01:45:51,029 --> 01:45:58,689
Now we are going to practice transforming
these from inequality notation to interval

1066
01:45:58,689 --> 01:46:05,569
notation, and vice versa. This is slightly
more difficult than our previous examples

1067
01:46:05,569 --> 01:46:11,939
of having two soft brackets or two hard brackets,
we'll have two different types. Now on the

1068
01:46:11,939 --> 01:46:18,559
left side will have a hard bracket because
it is including the three. However, on the

1069
01:46:18,560 --> 01:46:24,039
right side, it is a soft bracket because it
is it it is not including the one that we

1070
01:46:24,039 --> 01:46:31,880
put a comma in the middle, a lower key value
and our higher key value. For the next problem,

1071
01:46:31,880 --> 01:46:37,140
we're taking an equation already written an
interval notation and putting it back into

1072
01:46:37,140 --> 01:46:38,430
inequality notation.

1073
01:46:38,430 --> 01:46:47,600
For the second problem, we can see are key
values as being five and negative infinity.

1074
01:46:47,600 --> 01:46:54,340
Now, this sounds a little bit weird, but let's
just set up or an equality, we have our variable

1075
01:46:54,340 --> 01:47:02,470
x, which is less than because of the soft
bracket five, and greater than without the

1076
01:47:02,470 --> 01:47:09,409
or equal to because it has a soft bracket
there to infinity. But because x is always

1077
01:47:09,409 --> 01:47:16,319
greater than negative infinity, we can take
out this part, leaving us with x is less than

1078
01:47:16,319 --> 01:47:17,739
five.

1079
01:47:17,739 --> 01:47:25,269
It is important to note that a soft bracket
always accompanies infinity. This is because

1080
01:47:25,270 --> 01:47:32,010
infinity is not a real number, so we cannot
include it just go as far up to it as we can.

1081
01:47:32,010 --> 01:47:38,110
The next problem is a little tricky, because
we don't see another key value over here.

1082
01:47:38,109 --> 01:47:43,829
But let's just start off the equation as a
soft bracket on this side due to the absence

1083
01:47:43,829 --> 01:47:51,510
of an order equal to sign. And negative 15
is the lower key value.

1084
01:47:51,510 --> 01:47:57,190
On the other side, we put the highest possible
number infinity, and then close it off with

1085
01:47:57,189 --> 01:48:05,289
a soft bracket. We can see the relation of
this to the previous problem. And how it goes

1086
01:48:05,289 --> 01:48:12,119
into this problem when x is greater than a
key value instead of less than a key value.

1087
01:48:12,119 --> 01:48:19,189
Once again, we have our sauce bracket for
infinity, which is always true.

1088
01:48:19,189 --> 01:48:26,279
Part D brings up an important point about
which number goes on the left. For all of

1089
01:48:26,279 --> 01:48:30,130
our other problems, we have had the inequalities
pointing left

1090
01:48:30,130 --> 01:48:41,609
instead of rights. When seeing Part D, you
might think oh, four is on the left, so it

1091
01:48:41,609 --> 01:48:47,799
would go here, and zero is on the right who
would go here. But that is not true.

1092
01:48:47,800 --> 01:48:53,940
When writing inequalities, you must always
have the lower value on the left, which for

1093
01:48:53,939 --> 01:49:01,550
this problem is zero. To fix this, we can
simply flip this inequality. So it reads like

1094
01:49:01,551 --> 01:49:06,800
this, it is still identical just written in
an easier form to transform it into interval

1095
01:49:06,800 --> 01:49:13,070
notation, then we can see that zero has a
soft bracket, because it is not including

1096
01:49:13,069 --> 01:49:19,899
zero, but r comma four or other key value
and a hard backup because it is or equal to

1097
01:49:19,899 --> 01:49:21,859
four.

1098
01:49:21,859 --> 01:49:29,420
For interval notation, you must always have
the smaller number on the left side.

1099
01:49:29,420 --> 01:49:36,190
This was our video on interval notation, an
alternative way to write inequalities.

1100
01:49:36,189 --> 01:49:41,739
This video is about solving inequalities that
have absolute value signs in them.

1101
01:49:41,739 --> 01:49:45,979
Let's look at the inequality absolute value
of x is less than five on the number line.

1102
01:49:45,979 --> 01:49:52,529
Thinking of absolute value as distance. This
means that the distance between x and zero

1103
01:49:52,529 --> 01:50:01,380
is less than five units. So x has to live
somewhere in between negative five and five

1104
01:50:01,380 --> 01:50:06,560
We can express this as an inequality without
absolute value signs by saying negative five

1105
01:50:06,560 --> 01:50:12,850
is less than x, which is less than five. Or
we can use interval notation, soft bracket

1106
01:50:12,850 --> 01:50:16,890
negative five, five, soft bracket.

1107
01:50:16,890 --> 01:50:20,760
Both of these formulations are equivalent
to the original one, but don't involve the

1108
01:50:20,760 --> 01:50:23,650
absolute value signs.

1109
01:50:23,649 --> 01:50:28,670
In the second example, we're looking for the
values of x for which the absolute value of

1110
01:50:28,670 --> 01:50:32,649
x is greater than or equal to five.

1111
01:50:32,649 --> 01:50:39,299
on the number line, this means that the distance
of x from the zero, it's got to be bigger

1112
01:50:39,300 --> 01:50:42,070
than or equal to five units.

1113
01:50:42,069 --> 01:50:48,130
A distance bigger than five units means that
x has to live somewhere over here, or somewhere

1114
01:50:48,130 --> 01:50:54,730
over here, where it's farther than five units
away from zero. Course x could also have a

1115
01:50:54,729 --> 01:51:00,320
distance equal to five units. So I'll fill
in that dot and shade in the other parts of

1116
01:51:00,320 --> 01:51:04,569
the number line that satisfy my inequality.

1117
01:51:04,569 --> 01:51:10,130
Now I can rewrite the inequality without the
absolute value symbols by saying that x is

1118
01:51:10,130 --> 01:51:17,090
less than or equal to negative five, or x
is greater than or equal to five. I could

1119
01:51:17,090 --> 01:51:20,239
also write this in interval notation,

1120
01:51:20,239 --> 01:51:26,689
soft bracket, negative infinity, negative
five, hard bracket. And the second part is

1121
01:51:26,689 --> 01:51:33,829
hard bracket five infinity soft bracket, I
combine these with a u for union. Because

1122
01:51:33,829 --> 01:51:38,269
I'm trying to describe all these points on
the number line together with all these other

1123
01:51:38,270 --> 01:51:39,920
points.

1124
01:51:39,920 --> 01:51:44,520
Let's take this analysis a step further with
a slightly more complicated problem. Now I

1125
01:51:44,520 --> 01:51:49,010
want the absolute value of three minus two
t to be less than four.

1126
01:51:49,010 --> 01:51:56,020
And an absolute value less than four means
a distance less than four on the number line.

1127
01:51:56,020 --> 01:52:00,260
But it's not the variable t that lives in
here at a distance of less than four from

1128
01:52:00,260 --> 01:52:04,960
zero, it's the whole expression, three minus
two t.

1129
01:52:04,960 --> 01:52:13,980
So three minus two t, live somewhere in here.
And I can rewrite this as an inequality without

1130
01:52:13,979 --> 01:52:21,209
absolute value signs by saying negative four
is less than three minus two t is less than

1131
01:52:21,210 --> 01:52:27,730
four. Now I have a compound inequality that
I can solve the usual way. First, I subtract

1132
01:52:27,729 --> 01:52:33,739
three from all three sides to get negative
seven is less than negative two t is less

1133
01:52:33,739 --> 01:52:40,719
than one. And now I'll divide all three sides
by negative two. Since negative two is a negative

1134
01:52:40,720 --> 01:52:46,140
number, this reverses the directions of the
inequalities.

1135
01:52:46,140 --> 01:52:52,050
Simplifying, I get seven halves is greater
than t is greater than negative one half.

1136
01:52:52,050 --> 01:52:58,720
So my final answer on the number line looks
like all the stuff between negative a half

1137
01:52:58,720 --> 01:53:01,880
and seven halves.

1138
01:53:01,880 --> 01:53:07,340
But not including the endpoints, and an interval
notation, I can write this soft bracket negative

1139
01:53:07,340 --> 01:53:09,920
a half, seven, half soft bracket,

1140
01:53:09,920 --> 01:53:14,539
please pause the video and try the next problem
on your own.

1141
01:53:14,539 --> 01:53:19,909
thinking in terms of distance, this inequality
says that the distance between the expression

1142
01:53:19,909 --> 01:53:25,880
three minus two t and zero is always bigger
than four. Let me draw this on the number

1143
01:53:25,880 --> 01:53:27,609
line.

1144
01:53:27,609 --> 01:53:33,779
If three minus two t has a distance bigger
than four from zero, then it can't be in this

1145
01:53:33,779 --> 01:53:39,869
region that's near zero, it has to be on the
outside in one of these two regions.

1146
01:53:39,869 --> 01:53:48,569
That is three minus two t is either less than
negative four, or three minus two t is bigger

1147
01:53:48,569 --> 01:53:54,369
than four. I solve these two inequalities
separately, the first one, subtracting three

1148
01:53:54,369 --> 01:54:01,390
from both sides, I get negative two t is less
than negative seven divided by negative two,

1149
01:54:01,390 --> 01:54:08,350
I get T is bigger than seven halves. And then
on the other side, I get negative two t is

1150
01:54:08,350 --> 01:54:15,820
greater than one. So t is less than negative
one half, I had to flip inequalities in the

1151
01:54:15,819 --> 01:54:21,259
last step due to dividing by a negative number.
So let's check this out on the number line

1152
01:54:21,260 --> 01:54:28,750
again, the first piece says that t is greater
than seven halves. I'll draw that over here.

1153
01:54:28,750 --> 01:54:41,029
And the second piece says that t is less than
negative one half. I'll draw that down here.

1154
01:54:41,029 --> 01:54:46,380
Because these two statements are joined with
an or I'm looking for the t values that are

1155
01:54:46,380 --> 01:54:52,690
in this one, or in this one. That is I want
both of these regions put together. So an

1156
01:54:52,689 --> 01:55:00,029
interval notation, this reads negative infinity
to negative one half soft bracket union soft

1157
01:55:00,029 --> 01:55:01,029
bracket

1158
01:55:01,029 --> 01:55:03,130
Seven halves to infinity.

1159
01:55:03,130 --> 01:55:09,289
This last example looks more complicated.
But if I simplify first and isolate the absolute

1160
01:55:09,289 --> 01:55:14,500
value part, it looks pretty much like the
previous ones. So I'll start by subtracting

1161
01:55:14,500 --> 01:55:20,310
seven from both sides. And then I'll divide
both sides by two. Now I'll draw my number

1162
01:55:20,310 --> 01:55:21,310
line.

1163
01:55:21,310 --> 01:55:28,060
And I'm looking for this expression for x
plus five, to always have distance greater

1164
01:55:28,060 --> 01:55:33,471
than or equal to negative three from zero,
wait a second, distance greater than equal

1165
01:55:33,470 --> 01:55:36,759
to negative three, well, distance is always
greater than or equal to negative three is

1166
01:55:36,760 --> 01:55:43,340
always greater than equal to zero. So this,
in fact, is always true.

1167
01:55:43,340 --> 01:55:50,119
And so the answer to my inequality is all
numbers between negative infinity and infinity.

1168
01:55:50,119 --> 01:55:55,930
In other words, all real numbers.

1169
01:55:55,930 --> 01:56:01,400
Once solving absolute value inequalities,
it's good to think about distance. And absolute

1170
01:56:01,399 --> 01:56:08,500
value of something that's less than a number
means that whatever's inside the absolute

1171
01:56:08,500 --> 01:56:12,869
value signs is close to zero.

1172
01:56:12,869 --> 01:56:18,579
On the other hand, an absolute value is something
and being greater than a number means that

1173
01:56:18,579 --> 01:56:25,829
whatever's inside the absolute value sign
is far away from zero, because its distance

1174
01:56:25,829 --> 01:56:29,960
from zero is bigger than that certain number.

1175
01:56:29,960 --> 01:56:34,739
drawing these pictures on the number line
is a helpful way to rewrite the absolute value

1176
01:56:34,739 --> 01:56:40,899
and equality as an inequality that doesn't
contain an absolute value sign. In this case,

1177
01:56:40,899 --> 01:56:47,179
it would be negative three is less than x
plus two is less than three. And in the other

1178
01:56:47,180 --> 01:56:53,630
case, it would be either x plus two is less
than negative three, or x plus two is greater

1179
01:56:53,630 --> 01:56:55,880
than three.

1180
01:56:55,880 --> 01:57:01,670
This video is about solving linear inequalities.
Those are inequalities like this one that

1181
01:57:01,670 --> 01:57:09,170
involve a variable here x, but don't involve
any x squared or other higher power terms.

1182
01:57:09,170 --> 01:57:15,739
The good news is, we can solve linear inequalities,
just like we solve linear equations by distributing,

1183
01:57:15,739 --> 01:57:19,779
adding and subtracting terms to both sides
and multiplying and dividing by numbers on

1184
01:57:19,779 --> 01:57:26,719
both sides. The only thing that's different
is that if you multiply or divide by a negative

1185
01:57:26,720 --> 01:57:32,840
number, then you need to reverse the direction
of the inequality. For example, if we had

1186
01:57:32,840 --> 01:57:39,319
the inequality, negative x is less than negative
five, and we wanted to multiply both sides

1187
01:57:39,319 --> 01:57:45,710
by negative one to get rid of the negative
signs, we'd have to also switch or reverse

1188
01:57:45,710 --> 01:57:50,560
the inequality. With this caution in mind,
let's look at our first example.

1189
01:57:50,560 --> 01:57:56,170
Since our variable x is trapped in parentheses,
I'll distribute the negative five to free

1190
01:57:56,170 --> 01:57:58,569
it from the parentheses.

1191
01:57:58,569 --> 01:58:04,849
That gives me negative 5x minus 10 plus three
is greater than eight.

1192
01:58:04,850 --> 01:58:10,020
Negative 10 plus three is negative seven,
so I'll rewrite this as negative 5x minus

1193
01:58:10,020 --> 01:58:16,890
seven is greater than eight. Now I'll add
seven to both sides to get negative 5x is

1194
01:58:16,890 --> 01:58:23,100
greater than 15. Now I'd like to divide both
sides by negative five. Since negative five

1195
01:58:23,100 --> 01:58:31,260
is a negative number that reverses the inequality.
So I get x is less than 15 divided by negative

1196
01:58:31,260 --> 01:58:36,130
five. In other words, x is less than negative
three.

1197
01:58:36,130 --> 01:58:40,390
If I wanted to graph this on a number line,
I just need to put down a negative three,

1198
01:58:40,390 --> 01:58:44,400
an open circle around it, and shade in to
the left,

1199
01:58:44,399 --> 01:58:51,049
I use an open circle, because x is strictly
less than negative three and can't equal negative

1200
01:58:51,050 --> 01:58:56,070
three. If I wanted to write this in interval
notation, I've added a soft bracket negative

1201
01:58:56,069 --> 01:59:01,989
infinity, negative three soft bracket. Again,
the soft bracket is because the negative three

1202
01:59:01,989 --> 01:59:10,090
is not included. This next example is an example
of a compound inequality. It has two parts.

1203
01:59:10,090 --> 01:59:15,900
Either this statement is true, or this statement
is true. I want to find the values of x that

1204
01:59:15,899 --> 01:59:21,529
satisfy either one. I'll solve this by working
out each part separately, and then combining

1205
01:59:21,529 --> 01:59:28,079
them at the end. For the inequality on the
left, I'll copy it over here. I'm going to

1206
01:59:28,079 --> 01:59:32,149
add four to both sides.

1207
01:59:32,149 --> 01:59:36,189
then subtract x from both sides

1208
01:59:36,189 --> 01:59:40,149
and then divide both sides by two.

1209
01:59:40,149 --> 01:59:45,170
I didn't have to reverse the inequality sign
because I divided by a positive number

1210
01:59:45,170 --> 01:59:51,640
on the right side, I'll copy the equation
over, subtract one from both sides, and divide

1211
01:59:51,640 --> 01:59:59,329
both sides by 696 is the same as three halves.
Now I'm looking for the x values that make

1212
01:59:59,329 --> 02:00:00,329
this statement.

1213
02:00:00,329 --> 02:00:07,300
row four, make this statement true. Let me
graph this on a number line.

1214
02:00:07,300 --> 02:00:14,329
x is less than or equal to negative two, means
I put a filled in circle there and graph everything

1215
02:00:14,329 --> 02:00:22,670
to the left. X is greater than three halves
means I put a empty circle here and shaded

1216
02:00:22,670 --> 02:00:25,239
and everything to the right.

1217
02:00:25,239 --> 02:00:33,050
My final answer includes both of these pieces,
which I'll reshade in green, because x is

1218
02:00:33,050 --> 02:00:38,880
allowed to be an either one or the other.
Finally, I can write this in interval notation.

1219
02:00:38,880 --> 02:00:44,340
The first piece on the number line can be
described as soft bracket negative infinity,

1220
02:00:44,340 --> 02:00:50,900
negative two hard bracket. And the second
piece can be described as soft bracket three

1221
02:00:50,899 --> 02:00:57,379
halves, infinity soft bracket to indicate
that x can be in either one of these pieces,

1222
02:00:57,380 --> 02:01:01,590
I use the union side, which is a U.

1223
02:01:01,590 --> 02:01:09,340
That means that my answer includes all x values
in here together with all x values in here.

1224
02:01:09,340 --> 02:01:15,079
This next example is also a compound inequality.
This time I'm joined by an ad, the and means

1225
02:01:15,079 --> 02:01:22,029
I'm looking for all y values that satisfy
both this and this at the same time. Again,

1226
02:01:22,029 --> 02:01:25,340
I can solve each piece separately

1227
02:01:25,340 --> 02:01:34,090
on the left, to isolate the Y, I need to multiply
by negative three halves on both sides. So

1228
02:01:34,090 --> 02:01:40,460
that gives me Why is less than negative 12
times negative three halves, the greater than

1229
02:01:40,460 --> 02:01:46,649
flip to a less than because I was multiplying
by negative three halves, which is a negative

1230
02:01:46,649 --> 02:01:47,729
number.

1231
02:01:47,729 --> 02:01:54,469
By clean up the right side, I get why is less
than 18.

1232
02:01:54,470 --> 02:01:59,590
On the right side, I'll start by subtracting
two from both sides.

1233
02:01:59,590 --> 02:02:07,300
And now I'll divide by negative four, again,
a negative number, so that flips the inequality.

1234
02:02:07,300 --> 02:02:13,270
So that's why is less than three over negative
four. In other words, y is less than negative

1235
02:02:13,270 --> 02:02:19,710
three fourths. Again, I'm looking for the
y values that make both of these statements

1236
02:02:19,710 --> 02:02:22,230
true at the same time.

1237
02:02:22,229 --> 02:02:25,819
Let me graph this on the number line,

1238
02:02:25,819 --> 02:02:34,819
the Y is less than 18. I can graph that by
drawing the number 18.

1239
02:02:34,819 --> 02:02:41,519
I don't want to include it. So I use an empty
circle and I shaded everything to the left

1240
02:02:41,520 --> 02:02:47,490
the statement y is less than negative three
fourths, I need to draw a negative three fourths.

1241
02:02:47,489 --> 02:02:54,529
And again, I don't include it, but I do include
everything on the left. Since I want the y

1242
02:02:54,529 --> 02:02:59,380
values for which both of these statements
are true, I need the y values that are in

1243
02:02:59,380 --> 02:03:05,900
both colored blue and colored red. And so
that would be this part right here. I'll just

1244
02:03:05,899 --> 02:03:10,819
draw it above so you can see it easily. So
that would be all the numbers from negative

1245
02:03:10,819 --> 02:03:15,869
three fourths and lower, not including negative
three, four, since those are the parts of

1246
02:03:15,869 --> 02:03:21,380
the number line that have both red and blue
colors on them. In interval notation, my final

1247
02:03:21,380 --> 02:03:26,920
answer will be soft bracket negative infinity
to negative three fourths soft bracket.

1248
02:03:26,920 --> 02:03:34,720
As my final example, I have an inequality
that has two inequality signs in it negative

1249
02:03:34,720 --> 02:03:40,380
three is less than or equal to 6x minus two
is less than 10. I can think of this as being

1250
02:03:40,380 --> 02:03:47,750
a compound inequality with two parts, negative
3x is less than or equal to 6x minus two.

1251
02:03:47,750 --> 02:03:54,289
And at the same time 6x minus two is less
than 10. I could solve this into pieces as

1252
02:03:54,289 --> 02:04:00,460
before. But instead, it's a little more efficient
to just solve it all at once, by doing the

1253
02:04:00,460 --> 02:04:08,010
same thing to all three sides. So as a first
step, I'll add two to all three sides. That

1254
02:04:08,010 --> 02:04:14,400
gives me negative one is less than or equal
to 6x is less than 12. And now I'm going to

1255
02:04:14,399 --> 02:04:20,149
divide all three sides by six to isolate the
x. So that gives me negative one, six is less

1256
02:04:20,149 --> 02:04:24,189
than or equal to x is less than two.

1257
02:04:24,189 --> 02:04:29,129
If we solved it, instead, in two pieces above,
we'd end up with the same thing because we

1258
02:04:29,130 --> 02:04:35,150
get negative one six is less than or equal
to x from this piece, and we'd get x is less

1259
02:04:35,149 --> 02:04:40,299
than two on this piece. And because of the
and statement, that's the same thing as saying

1260
02:04:40,300 --> 02:04:45,000
negative one, six is less than or equal to
x, which is less than two.

1261
02:04:45,000 --> 02:04:50,050
Either way we do it. Let's see if what it
looks like on the number line. So on the number

1262
02:04:50,050 --> 02:04:58,829
line, we're looking for things that are between
two and negative one six, including the negative

1263
02:04:58,829 --> 02:04:59,970
one, six

1264
02:04:59,970 --> 02:05:05,470
But not including the to interval notation,
we can write this as hard bracket, negative

1265
02:05:05,470 --> 02:05:08,510
162 soft bracket.

1266
02:05:08,510 --> 02:05:13,600
In this video, we solve linear inequalities,
including some compound inequalities, joined

1267
02:05:13,600 --> 02:05:18,670
by the conjunctions and, and or,

1268
02:05:18,670 --> 02:05:22,789
remember when we're working with and we're
looking for places on the number line where

1269
02:05:22,789 --> 02:05:24,899
both statements are true.

1270
02:05:24,899 --> 02:05:30,769
That is, we're looking for the overlap on
the number line.

1271
02:05:30,770 --> 02:05:35,510
In this case, the points on the number line
that are colored both red and blue at the

1272
02:05:35,510 --> 02:05:36,650
same time.

1273
02:05:36,649 --> 02:05:43,089
When we're working with oral statements, we're
looking for places where either one or the

1274
02:05:43,090 --> 02:05:46,380
other statement is true or both

1275
02:05:46,380 --> 02:05:49,010
on the number line,

1276
02:05:49,010 --> 02:05:54,380
this corresponds to points that are colored
either red or blue, or both. And in this picture,

1277
02:05:54,380 --> 02:05:57,949
that will actually correspond to the entire
number line.

1278
02:05:57,949 --> 02:06:04,630
In this video, we'll solve inequalities involving
polynomials like this one, and inequalities

1279
02:06:04,630 --> 02:06:09,340
involving rational expressions like this one.

1280
02:06:09,340 --> 02:06:14,500
Let's start with a simple example. Maybe a
deceptively simple example. If you see the

1281
02:06:14,500 --> 02:06:19,550
inequality, x squared is less than four, you
might be very tempted to take the square root

1282
02:06:19,550 --> 02:06:26,220
of both sides and get something like x is
less than two as your answer. But in fact,

1283
02:06:26,220 --> 02:06:27,789
that doesn't work.

1284
02:06:27,789 --> 02:06:35,850
To see why it's not correct, consider the
x value of negative 10.

1285
02:06:35,850 --> 02:06:43,500
Negative 10 satisfies the inequality, x is
less than two since negative 10 is less than

1286
02:06:43,500 --> 02:06:51,130
two. But it doesn't satisfy the inequality
x squared is less than four, since negative

1287
02:06:51,130 --> 02:06:58,420
10 squared is 100, which is not less than
four. So these two inequalities are not the

1288
02:06:58,420 --> 02:07:04,230
same. And it doesn't work to solve a quadratic
inequality just to take the square root of

1289
02:07:04,229 --> 02:07:09,349
both sides, you might be thinking part of
why this reasoning is wrong, as we've ignored

1290
02:07:09,350 --> 02:07:15,420
the negative two option, right? If we had
the equation, x squared equals four, then

1291
02:07:15,420 --> 02:07:21,430
x equals two would just be one option, x equals
negative two would be another solution. So

1292
02:07:21,430 --> 02:07:27,869
somehow, our solution to this inequality should
take this into account. In fact, a good way

1293
02:07:27,869 --> 02:07:33,960
to solve an inequality involving x squares
or higher power terms, is to solve the associated

1294
02:07:33,960 --> 02:07:40,399
equation first. But before we even do that,
I like to pull everything over to one side,

1295
02:07:40,399 --> 02:07:44,449
so that my inequality has zero on the other
side.

1296
02:07:44,449 --> 02:07:49,569
So for our equation, I'll subtract four from
both sides to get x squared minus four is

1297
02:07:49,569 --> 02:07:50,590
less than zero.

1298
02:07:50,590 --> 02:07:58,039
Now, I'm going to actually solve the associated
equation, x squared minus four is equal to

1299
02:07:58,039 --> 02:08:06,420
zero, I can do this by factoring to x minus
two times x plus two is equal to zero. And

1300
02:08:06,420 --> 02:08:12,579
I'll set my factors equal to zero, and I get
x equals two and x equals minus two.

1301
02:08:12,579 --> 02:08:17,970
Now, I'm going to plot the solutions to my
equation on the number line. So I write down

1302
02:08:17,970 --> 02:08:23,240
negative two and two, those are the places
where my expression x squared minus four is

1303
02:08:23,239 --> 02:08:26,170
equal to zero.

1304
02:08:26,170 --> 02:08:30,550
Since I want to find where x squared minus
four is less than zero, I want to know where

1305
02:08:30,550 --> 02:08:36,029
this expression x squared minus four is positive
or negative, a good way to find that out is

1306
02:08:36,029 --> 02:08:43,489
to plug in test values. So first, a plug in
a test value in this area of the number line,

1307
02:08:43,489 --> 02:08:47,369
something less than negative to say x equals
negative three.

1308
02:08:47,369 --> 02:08:53,720
If I plug in negative three into x squared
minus four, I get negative three squared minus

1309
02:08:53,720 --> 02:09:02,090
four, which is nine minus four, which is five,
that's a positive number. So at negative three,

1310
02:09:02,090 --> 02:09:08,029
the expression x squared minus four is positive.
And in fact, everywhere on this region of

1311
02:09:08,029 --> 02:09:12,420
the number line, my expression is going to
be positive, because it can jump from positive

1312
02:09:12,420 --> 02:09:17,529
to negative without going through a place
where it's zero, I can figure out whether

1313
02:09:17,529 --> 02:09:21,859
x squared minus four is positive or negative
on this region, and on this region of the

1314
02:09:21,859 --> 02:09:25,109
number line by plugging in test value similar
way,

1315
02:09:25,109 --> 02:09:31,849
evaluate the plug in between negative two
and two, a nice value is x equals 00 squared

1316
02:09:31,850 --> 02:09:36,700
minus four, that's negative four and negative
number. So I know that my expression x squared

1317
02:09:36,699 --> 02:09:43,109
minus four is negative on this whole interval.
Finally, I can plug in something like x equals

1318
02:09:43,109 --> 02:09:49,349
10, something bigger than two, and I get 10
squared minus four. Without even computing

1319
02:09:49,350 --> 02:09:53,890
that I can tell that that's going to be a
positive number. And that's all that's important.

1320
02:09:53,890 --> 02:09:57,980
Again, since I want x squared minus four to
be less than zero, I'm looking for the places

1321
02:09:57,979 --> 02:09:59,719
on this number line where I'm getting

1322
02:09:59,720 --> 02:10:06,230
negatives. So I will share that in on my number
line. It's in here, not including the endpoints.

1323
02:10:06,229 --> 02:10:10,609
Because the endpoints are where my expression
x squared minus four is equal to zero, and

1324
02:10:10,609 --> 02:10:12,599
I want it strictly less than zero,

1325
02:10:12,600 --> 02:10:19,900
I can write my answer as an inequality, negative
two is less than x is less than two, or an

1326
02:10:19,899 --> 02:10:25,059
interval notation as soft bracket negative
two to soft bracket.

1327
02:10:25,060 --> 02:10:32,780
Our next example, we can solve similarly,
first, we'll move everything to one side,

1328
02:10:32,779 --> 02:10:39,769
so that our inequality is x cubed minus 5x
squared minus 6x is greater than or equal

1329
02:10:39,770 --> 02:10:47,420
to zero. Next, we'll solve the associated
equation by factoring. So first, I'll write

1330
02:10:47,420 --> 02:10:55,449
down the equation. Now I'll factor out an
x. And now I'll factor the quadratic. So the

1331
02:10:55,449 --> 02:11:01,239
solutions to my equation are x equals 0x equals
six and x equals negative one,

1332
02:11:01,239 --> 02:11:05,489
I'll write the solutions to the equation on
the number line.

1333
02:11:05,489 --> 02:11:17,359
So that's negative one, zero, and six. That's
where my expression x times x minus six times

1334
02:11:17,359 --> 02:11:21,049
x plus one is equal to zero.

1335
02:11:21,050 --> 02:11:28,110
But I want to find where it's greater than
or equal to zero. So again, I can use test

1336
02:11:28,109 --> 02:11:35,630
values, I can plug in, for example, x equals
negative two, either to this version of expression,

1337
02:11:35,630 --> 02:11:40,760
or to this factored version. Since I only
care whether my answer is positive or negative,

1338
02:11:40,760 --> 02:11:45,630
it's sometimes easier to use the factored
version. For example, when x is negative two,

1339
02:11:45,630 --> 02:11:48,069
this factor is negative.

1340
02:11:48,069 --> 02:11:55,539
But this factor, x minus six is also negative
when I plug in negative two for x.

1341
02:11:55,539 --> 02:12:03,269
Finally, x plus one, when I plug in negative
two for x, that's negative one, that's also

1342
02:12:03,270 --> 02:12:08,980
negative. And a negative times a negative
times a negative gives me a negative number.

1343
02:12:08,979 --> 02:12:16,750
If I plug in something, between negative one
and zero, say x equals negative one half,

1344
02:12:16,750 --> 02:12:21,600
then I'm going to get a negative for this
factor, a negative for this factor, but a

1345
02:12:21,600 --> 02:12:25,020
positive for this third factor.

1346
02:12:25,020 --> 02:12:28,860
Negative times negative times positive gives
me a positive

1347
02:12:28,859 --> 02:12:34,349
for a test value between zero and six, let's
try x equals one.

1348
02:12:34,350 --> 02:12:41,130
Now I'll get a positive for this factor a
negative for this factor, and a positive for

1349
02:12:41,130 --> 02:12:43,880
this factor.

1350
02:12:43,880 --> 02:12:50,920
positive times a negative times a positive
gives me a negative. Finally, for a test value

1351
02:12:50,920 --> 02:12:56,649
bigger than six, we could use say x equals
100. That's going to give me positive, positive

1352
02:12:56,649 --> 02:12:59,219
positive.

1353
02:12:59,220 --> 02:13:03,630
So my product will be positive.

1354
02:13:03,630 --> 02:13:08,060
Since I want values where my expression is
greater than or equal to zero, I want the

1355
02:13:08,060 --> 02:13:12,970
places where it equals zero.

1356
02:13:12,970 --> 02:13:16,980
And the places where it's positive.

1357
02:13:16,979 --> 02:13:24,709
So my final answer will be close bracket negative
one to zero, close bracket union, close bracket

1358
02:13:24,710 --> 02:13:25,909
six to infinity.

1359
02:13:25,909 --> 02:13:34,489
As our final example, let's consider the rational
inequality, x squared plus 6x plus nine divided

1360
02:13:34,489 --> 02:13:38,229
by x minus one is less than or equal to zero.

1361
02:13:38,229 --> 02:13:43,449
Although it might be tempting to clear the
denominator and multiply both sides by x minus

1362
02:13:43,449 --> 02:13:49,050
one, it's dangerous to do that, because x
minus one could be a positive number. But

1363
02:13:49,050 --> 02:13:53,699
it could also be a negative number. And when
you multiply both sides by a negative number,

1364
02:13:53,699 --> 02:13:59,260
you have to reverse the inequality. Although
it's possible to solve the inequality this

1365
02:13:59,260 --> 02:14:04,869
way, by thinking of cases where x minus one
is less than zero or bigger than zero, I think

1366
02:14:04,869 --> 02:14:10,369
it's much easier just to solve the same way
as we did before. So we'll start by rewriting

1367
02:14:10,369 --> 02:14:15,340
so that we move all terms to the left and
have zero on the right. Well, that's already

1368
02:14:15,340 --> 02:14:20,750
true. So the next step would be to solve the
associated equation.

1369
02:14:20,750 --> 02:14:29,470
That is x squared plus 6x plus nine over x
minus one is equal to zero.

1370
02:14:29,470 --> 02:14:35,240
That would be where the numerator is 0x squared
plus 6x plus nine is equal to zero. So we're

1371
02:14:35,239 --> 02:14:42,609
x plus three squared is zero, or x equals
negative three. There's one extra step we

1372
02:14:42,609 --> 02:14:48,069
have to do for rational expressions. And that's
we need to find where the expression does

1373
02:14:48,069 --> 02:14:55,559
not exist. That is, let's find where the denominator
is zero. And that said, x equals one.

1374
02:14:55,560 --> 02:14:59,780
I'll put all those numbers on the number line.
The places where am I

1375
02:14:59,779 --> 02:15:04,859
rational expression is equal to zero, and
the place where my rational expression doesn't

1376
02:15:04,859 --> 02:15:13,630
exist, then I can start in with test values.
For example, x equals minus four, zero and

1377
02:15:13,630 --> 02:15:20,859
to work. If I plug those values into this
expression here, I get a negative answer a

1378
02:15:20,859 --> 02:15:26,109
negative answer and a positive answer. The
reason I need to conclude the values on my

1379
02:15:26,109 --> 02:15:32,009
number line where my denominators zero is
because I can my expression can switch from

1380
02:15:32,010 --> 02:15:36,630
negative to positive by passing through a
place where my rational expression doesn't

1381
02:15:36,630 --> 02:15:41,859
exist, as well as passing by passing flew
through a place where my rational expression

1382
02:15:41,859 --> 02:15:43,219
is equal to zero.

1383
02:15:43,220 --> 02:15:49,570
Now I'm looking for where my original expression
was less than or equal to zero. So that means

1384
02:15:49,569 --> 02:15:56,109
I want the places on the number line where
my expression is equal to zero, and also the

1385
02:15:56,109 --> 02:15:59,589
places where it's negative.

1386
02:15:59,590 --> 02:16:06,090
So my final answer is x is less than one,
or an interval notation, negative infinity

1387
02:16:06,090 --> 02:16:08,550
to one.

1388
02:16:08,550 --> 02:16:14,630
In this video, we solved polynomial and rational
inequalities by making a number line and you

1389
02:16:14,630 --> 02:16:19,869
think test values to make a sine chart.

1390
02:16:19,869 --> 02:16:26,109
The distance formula can be used to find the
distance between two points, if we know their

1391
02:16:26,109 --> 02:16:27,109
coordinates,

1392
02:16:27,109 --> 02:16:34,039
if this first point has coordinates, x one,
y one, and the second point has coordinates

1393
02:16:34,040 --> 02:16:42,010
x two, y two, then the distance between them
is given by the formula, the square root of

1394
02:16:42,010 --> 02:16:49,200
x two minus x one squared plus y two minus
y one squared.

1395
02:16:49,200 --> 02:16:55,739
This formula actually comes from the Pythagorean
Theorem, let me draw a right triangle with

1396
02:16:55,738 --> 02:17:01,399
these two points as two of its vertices,

1397
02:17:01,399 --> 02:17:09,148
then the length of this side is the difference
between the 2x coordinates. Similarly, the

1398
02:17:09,148 --> 02:17:14,629
length of this vertical side is the difference
between the two y coordinates.

1399
02:17:14,629 --> 02:17:20,170
Now that Pythagoras theorem says that for
any right triangle, with sides labeled A and

1400
02:17:20,170 --> 02:17:27,019
B, and hypotony is labeled C, we have that
a squared plus b squared equals c squared.

1401
02:17:27,019 --> 02:17:31,728
Well, if we apply that to this triangle here,

1402
02:17:31,728 --> 02:17:38,010
this hypotony News is the distance between
the two points that we're looking for.

1403
02:17:38,010 --> 02:17:45,818
So but Tyrion theorem says, the square of
this side length, that is x two minus x one

1404
02:17:45,818 --> 02:17:55,149
squared, plus the square of this side length,
which is y two minus y one squared has to

1405
02:17:55,149 --> 02:18:01,658
equal the square of the hypothesis, that is
d squared.

1406
02:18:01,658 --> 02:18:07,148
taking the square root of both sides of this
equation, we get the square root of x two

1407
02:18:07,148 --> 02:18:13,769
minus x one squared plus y two minus y one
squared is equal to d,

1408
02:18:13,769 --> 02:18:18,439
we don't have to worry about using plus or
minus signs when we take the square root because

1409
02:18:18,439 --> 02:18:21,260
distance is always positive.

1410
02:18:21,260 --> 02:18:26,639
So we've derived our distance formula. Now
let's use it as an example.

1411
02:18:26,638 --> 02:18:34,589
Let's find the distance between the two points,
negative one five, for negative one, five,

1412
02:18:34,590 --> 02:18:39,079
maybe over here, and for two,

1413
02:18:39,079 --> 02:18:48,359
this notation just means that P is the point
with coordinates negative one five, and q

1414
02:18:48,359 --> 02:18:52,699
is the point with coordinates for two.

1415
02:18:52,699 --> 02:18:54,590
We have the distance formula,

1416
02:18:54,590 --> 02:19:02,299
let's think of P being the point with coordinates
x one y one, and Q being the point with coordinates

1417
02:19:02,299 --> 02:19:04,278
x two y two.

1418
02:19:04,279 --> 02:19:10,179
But as we'll see, it really doesn't matter
which one is which, plugging into our formula,

1419
02:19:10,179 --> 02:19:23,609
we get d is the square root of n x two minus
x one. So that's four minus negative one squared,

1420
02:19:23,609 --> 02:19:26,769
and then we add

1421
02:19:26,769 --> 02:19:37,349
y two minus y one, so that's two minus five
squared.

1422
02:19:37,349 --> 02:19:41,710
Working out the arithmetic a little bit for
minus negative one, that's four plus one or

1423
02:19:41,709 --> 02:19:49,139
five squared, plus negative three squared.
So that's the square root of 25 plus nine

1424
02:19:49,139 --> 02:19:55,659
are the square root of 34. Let's see what
would have happened if we'd call this first

1425
02:19:55,659 --> 02:20:00,200
point, x two y two instead, and a second.

1426
02:20:00,200 --> 02:20:06,390
Why'd x one y one, then we would have gotten
the same distance formula, but we would have

1427
02:20:06,389 --> 02:20:12,180
taken negative one minus four

1428
02:20:12,180 --> 02:20:22,130
and added the difference of y's squared. So
that's why two five minus two squared.

1429
02:20:22,129 --> 02:20:27,139
That gives us the square root of negative
five squared plus three squared, which works

1430
02:20:27,139 --> 02:20:31,180
out to the same thing.

1431
02:20:31,180 --> 02:20:36,771
In this video, we use the distance formula
to find the distance between two points. When

1432
02:20:36,771 --> 02:20:40,579
writing down the distance formula, sometimes
students forget whether this is a plus or

1433
02:20:40,578 --> 02:20:45,488
a minus and whether these are pluses or minuses.
One way to remember is to think that the distance

1434
02:20:45,488 --> 02:20:51,629
formula comes from the Pythagorean Theorem.
That's why there's a plus in here.

1435
02:20:51,629 --> 02:20:56,239
And then a minus in here, because we're finding
the difference between the coordinates to

1436
02:20:56,239 --> 02:20:59,421
find the lengths of the sides.

1437
02:20:59,421 --> 02:21:05,699
The midpoint formula helps us find the coordinates
of the midpoint of a line segment, as long

1438
02:21:05,699 --> 02:21:09,819
as we know the coordinates of the endpoints.

1439
02:21:09,818 --> 02:21:15,328
So let's call the coordinates of this endpoint
x one, y one, and the coordinates of this

1440
02:21:15,328 --> 02:21:17,600
other endpoint x two, y two,

1441
02:21:17,600 --> 02:21:25,750
the x coordinate of the midpoint is going
to be exactly halfway in between the x coordinates

1442
02:21:25,750 --> 02:21:31,068
of the endpoints. To get a number halfway
in between two other numbers, we just take

1443
02:21:31,068 --> 02:21:34,340
the average.

1444
02:21:34,340 --> 02:21:42,238
Similarly, the y coordinate of this midpoint
is going to be exactly halfway in between

1445
02:21:42,238 --> 02:21:47,430
the y coordinates of the endpoints. So the
y coordinate of the midpoint is going to be

1446
02:21:47,430 --> 02:21:53,050
the average of those y coordinates.

1447
02:21:53,049 --> 02:22:00,938
So we see that the coordinates of the midpoint
are x one plus x two over two, y one plus

1448
02:22:00,939 --> 02:22:04,059
y two over two.

1449
02:22:04,059 --> 02:22:07,028
Let's use this midpoint formula in an example,

1450
02:22:07,029 --> 02:22:17,069
we want to find the midpoint of the segment
between the points, negative one, five,

1451
02:22:17,068 --> 02:22:20,988
and four

1452
02:22:20,988 --> 02:22:24,090
to

1453
02:22:24,090 --> 02:22:29,510
me draw the line segment between them. So
visually, the midpoint is going to be somewhere

1454
02:22:29,510 --> 02:22:36,248
around here. But to find its exact coordinates,
we're going to use x one plus x two over two,

1455
02:22:36,248 --> 02:22:44,209
and y one plus y two over two, where this
first point is coordinates, x one, y one,

1456
02:22:44,209 --> 02:22:49,788
and the second point has coordinates x two
y two, it doesn't actually matter which point

1457
02:22:49,789 --> 02:22:54,600
you decide is x one y one, which one is x
two, y two, the formula will still give you

1458
02:22:54,600 --> 02:22:56,460
the same answer for the midpoint.

1459
02:22:56,459 --> 02:23:06,349
So let's see, I take my average of my x coordinates.
So that's negative one, plus four over two,

1460
02:23:06,350 --> 02:23:15,369
and the average of my Y coordinates, so that's
five plus two over two, and I get three halves,

1461
02:23:15,369 --> 02:23:19,069
seven halves, as the coordinates of my midpoint.

1462
02:23:19,068 --> 02:23:25,449
In this video, we use the midpoint formula
to find the midpoint of a line segment, just

1463
02:23:25,450 --> 02:23:30,351
by taking the average of the x coordinates
and the average of the y coordinates. This

1464
02:23:30,351 --> 02:23:36,149
video is about graphs and equations of circles.
Suppose we want to find the equation of a

1465
02:23:36,148 --> 02:23:42,439
circle of radius five, centered at the point
three, two

1466
02:23:42,439 --> 02:23:47,510
will look something like this.

1467
02:23:47,510 --> 02:23:53,420
For any point, x, y on the circle, we know
that

1468
02:23:53,420 --> 02:24:00,930
the distance of that point xy from the center
is equal to the radius that is five from the

1469
02:24:00,930 --> 02:24:07,398
distance formula, that distance of five is
equal to the square root of the difference

1470
02:24:07,398 --> 02:24:11,498
of the x coordinates. That's x minus three

1471
02:24:11,498 --> 02:24:20,199
squared plus the difference of the y coordinates,
that's y minus two squared.

1472
02:24:20,199 --> 02:24:31,239
And if I square both sides of that equation,
I get five squared is this square root squared.

1473
02:24:31,239 --> 02:24:40,238
In other words, 25 is equal to x minus three
squared plus y minus two squared. Since the

1474
02:24:40,238 --> 02:24:43,478
square root and the squared undo each other.

1475
02:24:43,478 --> 02:24:49,519
A lot of times people will write the X minus
three squared plus y minus two squared on

1476
02:24:49,520 --> 02:24:53,899
the left side of the equation and the 25 on
the right side of the equation. This is the

1477
02:24:53,898 --> 02:24:58,099
standard form for the equation of the circle.

1478
02:24:58,100 --> 02:24:59,998
The same reasoning can be generalized

1479
02:24:59,998 --> 02:25:07,119
Find the general equation for a circle with
radius R centered at a point HK.

1480
02:25:07,120 --> 02:25:14,609
For any point with coordinates, x, y on the
circle, the distance between the point with

1481
02:25:14,609 --> 02:25:22,130
coordinates x, y and the point with coordinates
HK is equal to the radius r. So we have that

1482
02:25:22,129 --> 02:25:28,709
the distance is equal to r, but by the distance
formula, that's the square root of the difference

1483
02:25:28,709 --> 02:25:36,648
between the x coordinates x minus h squared
plus the difference in the y coordinates y

1484
02:25:36,648 --> 02:25:48,559
minus k squared. squaring both sides as before,
we get r squared is the square root squared,

1485
02:25:48,559 --> 02:25:53,078
canceling the square root and the squared,
and rearranging the equation, that gives us

1486
02:25:53,078 --> 02:26:01,090
x minus h squared plus y minus k squared equals
r squared. That's the general formula for

1487
02:26:01,090 --> 02:26:06,969
a circle with radius r, and center HK.

1488
02:26:06,969 --> 02:26:13,359
Notice that the coordinates h and k are subtracted
here, the two squares are added because they

1489
02:26:13,360 --> 02:26:18,760
are in the distance formula, and the radius
is squared on the other side, if you remember

1490
02:26:18,760 --> 02:26:23,960
this general formula, that makes it easy to
write down the equation for a circle, for

1491
02:26:23,959 --> 02:26:29,238
example, if we want the equation for a circle,

1492
02:26:29,238 --> 02:26:35,618
of radius six, and center at zero, negative
three,

1493
02:26:35,619 --> 02:26:45,670
then we have our equal six, and h k is our
zero negative three. So plugging into the

1494
02:26:45,670 --> 02:26:47,040
formula,

1495
02:26:47,040 --> 02:26:57,130
we get x minus zero squared, plus y minus
negative three squared equals six squared,

1496
02:26:57,129 --> 02:27:04,299
or simplified this is x squared plus y plus
three squared equals 36.

1497
02:27:04,299 --> 02:27:09,748
Suppose we're given an equation like this
one, and we want to decide if it's the equation

1498
02:27:09,748 --> 02:27:13,760
of a circle, and if so what's the center was
the radius.

1499
02:27:13,760 --> 02:27:21,728
Well, this equation matches the form for a
circle, x minus h squared plus y minus k squared

1500
02:27:21,728 --> 02:27:30,129
equals r squared. If we let h, b five, Why
be negative sex since that way, subtracting

1501
02:27:30,129 --> 02:27:34,349
a negative six is like adding a six,

1502
02:27:34,350 --> 02:27:41,140
five must be our r squared. So that means
that R is the square root of five. So this

1503
02:27:41,139 --> 02:27:49,500
is our radius. And our center is the point
five, negative six. And this is indeed the

1504
02:27:49,500 --> 02:27:52,109
equation for a circle,

1505
02:27:52,109 --> 02:27:57,930
which we could then graph by putting down
the center and estimating the radius, which

1506
02:27:57,930 --> 02:28:03,000
is a little bit more than two.

1507
02:28:03,000 --> 02:28:08,859
This equation might not look like the equation
of a circle, but it actually can be transformed.

1508
02:28:08,859 --> 02:28:13,498
To look like the equation of a circle, we're
trying to make it look like something of the

1509
02:28:13,498 --> 02:28:19,270
form x minus h squared plus y minus k squared
equals r squared.

1510
02:28:19,270 --> 02:28:24,020
First, I'd like to get rid of the coefficients
in front of the x squared and the y squared,

1511
02:28:24,020 --> 02:28:30,488
so I'm going to divide everything by nine.
This gives me x squared plus y squared plus

1512
02:28:30,488 --> 02:28:38,609
8x minus two y plus four equals zero. Next,
I'm going to group the x terms together the

1513
02:28:38,609 --> 02:28:45,238
x squared and the 8x. So write x squared plus
8x. I'll group the y terms together the Y

1514
02:28:45,238 --> 02:28:49,119
squared minus two y.

1515
02:28:49,120 --> 02:28:54,750
And I'll subtract the four over to the other
side.

1516
02:28:54,750 --> 02:28:59,068
This still doesn't look much like the equation
of a circle. But as the next step, I'm going

1517
02:28:59,068 --> 02:29:05,288
to do something called completing the square,
I'm going to take this coefficient of x eight

1518
02:29:05,289 --> 02:29:12,210
divided by two and then square it. In other
words, eight over two squared, that's four

1519
02:29:12,209 --> 02:29:18,608
squared, which is 16. I'm going to add 16
to both sides of my equation.

1520
02:29:18,609 --> 02:29:22,318
So the 16 here,

1521
02:29:22,318 --> 02:29:24,439
and on the other side,

1522
02:29:24,439 --> 02:29:31,488
now I'm going to do the same thing to the
coefficient of y, negative two divided by

1523
02:29:31,488 --> 02:29:38,260
two is negative one, square that and I get
one. So now I'm going to add a one to both

1524
02:29:38,260 --> 02:29:44,068
sides, but I'll put it near the y terms. So
add a one there. And then to keep things balanced,

1525
02:29:44,068 --> 02:29:50,639
I have to add it to the other side. On the
right side, I have the number 13. On the left

1526
02:29:50,639 --> 02:29:59,439
side, I can wrap up this expression x squared
plus 8x plus 16 into x plus four squared.

1527
02:29:59,439 --> 02:30:07,460
To convince you, that's correct. If we multiply
out x plus four times x plus four, we get

1528
02:30:07,459 --> 02:30:16,919
x squared plus 4x plus 4x plus 16, or x squared
plus 8x plus 16, the same thing we have right

1529
02:30:16,920 --> 02:30:25,260
here. Similarly, we can wrap up y squared
minus two y plus one as

1530
02:30:25,260 --> 02:30:32,340
y minus one squared. Again, I'll work out
the distribution down here, that's y squared

1531
02:30:32,340 --> 02:30:40,030
minus y minus y plus one, or y squared minus
two y plus one, just like we have up here.

1532
02:30:40,030 --> 02:30:46,989
If you're wondering how I knew to use four
and minus one, the four came from taking half

1533
02:30:46,988 --> 02:30:53,988
of eight, and the minus one came from taking
half the negative two. Now we have an equation

1534
02:30:53,988 --> 02:31:01,629
for a circle and standard form. And we can
read off the center, which is negative four,

1535
02:31:01,629 --> 02:31:10,429
negative one, and the radius, which is the
square root of 13.

1536
02:31:10,430 --> 02:31:14,979
It might seem like magic that this trick of
taking half of this coefficient and squaring

1537
02:31:14,978 --> 02:31:20,568
it and adding it to both sides lets us wrap
up this expression into this expression a

1538
02:31:20,568 --> 02:31:26,988
perfect square. But to see why that works,
let's take a look at what happens if we expand

1539
02:31:26,988 --> 02:31:33,760
out x minus h squared, the thing that we're
trying to get to. So if we expand that out,

1540
02:31:33,760 --> 02:31:43,609
x minus h squared, which is x minus h times
x minus h is, which multiplies out to x squared

1541
02:31:43,609 --> 02:31:53,430
minus HX minus h x plus h squared, or x squared
minus two h x plus h squared. Now if we start

1542
02:31:53,430 --> 02:32:01,090
out with this part, with some x squared term
and some coefficient of x, we're trying to

1543
02:32:01,090 --> 02:32:06,409
decide what to add on in order to wrap it
up into x minus h squared. The thing to add

1544
02:32:06,409 --> 02:32:13,430
on is h squared here, which comes from half
of this coefficient squared, right, because

1545
02:32:13,430 --> 02:32:18,689
half of negative two H is a negative H, square
that you get the H squared. And then when

1546
02:32:18,689 --> 02:32:26,738
we do wrap it up, it's half of this coefficient
of x, that appears right here.

1547
02:32:26,738 --> 02:32:30,988
This trick of completing the square is really
handy for turning an equation for a circle

1548
02:32:30,988 --> 02:32:37,260
in disguise, into the standard equation for
the circle. In this video, we found the standard

1549
02:32:37,260 --> 02:32:44,728
equation for a circle x minus h squared plus
y minus k squared equals r squared, where

1550
02:32:44,728 --> 02:32:52,159
r is the radius, and h k is the center.

1551
02:32:52,159 --> 02:32:56,648
We also showed a method of completing the
square.

1552
02:32:56,648 --> 02:33:01,840
When you have an equation for a circle in
disguise, completing the square will help

1553
02:33:01,840 --> 02:33:06,279
you rewrite it into the standard form.

1554
02:33:06,279 --> 02:33:10,979
This video is about graphs and equations of
lines.

1555
02:33:10,978 --> 02:33:16,799
Here we're given the graph of a line, we want
to find the equation, one standard format

1556
02:33:16,799 --> 02:33:25,059
for the equation of a line is y equals mx
plus b. here am represents the slope, and

1557
02:33:25,059 --> 02:33:34,549
B represents the y intercept, the y value
where the line crosses the y axis, the slope

1558
02:33:34,549 --> 02:33:41,228
is equal to the rise over the run. Or sometimes
this is written as the change in y values

1559
02:33:41,228 --> 02:33:49,549
over the change in x values. Or in other words,
y two minus y one over x two minus x one,

1560
02:33:49,549 --> 02:33:56,509
where x one y one and x two y two are points
on the line.

1561
02:33:56,510 --> 02:34:01,600
While we could use any two points on the line,
to find the slope, it's convenient to use

1562
02:34:01,600 --> 02:34:08,590
points where the x and y coordinates are integers,
that is points where the line passes through

1563
02:34:08,590 --> 02:34:13,630
grid points. So here would be one convenient
point to use. And here's another convenient

1564
02:34:13,629 --> 02:34:15,639
point to use.

1565
02:34:15,639 --> 02:34:22,068
The coordinates of the first point are one,
two, and the next point, this is let's say,

1566
02:34:22,068 --> 02:34:29,319
five, negative one. Now I can find the slope
by looking at the rise over the run. So as

1567
02:34:29,319 --> 02:34:35,629
I go through a run of this distance, I go
through a rise of that distance especially

1568
02:34:35,629 --> 02:34:41,078
gonna be a negative rise or a fall because
my line is pointing down. So let's see counting

1569
02:34:41,078 --> 02:34:48,219
off squares. This is a run of 1234 squares
and a rise of 123. So negative three, so my

1570
02:34:48,219 --> 02:34:52,658
slope is going to be

1571
02:34:52,658 --> 02:34:56,270
negative three over four.

1572
02:34:56,270 --> 02:35:00,510
I got that answer by counting squares, but
I could have also gotten it by

1573
02:35:00,510 --> 02:35:05,630
Looking at the difference in my y values over
the difference in my x values, that is, I

1574
02:35:05,629 --> 02:35:08,179
could have done

1575
02:35:08,180 --> 02:35:15,898
negative one minus two, that's from my difference
in Y values, and divide that by my difference

1576
02:35:15,898 --> 02:35:18,459
in x values,

1577
02:35:18,459 --> 02:35:26,188
which is five minus one, that gives me negative
three over four, as before.

1578
02:35:26,189 --> 02:35:30,750
So my M is negative three fourths.

1579
02:35:30,750 --> 02:35:36,219
Now I need to figure out the value of b, my
y intercept, well, I could just read it off

1580
02:35:36,219 --> 02:35:41,868
the graph, it looks like approximately 2.75.
But if I want to be more accurate, I can again

1581
02:35:41,869 --> 02:35:47,539
use a point that has integer coordinates that
I know it's exact coordinates. So either this

1582
02:35:47,539 --> 02:35:53,689
point or that point, let's try this point.
And I can start off with my equation y equals

1583
02:35:53,689 --> 02:36:02,389
mx plus b, that is y equals negative three
fourths x plus b. And I can plug in the point

1584
02:36:02,389 --> 02:36:12,118
one, two, for my x and y. So that gives me
two equals negative three fourths times one

1585
02:36:12,119 --> 02:36:20,078
plus b, solving for B. Let's see that's two
equals negative three fourths plus b. So add

1586
02:36:20,078 --> 02:36:28,090
three fourths to both sides, that's two plus
three fourths equals b. So b is eight fourths

1587
02:36:28,090 --> 02:36:33,500
plus three fourths, which is 11. Four switches
actually, just what I eyeballed it today.

1588
02:36:33,500 --> 02:36:40,559
So now I can write out my final equation for
my line y equals negative three fourths x

1589
02:36:40,559 --> 02:36:48,289
plus 11 fourths by plugging in for m and b.
Next, let's find the equation for this horizontal

1590
02:36:48,290 --> 02:36:49,420
line.

1591
02:36:49,420 --> 02:36:57,158
a horizontal line has slope zero. So if we
think of it as y equals mx plus b, m is going

1592
02:36:57,158 --> 02:37:03,898
to be zero. In other words, it's just y equals
b, y is some constant. So if we can figure

1593
02:37:03,898 --> 02:37:08,139
out what that that constant y value is, it
looks like it's

1594
02:37:08,139 --> 02:37:14,818
two, let's see this three, three and a half,
we can just write down the equation directly,

1595
02:37:14,818 --> 02:37:18,998
y equals 3.5.

1596
02:37:18,998 --> 02:37:24,898
For a vertical line, like this one, it doesn't
really have a slope. I mean, if you tried

1597
02:37:24,898 --> 02:37:26,449
to do the rise over the run,

1598
02:37:26,450 --> 02:37:32,340
there's no run. So you'd I guess you'd be
dividing by zero and get an infinite slope.

1599
02:37:32,340 --> 02:37:39,859
But But instead, we just think of it as an
equation of the form x equals something. And

1600
02:37:39,859 --> 02:37:45,199
in this case, x equals negative two, notice
that all of the points on our line have the

1601
02:37:45,199 --> 02:37:50,380
same x coordinate of negative two and the
y coordinate can be anything.

1602
02:37:50,379 --> 02:37:54,879
So this is how we write the equation for a
vertical line.

1603
02:37:54,879 --> 02:37:59,299
In this example, we're not shown a graph of
the line, we're just get told that it goes

1604
02:37:59,299 --> 02:38:04,629
through two points. But knowing that I go
through two points is enough to find the equation

1605
02:38:04,629 --> 02:38:12,608
for the line. First, we can find the slope
by taking the difference in Y values over

1606
02:38:12,609 --> 02:38:20,090
the difference in x values. So that's negative
three minus two over four minus one, which

1607
02:38:20,090 --> 02:38:24,319
is negative five thirds.

1608
02:38:24,319 --> 02:38:26,329
So we can use the

1609
02:38:26,329 --> 02:38:34,648
standard equation for the line, this is called
the slope intercept form.

1610
02:38:34,648 --> 02:38:42,719
And we can plug in negative five thirds. And
we can use one point, either one will do will

1611
02:38:42,719 --> 02:38:48,488
still get the same final answer. So let's
use one two and plug that in to get two equals

1612
02:38:48,488 --> 02:38:57,779
negative five thirds times one plus b. And
so B is two plus five thirds, which is six

1613
02:38:57,779 --> 02:39:03,550
thirds plus five thirds, which is 11 thirds.
So our equation is y equals negative five

1614
02:39:03,549 --> 02:39:07,389
thirds x plus 11 thirds.

1615
02:39:07,389 --> 02:39:11,728
This is method one.

1616
02:39:11,728 --> 02:39:17,119
method two uses a slightly different form
of the equation, it's called the point slope

1617
02:39:17,120 --> 02:39:25,680
form and it goes y minus y naught is equal
to m times x minus x naught, where x naught

1618
02:39:25,680 --> 02:39:33,380
y naught is a point on the line and again
is the slope. So we calculate the slope the

1619
02:39:33,379 --> 02:39:40,309
same way by taking a difference in Y values
over a difference in x values. But then, we

1620
02:39:40,309 --> 02:39:43,698
can simply plug in any point.

1621
02:39:43,699 --> 02:39:53,039
For example, the point one two will work we
can plug one in for x naught and two in for

1622
02:39:53,039 --> 02:39:59,420
Y not in this point slope form. That gives
us y

1623
02:39:59,420 --> 02:40:00,420
minus

1624
02:40:00,420 --> 02:40:06,719
Two is equal to minus five thirds x minus
one.

1625
02:40:06,719 --> 02:40:11,090
Notice that these two equations, while they
may look different, are actually equivalent,

1626
02:40:11,090 --> 02:40:25,380
because if I distribute the negative five
thirds, and then add the two to both sides,

1627
02:40:25,379 --> 02:40:28,529
I get the same equation as above.

1628
02:40:28,530 --> 02:40:37,100
So we've seen two ways of finding the equation
for the line using the slope intercept form.

1629
02:40:37,100 --> 02:40:40,199
And using the point slope form.

1630
02:40:40,199 --> 02:40:48,920
In this video, we saw that you can find the
equation for a line, if you know the slope.

1631
02:40:48,920 --> 02:40:52,158
And you know one point,

1632
02:40:52,158 --> 02:40:57,760
you can also find the equation for the line
if you know two points, because you can use

1633
02:40:57,760 --> 02:41:03,529
the two points to get the slope and then plug
in one of those points. To figure out the

1634
02:41:03,529 --> 02:41:06,430
rest of the equation.

1635
02:41:06,430 --> 02:41:08,898
We saw two standard forms for the equation
of a line

1636
02:41:08,898 --> 02:41:19,738
the slope intercept form y equals mx plus
b, where m is the slope, and B is the y intercept.

1637
02:41:19,738 --> 02:41:27,398
And the point slope form y minus y naught
equals m times x minus x naught, where m again

1638
02:41:27,398 --> 02:41:35,328
is the slope and x naught y naught is a point
on the line.

1639
02:41:35,328 --> 02:41:43,219
This video is about finding parallel and perpendicular
lines. Suppose we have a line of slope three

1640
02:41:43,219 --> 02:41:47,469
fourths, in other words, the rise

1641
02:41:47,469 --> 02:41:50,590
over the run

1642
02:41:50,590 --> 02:41:52,529
is three over four,

1643
02:41:52,529 --> 02:42:01,119
than any other line that's parallel to this
line will have the same slope, the same fries

1644
02:42:01,119 --> 02:42:03,939
over run.

1645
02:42:03,939 --> 02:42:09,909
So that's our first fact to keep in mind,
parallel lines have the same slopes. Suppose

1646
02:42:09,909 --> 02:42:15,000
on the other hand, we want to find a line
perpendicular to our original line with its

1647
02:42:15,000 --> 02:42:18,129
original slope of three fourths.

1648
02:42:18,129 --> 02:42:25,358
A perpendicular line, in other words, align
at a 90 degree angle to our original line

1649
02:42:25,359 --> 02:42:32,039
will have a slope, that's the negative reciprocal
of the opposite reciprocal of our original

1650
02:42:32,039 --> 02:42:37,579
slope. So we take the reciprocal of three
for us, that's four thirds, and we make it

1651
02:42:37,578 --> 02:42:41,359
its opposite by changing it from positive
to negative.

1652
02:42:41,359 --> 02:42:49,738
So I'll write this as a principle that perpendicular
lines have opposite reciprocal slopes, to

1653
02:42:49,738 --> 02:42:54,719
get the hang of what it means to be an opposite
reciprocal, let's look at a few examples.

1654
02:42:54,719 --> 02:43:01,288
So here's the original slope, and this will
be the opposite reciprocal, which would represent

1655
02:43:01,289 --> 02:43:07,610
the slope of our perpendicular line. So for
example, if one was to the opposite reciprocal,

1656
02:43:07,610 --> 02:43:14,319
reciprocal of two is one half opposite means
I change the sign, if the first slope turned

1657
02:43:14,318 --> 02:43:23,260
out to be, say, negative 1/3, the reciprocal
of 1/3 is three over one or just three. And

1658
02:43:23,260 --> 02:43:29,359
the opposite means I change it from a negative
to a positive, so my perpendicular slope would

1659
02:43:29,359 --> 02:43:32,189
be would be three.

1660
02:43:32,189 --> 02:43:38,609
One more example, if I started off with a
slope of, say, seven halves, then the reciprocal

1661
02:43:38,609 --> 02:43:43,890
of that would be two sevenths. And I change
it to an opposite reciprocal, by changing

1662
02:43:43,889 --> 02:43:47,590
the positive to a negative.

1663
02:43:47,590 --> 02:43:52,469
Let's use these two principles in some examples.

1664
02:43:52,469 --> 02:43:56,488
In our first example, we need to find the
equation of a line that's parallel to this

1665
02:43:56,488 --> 02:44:02,908
slump, and go through the point negative three
two. Well, in order to get started, I need

1666
02:44:02,908 --> 02:44:07,538
to figure out the slope of this line. So let
me put this line into a more standard form

1667
02:44:07,539 --> 02:44:14,459
the the slope intercept form. So I'll start
with three y minus 4x plus six equals zero,

1668
02:44:14,459 --> 02:44:19,829
I'm going to solve for y to put it in this
this more standard form. So I'll say three

1669
02:44:19,829 --> 02:44:29,439
y equals 4x minus six and then divide by three
to get y equals four thirds x minus six thirds

1670
02:44:29,439 --> 02:44:34,850
is divided all my turns by three, I can simplify
that a little bit y equals four thirds x minus

1671
02:44:34,850 --> 02:44:43,439
two. Now I can read off the slope of my original
line, and one is four thirds. That means my

1672
02:44:43,439 --> 02:44:51,279
slope of my new line, my parallel line will
also be four thirds. So my new line will have

1673
02:44:51,279 --> 02:44:58,060
the equation y equals four thirds ax plus
b for some B, of course, B will probably not

1674
02:44:58,059 --> 02:45:00,029
be negative two like it was for the

1675
02:45:00,030 --> 02:45:04,189
first line, it'll be something else determined
by the fact that it goes through this point

1676
02:45:04,189 --> 02:45:10,359
negative three to, to figure out what b is,
I plug in the point negative three to four,

1677
02:45:10,359 --> 02:45:16,408
negative three for x, and two for y. So that
gives me two equals four thirds times negative

1678
02:45:16,408 --> 02:45:25,449
three plus b, and I'll solve for b. So let's
see. First, let me just simplify. So two equals,

1679
02:45:25,449 --> 02:45:32,440
this is negative 12 thirds plus b, or in other
words, two is negative four plus b. So that

1680
02:45:32,440 --> 02:45:38,479
means that B is going to be six, and by my
parallel line will have the equation y equals

1681
02:45:38,478 --> 02:45:42,340
four thirds x plus six.

1682
02:45:42,340 --> 02:45:49,398
Next, let's find the equation of a line that's
perpendicular to a given line through and

1683
02:45:49,398 --> 02:45:54,019
go through a given point. Again, in order
to get started, I need to find what the slope

1684
02:45:54,020 --> 02:45:59,869
of this given line is. So I'll rewrite it,
well, first, I'll just copy it over 6x plus

1685
02:45:59,869 --> 02:46:06,670
three, y equals four. And then I can put it
in the standard slope intercept form by solving

1686
02:46:06,670 --> 02:46:14,879
for y. So three, y is negative 6x plus four,
divide everything by three, I get y equals

1687
02:46:14,879 --> 02:46:22,248
negative six thirds x plus four thirds. So
y is negative 2x, plus four thirds. So my

1688
02:46:22,248 --> 02:46:28,068
original slope of my original line is negative
two, which means my perpendicular slope is

1689
02:46:28,068 --> 02:46:34,028
the opposite reciprocal, so I take the reciprocal
of of negative two, that's negative one half

1690
02:46:34,029 --> 02:46:39,380
and I change the sign so that gives me one
half as the slope of my perpendicular line.

1691
02:46:39,379 --> 02:46:47,209
Now, my new line, I know is going to be y
equals one half x plus b for some B, and I

1692
02:46:47,209 --> 02:46:57,148
can plug in my point on my new line, so one
for y and four for x, and solve for b, so

1693
02:46:57,148 --> 02:47:04,778
I get one equals two plus b, So b is negative
one. And so my new line has equation one half

1694
02:47:04,779 --> 02:47:07,520
x minus one.

1695
02:47:07,520 --> 02:47:14,439
These next two examples are a little bit different,
because now we're looking at a line parallel

1696
02:47:14,439 --> 02:47:23,359
to a completely horizontal line, let me draw
this line y equals three, so the y coordinate

1697
02:47:23,359 --> 02:47:29,270
is always three, which means that my line
will be a horizontal line through that height

1698
02:47:29,270 --> 02:47:36,329
at height y equals three, if I want something
parallel to this line, it will also be a horizontal

1699
02:47:36,329 --> 02:47:42,799
line. Since it goes through the point negative
two, one, C, negative two, one has to go through

1700
02:47:42,799 --> 02:47:45,590
that point there,

1701
02:47:45,590 --> 02:47:51,309
it's gonna have always have a y coordinate
of one the same as the y coordinate of the

1702
02:47:51,309 --> 02:47:55,510
point it goes through, so my answer will just
be y equals one.

1703
02:47:55,510 --> 02:48:06,260
In the next example, we want a line that's
perpendicular to the line y equals four another

1704
02:48:06,260 --> 02:48:12,328
horizontal line y equals four, but perpendicular
means I'm going to need a vertical line. So

1705
02:48:12,328 --> 02:48:16,818
I need a vertical line that goes through the
point three, four.

1706
02:48:16,818 --> 02:48:27,748
Okay, and so I'm going to draw a vertical
line there. Now vertical lines have the form

1707
02:48:27,748 --> 02:48:32,949
x equals something for the equation. And to
find what x equals, I just need to look at

1708
02:48:32,950 --> 02:48:38,359
the x coordinate of the point I'm going through
this point three four, so that x coordinate

1709
02:48:38,359 --> 02:48:45,059
is three and all the points on this, this
perpendicular and vertical line have X coordinate

1710
02:48:45,059 --> 02:48:53,748
three so my answer is x equals three.

1711
02:48:53,748 --> 02:48:58,148
In this video will use the fact that parallel
lines have the same slope and perpendicular

1712
02:48:58,148 --> 02:49:03,250
lines have opposite reciprocal slopes, to
find the equations of parallel and perpendicular

1713
02:49:03,250 --> 02:49:05,510
lines.

1714
02:49:05,510 --> 02:49:09,988
This video introduces functions and their
domains and ranges.

1715
02:49:09,988 --> 02:49:17,930
A function is a correspondence between input
numbers usually the x values and output numbers,

1716
02:49:17,930 --> 02:49:25,079
usually the y values, that sends each input
number to exactly one output number. Sometimes

1717
02:49:25,079 --> 02:49:28,260
a function is thought of as a rule or machine

1718
02:49:28,260 --> 02:49:38,880
in which you can feed in x values as input
and get out y values as output. So, non mathematical

1719
02:49:38,879 --> 02:49:44,658
example of a function might be the biological
mother function, which takes as input any

1720
02:49:44,658 --> 02:49:49,959
person and it gives us output their biological
mother.

1721
02:49:49,959 --> 02:49:57,349
This function satisfies the condition that
each input number object person in this case,

1722
02:49:57,350 --> 02:50:01,068
get sent to exactly one output per

1723
02:50:01,068 --> 02:50:07,059
Because if you take any person, they just
have one biological mother. So that rule does

1724
02:50:07,059 --> 02:50:14,049
give you a function. But if I change things
around and just use the the mother function,

1725
02:50:14,049 --> 02:50:19,139
which sends to each person, their mother,
that's no longer going to be a function because

1726
02:50:19,139 --> 02:50:23,250
there are some people who have more than one
mother, right, they could have a biological

1727
02:50:23,250 --> 02:50:28,318
mother and adopted mother, or a mother and
a stepmother, or any number of situations.

1728
02:50:28,318 --> 02:50:33,590
So since there's, there's at least some people
who you would put it as input, and then you'd

1729
02:50:33,590 --> 02:50:39,029
get like more than one possible output that
violates this, this rule of functions, that

1730
02:50:39,029 --> 02:50:43,810
would not be a function. Now, most of the
time, we'll work with functions that are described

1731
02:50:43,809 --> 02:50:48,948
with equations, not in terms of mothers. So
for example, we can have the function y equals

1732
02:50:48,949 --> 02:50:51,329
x squared plus one.

1733
02:50:51,329 --> 02:50:57,409
This can also be written as f of x equals
x squared plus one. Here, f of x is function

1734
02:50:57,409 --> 02:51:02,510
notation that stands for the output value
of y.

1735
02:51:02,510 --> 02:51:07,760
Notice that this notation is not representing
multiplication, we're not multiplying f by

1736
02:51:07,760 --> 02:51:14,309
x, instead, we're going to be putting in a
value for x as input and getting out a value

1737
02:51:14,309 --> 02:51:21,670
of f of x or y. For example, if we want to
evaluate f of two, we're plugging in two as

1738
02:51:21,670 --> 02:51:30,248
input for x either in this equation, or in
that equation. Since F of two means two squared

1739
02:51:30,248 --> 02:51:38,629
plus one, f of two is going to equal five.
Similarly, f of five means I plug in five

1740
02:51:38,629 --> 02:51:46,719
for x, so that's gonna be five squared plus
one, or 26. Sometimes it's useful to evaluate

1741
02:51:46,719 --> 02:51:52,420
a function on a more complicated expression
involving other variables. In this case, remember,

1742
02:51:52,420 --> 02:51:57,748
the functions value on any expression, it's
just what you what you get when you plug in

1743
02:51:57,748 --> 02:52:00,680
that whole expression for x.

1744
02:52:00,680 --> 02:52:09,389
So f of a plus three is going to be the quantity
a plus three squared plus one,

1745
02:52:09,389 --> 02:52:17,029
we could rewrite that as a squared plus six
a plus nine plus one, or a squared plus six

1746
02:52:17,030 --> 02:52:19,649
a plus 10.

1747
02:52:19,648 --> 02:52:25,090
When evaluating a function on a complex expression,
it's important to keep the parentheses when

1748
02:52:25,090 --> 02:52:32,119
you plug in for x. That way, you evaluate
the function on the whole expression. For

1749
02:52:32,119 --> 02:52:35,209
example, it would be wrong

1750
02:52:35,209 --> 02:52:42,590
to write f of a plus three equals a plus three
squared plus one without the parentheses,

1751
02:52:42,590 --> 02:52:47,978
because that would imply that we were just
squaring the three and not the whole expression,

1752
02:52:47,978 --> 02:52:52,379
a plus three.

1753
02:52:52,379 --> 02:52:58,009
Sometimes a function is described with a graph
instead of an equation. In this example, this

1754
02:52:58,010 --> 02:53:04,939
graph is supposed to represent the function
g of x. Not all graphs actually represent

1755
02:53:04,939 --> 02:53:11,050
functions. For example, the graph of a circle
doesn't represent a function. That's because

1756
02:53:11,049 --> 02:53:15,948
the graph of a circle violates the vertical
line test, you can draw a vertical line and

1757
02:53:15,949 --> 02:53:18,289
intersect the graph in more than one point.

1758
02:53:18,289 --> 02:53:24,130
But our graph, it left satisfies the vertical
line test, any vertical line intersects the

1759
02:53:24,129 --> 02:53:29,028
graph, and at most one point, that means is
a function, because every x value will have

1760
02:53:29,029 --> 02:53:37,439
at most one y value that corresponds to it.
Let's evaluate g have to note that two is

1761
02:53:37,439 --> 02:53:44,380
an x value. And we'll use the graph to find
the corresponding y value. So I look for two

1762
02:53:44,379 --> 02:53:51,278
on the x axis, and find the point on the graph
that has that x value that's right here. Now

1763
02:53:51,279 --> 02:53:58,630
I can look at the y value of that point looks
like three, and therefore gf two is equal

1764
02:53:58,629 --> 02:54:05,559
to three. If I try to do the same thing to
evaluate g of five, I run into trouble. Five

1765
02:54:05,559 --> 02:54:11,648
is an x value, I look for it on the x axis,
but there's no point on the graph that has

1766
02:54:11,648 --> 02:54:13,778
that x value.

1767
02:54:13,779 --> 02:54:21,050
Therefore, g of five is undefined, or we can
say it does not exist. The question of what

1768
02:54:21,049 --> 02:54:27,670
x values and y values make sense for a function
leads us to a discussion of domain and range.

1769
02:54:27,670 --> 02:54:32,939
The domain of a function is all possible x
values that makes sense for that function.

1770
02:54:32,939 --> 02:54:36,898
The range is the y values that make sense
for the function.

1771
02:54:36,898 --> 02:54:42,180
In this example, we saw that the x value of
five didn't have a corresponding y value for

1772
02:54:42,180 --> 02:54:49,090
this function. So the x value of five is not
in the domain of our function J. To find the

1773
02:54:49,090 --> 02:54:54,389
x values in the domain, we have to look at
the x values that correspond to points on

1774
02:54:54,389 --> 02:55:00,228
the graph. One way to do that is to take the
shadow or projection of the graph

1775
02:55:00,228 --> 02:55:07,969
onto the x axis and see what x values are
hit. It looks like we're hitting all x values

1776
02:55:07,969 --> 02:55:15,538
starting at negative eight, and continuing
up to four. So our domain

1777
02:55:15,539 --> 02:55:21,289
is the x's between negative eight, and four,
including those endpoints. Or we can write

1778
02:55:21,289 --> 02:55:27,248
this in interval notation as negative eight
comma four with square brackets.

1779
02:55:27,248 --> 02:55:31,559
To find the range of the function, we look
for the y values corresponding to points on

1780
02:55:31,559 --> 02:55:35,278
this graph, we can do that.

1781
02:55:35,279 --> 02:55:42,979
By taking the shadow or projection of the
graph onto the y axis,

1782
02:55:42,978 --> 02:55:51,398
we seem to be hitting our Y values from negative
five, up through three.

1783
02:55:51,398 --> 02:55:59,340
So our range is wise between negative five
and three are an interval notation negative

1784
02:55:59,340 --> 02:56:05,369
five, three with square brackets. If we meet
a function that's described as an equation

1785
02:56:05,369 --> 02:56:11,498
instead of a graph, one way to find the domain
and range are to graph the function. But it's

1786
02:56:11,498 --> 02:56:17,478
often possible to find the domain at least
more quickly, by using algebraic considerations.

1787
02:56:17,478 --> 02:56:22,328
We think about what x values that makes sense
to plug into this expression, and what x values

1788
02:56:22,328 --> 02:56:27,049
need to be excluded, because they make the
algebraic expression impossible to evaluate.

1789
02:56:27,049 --> 02:56:35,629
Specifically, to find the domain of a function,
we need to exclude x values that make the

1790
02:56:35,629 --> 02:56:38,698
denominator zero. Since we can't divide by
zero,

1791
02:56:38,699 --> 02:56:47,470
we also need to exclude x values that make
an expression inside a square root sign negative.

1792
02:56:47,469 --> 02:56:52,108
Since we can't take the square root of a negative
number. In fact, we need to exclude values

1793
02:56:52,109 --> 02:56:58,199
that make the expression inside any even root
negative because we can't take an even root

1794
02:56:58,199 --> 02:57:02,520
of a negative number, even though we can take
an odd root like a cube root of a negative

1795
02:57:02,520 --> 02:57:07,488
number. Later, when we look at logarithmic
functions, we'll have some additional exclusions

1796
02:57:07,488 --> 02:57:12,228
that we have to make. But for now, these two
principles should handle all functions We'll

1797
02:57:12,228 --> 02:57:16,639
see. So let's apply them to a couple examples.

1798
02:57:16,639 --> 02:57:23,170
For the function in part A, we don't have
any square root signs, but we do have a denominator.

1799
02:57:23,170 --> 02:57:28,389
So we need to exclude x values that make the
denominator zero. In other words, we need

1800
02:57:28,389 --> 02:57:34,458
x squared minus 4x plus three to not be equal
to zero.

1801
02:57:34,459 --> 02:57:42,239
If we solve x squared minus 4x plus three
equal to zero, we can do that by factoring.

1802
02:57:42,238 --> 02:57:48,609
And that gives us x equals three or x equals
one, so we need to exclude these values.

1803
02:57:48,609 --> 02:57:54,359
All other x values should be fine. So if I
draw the number line, I can put on one and

1804
02:57:54,359 --> 02:58:00,689
three and just dig out a hole at both of those.
And my domain includes everything else on

1805
02:58:00,689 --> 02:58:06,828
the number line. In interval notation, this
means my domain is everything from negative

1806
02:58:06,828 --> 02:58:12,969
infinity to one, together with everything
from one to three, together with everything

1807
02:58:12,969 --> 02:58:15,809
from three to infinity.

1808
02:58:15,809 --> 02:58:21,219
In the second example, we don't have any denominator
to worry about, but we do have a square root

1809
02:58:21,219 --> 02:58:29,688
sign. So we need to exclude any x values that
make three minus 2x less than zero. In other

1810
02:58:29,689 --> 02:58:35,760
words, we can include all x values for which
three minus 2x is greater than or equal to

1811
02:58:35,760 --> 02:58:43,279
zero. Solving that inequality gives us three
is greater than equal to 2x. In other words,

1812
02:58:43,279 --> 02:58:44,789
x

1813
02:58:44,789 --> 02:58:48,209
is less than or equal to three halves,

1814
02:58:48,209 --> 02:58:51,010
I can draw this on the number line,

1815
02:58:51,010 --> 02:58:54,898
or write it in interval notation.

1816
02:58:54,898 --> 02:59:00,719
Notice that three halves is included, and
that's because three minus 2x is allowed to

1817
02:59:00,719 --> 02:59:05,629
be zero, I can take the square root of zero,
that's just zero. And that's not a problem.

1818
02:59:05,629 --> 02:59:10,238
Finally, let's look at a more complicated
function that involves both the square root

1819
02:59:10,238 --> 02:59:14,969
and denominator. Now there are two things
I need to worry about.

1820
02:59:14,969 --> 02:59:18,099
I need the denominator

1821
02:59:18,100 --> 02:59:24,300
to not be equal to zero, and I need the stuff
inside the square root side to be greater

1822
02:59:24,299 --> 02:59:30,309
than or equal to zero. from our earlier work,
we know the first condition means that x is

1823
02:59:30,309 --> 02:59:36,488
not equal to three, and x is not equal to
one. And the second condition means that x

1824
02:59:36,488 --> 02:59:41,879
is less than or equal to three halves. Let's
try both of those conditions on the number

1825
02:59:41,879 --> 02:59:44,179
line.

1826
02:59:44,180 --> 02:59:51,039
x is not equal to three and x is not equal
to one means we've got everything except those

1827
02:59:51,039 --> 02:59:56,680
two dug out points. And the other condition
x is less than or equal to three halves means

1828
02:59:56,680 --> 03:00:00,209
we can have three halves and everything

1829
03:00:00,209 --> 03:00:05,060
To the left of it. Now to be in our domain
and to be legit for our function, we need

1830
03:00:05,059 --> 03:00:09,219
both of these conditions to be true. So I'm
going to connect these conditions with N and

1831
03:00:09,219 --> 03:00:13,708
N, that means we're looking for a numbers
on the number line that are colored both red

1832
03:00:13,709 --> 03:00:19,939
and blue. So I'll draw that above in purple.
So that's everything from three halves to

1833
03:00:19,939 --> 03:00:24,709
one, I have to dig out one because one was
a problem for the denominator. And then I

1834
03:00:24,709 --> 03:00:30,300
can continue for all the things that are colored
both both colors red and blue. So my final

1835
03:00:30,299 --> 03:00:37,679
domain is going to be, let's see negative
infinity, up to but not including one together

1836
03:00:37,680 --> 03:00:44,540
with one, but not including it to three halves,
and I include three half since that was colored,

1837
03:00:44,540 --> 03:00:46,199
both red and blue. Also,

1838
03:00:46,199 --> 03:00:51,340
in this video, we talked about functions,
how to evaluate them, and how to find domains

1839
03:00:51,340 --> 03:00:57,130
and ranges. This video gives the graphs of
some commonly used functions that I call the

1840
03:00:57,129 --> 03:01:03,849
toolkit functions. The first function is the
function y equals x, let's plot a few points

1841
03:01:03,850 --> 03:01:13,248
on the graph of this function. If x is zero,
y zero, if x is one, y is one, and so on y

1842
03:01:13,248 --> 03:01:19,059
is always equal to x doesn't have to just
be an integer, it can be any real number,

1843
03:01:19,059 --> 03:01:24,398
and we'll connecting the dots, we get a straight
line through the origin.

1844
03:01:24,398 --> 03:01:30,090
Let's look at the graph of y equals x squared.
If x is zero, y is zero. So we go through

1845
03:01:30,090 --> 03:01:40,398
the origin again, if x is one, y is one, and
x is negative one, y is also one,

1846
03:01:40,398 --> 03:01:45,248
the x value of two gives a y value of four
and the x value of negative two gives a y

1847
03:01:45,248 --> 03:01:51,059
value of four also connecting the dots, we
get a parabola.

1848
03:01:51,059 --> 03:01:56,828
That is this, this function is an even function.

1849
03:01:56,828 --> 03:02:02,840
That means it has mirror symmetry across the
y axis, the left side it looks like exactly

1850
03:02:02,840 --> 03:02:08,139
like the mirror image of the right side. That
happens because when you square a positive

1851
03:02:08,139 --> 03:02:14,760
number, like two, you get the exact same y
value as when you square it's a mirror image

1852
03:02:14,760 --> 03:02:17,699
x value of negative two.

1853
03:02:17,699 --> 03:02:24,119
The next function y equals x cubed. I'll call
that a cubic. Let's plot a few points when

1854
03:02:24,119 --> 03:02:31,370
x is zero, y is zero. When x is one, y is
one, when x is negative one, y is negative

1855
03:02:31,370 --> 03:02:37,319
one, two goes with the point eight way up
here and an x value of negative two is going

1856
03:02:37,318 --> 03:02:42,799
to give us negative eight. If I connect the
dots, I get something that looks kind of like

1857
03:02:42,799 --> 03:02:45,049
this.

1858
03:02:45,049 --> 03:02:53,728
This function is what's called an odd function,
because it has 180 degree rotational symmetry

1859
03:02:53,728 --> 03:02:58,688
occur around the origin. If I rotate this
whole graph by 180 degrees, or in other words,

1860
03:02:58,689 --> 03:03:04,908
turn the paper upside down, I'll get exactly
the same shape. The reason this function has

1861
03:03:04,908 --> 03:03:12,199
this odd symmetry is because when I cube a
positive number, and get its y value, I get

1862
03:03:12,199 --> 03:03:17,670
n cube the corresponding negative number to
get its y value, the negative number cubed

1863
03:03:17,670 --> 03:03:25,109
gives us exactly the negative of the the y
value we get with cubing the positive number.

1864
03:03:25,109 --> 03:03:29,828
Let's look at the next example. Y equals the
square root of x.

1865
03:03:29,828 --> 03:03:37,010
Notice that the domain of this function is
just x values bigger than or equal to zero

1866
03:03:37,010 --> 03:03:42,478
because we can't take the square root of a
negative number. Let's plug in a few x values.

1867
03:03:42,478 --> 03:03:49,559
X is zero gives y is 0x is one square root
of one is one, square root of four is two,

1868
03:03:49,559 --> 03:03:55,448
and connecting the dots, I get a function
that looks like this.

1869
03:03:55,449 --> 03:04:01,810
The absolute value function is next. Again,
if we plug in a few points, x is zero goes

1870
03:04:01,809 --> 03:04:11,079
with y equals 0x is one gives us one, the
absolute value of negative one is 122 is on

1871
03:04:11,079 --> 03:04:17,930
the graph, and the absolute value of negative
two is to I'm ending up getting a V shaped

1872
03:04:17,930 --> 03:04:24,470
graph. It also has even or a mirror symmetry.

1873
03:04:24,469 --> 03:04:30,379
Y equals two to the x is what's known as an
exponential function. That's because the variable

1874
03:04:30,379 --> 03:04:35,648
x is in the exponent. If I plot a few points,

1875
03:04:35,648 --> 03:04:41,930
two to the zero is one,

1876
03:04:41,930 --> 03:04:49,709
two to the one is two, two squared is four,
two to the minus one is one half. I'll plot

1877
03:04:49,709 --> 03:04:55,850
these on my graph, you know, let me fill in
a few more points. So let's see. two cubed

1878
03:04:55,850 --> 03:04:59,908
is eight. That's way up here and negative
two gives

1879
03:04:59,908 --> 03:05:07,318
Maybe 1/4 1/8 connecting the dots, I get something
that's shaped like this.

1880
03:05:07,318 --> 03:05:11,988
You might have heard the expression exponential
growth, when talking about, for example, population

1881
03:05:11,988 --> 03:05:19,359
growth, this is function is represents exponential
growth, it grows very, very steeply. Every

1882
03:05:19,359 --> 03:05:26,689
time we increase the x coordinate by one,
we double the y coordinate.

1883
03:05:26,689 --> 03:05:32,970
We could also look at a function like y equals
three dx, or sometimes y equals e to the x,

1884
03:05:32,969 --> 03:05:40,129
where E is just a number about 2.7. These
functions look very similar. It's just the

1885
03:05:40,129 --> 03:05:45,948
bigger bass makes us rise a little more steeply.

1886
03:05:45,949 --> 03:05:52,211
Now let's look at the function y equals one
over x. It's not defined when x is zero, but

1887
03:05:52,210 --> 03:05:58,228
I can plug in x equals one half, one over
one half is to

1888
03:05:58,228 --> 03:06:06,528
whenever one is one, and one or two is one
half. By connect the dots, I get this shape

1889
03:06:06,529 --> 03:06:11,090
in the first quadrant, but I haven't looked
at negative values of x, when x is negative,

1890
03:06:11,090 --> 03:06:18,639
whenever negative one is negative one, whenever
negative half is negative two, and I get a

1891
03:06:18,639 --> 03:06:22,318
similar looking piece in the third quadrant.

1892
03:06:22,318 --> 03:06:27,639
This is an example of a hyperbola.

1893
03:06:27,639 --> 03:06:34,788
And it's also an odd function, because it
has that 180 degree rotational symmetry, if

1894
03:06:34,789 --> 03:06:40,369
I turn the page upside down, it'll look exactly
the same. Finally, let's look at y equals

1895
03:06:40,369 --> 03:06:42,079
one over x squared.

1896
03:06:42,079 --> 03:06:48,689
Again, it's not defined when x is zero, but
I can plug in points like one half or X, let's

1897
03:06:48,689 --> 03:06:55,498
see one over one half squared is one over
1/4, which is four.

1898
03:06:55,498 --> 03:07:03,369
And one over one squared, one over two squared
is a fourth, it looks pretty much like the

1899
03:07:03,369 --> 03:07:08,399
previous function is just a little bit more
extreme rises a little more steeply, falls

1900
03:07:08,398 --> 03:07:12,959
a little more dramatically. But for negative
values of x, something a little bit different

1901
03:07:12,959 --> 03:07:19,039
goes on. For example, one over negative two
squared is just one over four, which is a

1902
03:07:19,040 --> 03:07:27,060
fourth. So I can plot that point there, and
one over a negative one squared, that's just

1903
03:07:27,059 --> 03:07:32,260
one. So my curve for negative values of x
is gonna lie in the second quadrant, not the

1904
03:07:32,260 --> 03:07:38,270
third quadrant. This is an example of an even
function

1905
03:07:38,270 --> 03:07:42,359
because it has perfect mirror symmetry across
the y axis.

1906
03:07:42,359 --> 03:07:49,809
These are the toolkit functions, and I recommend
you memorize the shape of each of them.

1907
03:07:49,809 --> 03:07:56,180
That way, you can draw at least a rough sketch
very quickly without having to plug in points.

1908
03:07:56,180 --> 03:07:58,850
That's all for the graphs of the toolkit functions.

1909
03:07:58,850 --> 03:08:07,029
If we change the equation of a function, then
the graph of the function changes or transforms

1910
03:08:07,029 --> 03:08:09,459
in predictable ways.

1911
03:08:09,459 --> 03:08:14,460
This video gives some rules and examples for
transformations of functions. To get the most

1912
03:08:14,459 --> 03:08:19,108
out of this video, it's helpful if you're
already familiar with the graph of some typical

1913
03:08:19,109 --> 03:08:25,439
functions, I call them toolkit functions like
y equals square root of x, y equals x squared,

1914
03:08:25,439 --> 03:08:30,350
y equals the absolute value of x and so on.
If you're not familiar with those graphs,

1915
03:08:30,350 --> 03:08:36,930
I encourage you to watch my video called toolkit
functions first, before watching this one.

1916
03:08:36,930 --> 03:08:39,520
I want to start by reviewing function notation.

1917
03:08:39,520 --> 03:08:46,340
If g of x represents the function, the square
root of x, then we can rewrite these expressions

1918
03:08:46,340 --> 03:08:53,199
in terms of square roots. For example, g of
x minus two is the same thing as the square

1919
03:08:53,199 --> 03:08:55,569
root of x minus two.

1920
03:08:55,568 --> 03:09:05,889
g of quantity x minus two means we plug in
x minus two, everywhere we see an x, so that

1921
03:09:05,889 --> 03:09:11,978
would be the same thing as the square root
of quantity x minus two.

1922
03:09:11,978 --> 03:09:16,840
In this second example, I say that we're subtracting
two on the inside of the function, because

1923
03:09:16,840 --> 03:09:21,930
we're subtracting two before we apply the
square root function. Whereas on the first

1924
03:09:21,930 --> 03:09:26,990
example, I say that the minus two is on the
outside of the function, we're doing the square

1925
03:09:26,990 --> 03:09:33,359
root function first and then subtracting two.
In this third example, g of 3x, we're multiplying

1926
03:09:33,359 --> 03:09:38,590
by three on the inside of the function. To
evaluate this in terms of square root, we

1927
03:09:38,590 --> 03:09:44,549
plug in the entire 3x for x and the square
root function. That gives us the square root

1928
03:09:44,549 --> 03:09:45,568
of 3x.

1929
03:09:45,568 --> 03:09:51,609
In the next example, we're multiplying by
three on the outside of the function. This

1930
03:09:51,609 --> 03:09:58,720
is just three times the square root of x.
Finally, g of minus x means the square root

1931
03:09:58,719 --> 03:10:00,059
of minus x.

1932
03:10:00,059 --> 03:10:03,769
Now, this might look a little odd because
we're not used to taking the square root of

1933
03:10:03,770 --> 03:10:09,760
a negative number. But remember that if x
itself is negative, like negative to the negative

1934
03:10:09,760 --> 03:10:13,850
x will be negative negative two or positive
two. So we're really be taking the square

1935
03:10:13,850 --> 03:10:18,899
root of a positive number in that case, let
me record which of these are inside and which

1936
03:10:18,898 --> 03:10:21,748
of these are outside of my function.

1937
03:10:21,748 --> 03:10:26,408
In this next set of examples, we're using
the same function g of x squared of x. But

1938
03:10:26,408 --> 03:10:29,958
this time, we're starting with an expression
with square roots in it, and trying to write

1939
03:10:29,959 --> 03:10:37,699
it in terms of g of x. So the first example,
the plus 17 is on the outside of my function,

1940
03:10:37,699 --> 03:10:43,248
because I'm taking the square root of x first,
and then adding 17. So I can write this as

1941
03:10:43,248 --> 03:10:46,418
g of x plus 17.

1942
03:10:46,418 --> 03:10:53,520
In the second example, I'm taking x and adding
12 first, then I take the square root of the

1943
03:10:53,520 --> 03:10:58,550
whole thing. Since I'm adding the 12 to x
first that's on the inside of my function.

1944
03:10:58,549 --> 03:11:04,049
So I write that as g of the quantity x plus
12.

1945
03:11:04,049 --> 03:11:09,559
Remember that this notation means I plug in
the entire x plus 12, into the square root

1946
03:11:09,559 --> 03:11:14,738
sign, which gives me exactly square root of
x plus 12. And this next example, undoing

1947
03:11:14,738 --> 03:11:20,539
the square root first and then multiplying
by negative 36. So i minus 36, multiplications,

1948
03:11:20,540 --> 03:11:27,760
outside my function, I can rewrite this as
minus 36 times g of x. Finally, in this last

1949
03:11:27,760 --> 03:11:33,130
example, I take x multiplied by a fourth and
then apply the square root of x. So that's

1950
03:11:33,129 --> 03:11:40,938
the same thing as g of 1/4 X, my 1/4 X is
on the inside of my function. In other words,

1951
03:11:40,939 --> 03:11:44,880
it's inside the parentheses when I use function
notation.

1952
03:11:44,879 --> 03:11:50,809
Let's graph the square root of x and two transformations
have this function,

1953
03:11:50,809 --> 03:11:56,019
y equals the square root of x goes to the
point 0011. And for two, since the square

1954
03:11:56,020 --> 03:12:01,270
root of four is two, it looks something like
this.

1955
03:12:01,270 --> 03:12:06,248
In order to graph y equals the square root
of x minus two, notice that the minus two

1956
03:12:06,248 --> 03:12:10,629
is on the outside of the function, that means
we're going to take the square root of x first

1957
03:12:10,629 --> 03:12:17,299
and then subtract two. So for example, if
we start with the x value of zero, and compute

1958
03:12:17,299 --> 03:12:21,938
the square root of zero, that's zero, then
we subtract two to give us a y value of negative

1959
03:12:21,939 --> 03:12:23,680
two,

1960
03:12:23,680 --> 03:12:29,510
an x value of one, which under the square
root function had a y value of one now has

1961
03:12:29,510 --> 03:12:36,898
a y value that's decreased by two, one minus
two is negative one. And finally, an x value

1962
03:12:36,898 --> 03:12:42,469
of four, which under the square root function
had a y value of two now has a y value of

1963
03:12:42,469 --> 03:12:49,840
two minus two or zero, its y value is also
decreased by two. If I plot these new points,

1964
03:12:49,840 --> 03:12:56,469
zero goes with negative two, one goes with
negative one, and four goes with zero, I have

1965
03:12:56,469 --> 03:13:00,078
my transformed graph.

1966
03:13:00,078 --> 03:13:04,978
Because I subtracted two on the outside of
my function, my y values were decreased by

1967
03:13:04,978 --> 03:13:12,118
two, which brought my graph down by two units.
Next, let's look at y equals the square root

1968
03:13:12,119 --> 03:13:17,359
of quantity x minus two. Now we're subtracting
two on the inside of our function, which means

1969
03:13:17,359 --> 03:13:21,899
we subtract two from x first and then take
the square root. In order to get the same

1970
03:13:21,898 --> 03:13:28,279
y value of zero as we had in our blue graph,
we need our x minus two to be zero, so we

1971
03:13:28,280 --> 03:13:30,430
need our original x to be two.

1972
03:13:30,430 --> 03:13:36,689
In order to get the y value of one that we
had in our blue graph, we need to be taking

1973
03:13:36,689 --> 03:13:41,659
the square root of one, so we need x minus
two to be one, which means that we need to

1974
03:13:41,659 --> 03:13:43,459
start with an x value of three.

1975
03:13:43,459 --> 03:13:50,819
And in order to reproduce our y value of two
from our original graph, we need the square

1976
03:13:50,819 --> 03:13:55,408
root of x minus two to be two, which means
we need to start out by taking the square

1977
03:13:55,408 --> 03:14:04,168
root of four, which means our x minus two
is four, so our x should be up there at six.

1978
03:14:04,168 --> 03:14:10,510
If I plot my x values, with my corresponding
y values of square root of x minus two, I

1979
03:14:10,510 --> 03:14:18,068
get the following graph. Notice that the graph
has moved horizontally to the right by two

1980
03:14:18,068 --> 03:14:19,068
units.

1981
03:14:19,068 --> 03:14:25,629
To me, moving down by two units, makes sense
because we're subtracting two, so we're decreasing

1982
03:14:25,629 --> 03:14:30,519
y's by two units, but the minus two on the
inside that kind of does the opposite of what

1983
03:14:30,520 --> 03:14:35,439
I expect, I might expected to to move the
graph left, I might expect the x values to

1984
03:14:35,439 --> 03:14:41,389
be going down by two units, but instead, it
moves the graph to the right,

1985
03:14:41,389 --> 03:14:46,869
because the x units have to go up by two units
in order to get the right square root when

1986
03:14:46,870 --> 03:14:52,919
I then subtract two units again, the observations
we made for these transformations of functions

1987
03:14:52,918 --> 03:14:58,590
hold in general, according to the following
rules. First of all, numbers on the outside

1988
03:14:58,590 --> 03:14:59,850
of the function like

1989
03:14:59,850 --> 03:15:06,399
In our example, y equals a squared of x minus
two, those numbers affect the y values. And

1990
03:15:06,398 --> 03:15:12,148
a result in vertical motions, like we saw,
these motions are in the direction you expect.

1991
03:15:12,148 --> 03:15:18,119
So subtracting two was just down by two. If
we were adding two instead, that would move

1992
03:15:18,120 --> 03:15:19,880
us up by two

1993
03:15:19,879 --> 03:15:24,829
numbers on the inside of the function. That's
like our example, y equals the square root

1994
03:15:24,829 --> 03:15:31,379
of quantity x minus two, those affect the
x values and results in a horizontal motion,

1995
03:15:31,379 --> 03:15:35,858
these motions go in the opposite direction
from what you expect. Remember, the minus

1996
03:15:35,859 --> 03:15:40,658
two on the inside actually shifted our graph
to the right, if it had had a plus two on

1997
03:15:40,658 --> 03:15:45,918
the inside, that would actually shift our
graph to the left.

1998
03:15:45,918 --> 03:15:51,738
Adding results in a shift those are called
translations, but multiplying like something

1999
03:15:51,738 --> 03:15:59,568
like y equals three times the square root
of x, that would result in a stretch or shrink.

2000
03:15:59,568 --> 03:16:03,748
In other words, if I start with the square
root of x,

2001
03:16:03,748 --> 03:16:08,539
and then when I graph y equals three times
the square root of x, that stretches my graph

2002
03:16:08,539 --> 03:16:12,590
vertically by a factor of three.

2003
03:16:12,590 --> 03:16:13,590
Like this,

2004
03:16:13,590 --> 03:16:20,789
if I want to graph y equals 1/3, times the
square root of x, that shrinks my graph vertically

2005
03:16:20,789 --> 03:16:22,909
by a factor of 1/3.

2006
03:16:22,908 --> 03:16:29,469
Finally, a negative sign results in reflection.
For example, if I start with the graph of

2007
03:16:29,469 --> 03:16:34,750
y equals the square root of x, and then when
I graph y equals the square root of negative

2008
03:16:34,750 --> 03:16:39,859
x, that's going to do a reflection in the
horizontal direction because the negative

2009
03:16:39,859 --> 03:16:42,130
is on the inside of the square root sign.

2010
03:16:42,129 --> 03:16:49,318
A reflection in the horizontal direction means
a reflection across the y axis.

2011
03:16:49,318 --> 03:16:53,939
If instead, I want to graph y equals negative
A squared of x, that negative sign on the

2012
03:16:53,939 --> 03:17:00,039
outside means a vertical reflection, a reflection
across the x axis.

2013
03:17:00,039 --> 03:17:06,529
Pause the video for a moment and see if you
could describe what happens in these four

2014
03:17:06,529 --> 03:17:07,779
transformations.

2015
03:17:07,779 --> 03:17:12,859
In the first example, we're subtracting four
on the outside of the function. adding or

2016
03:17:12,859 --> 03:17:18,380
subtracting means a translation or shift.
And since we're on the outside of the function

2017
03:17:18,379 --> 03:17:24,049
affects the y value, so that's moving us vertically.
So this transformation should take the square

2018
03:17:24,049 --> 03:17:31,358
root of graph and move it down by four units,
that would look something like this.

2019
03:17:31,359 --> 03:17:39,090
In the next example, we're adding 12 on the
inside, that's still a translation. But now

2020
03:17:39,090 --> 03:17:44,719
we're moving horizontally. And so since we
go the opposite direction, we expect we are

2021
03:17:44,719 --> 03:17:51,148
going to go to the left by 12 units, that's
going to look something like this.

2022
03:17:51,148 --> 03:17:55,278
And the next example, we're multiplying by
three and introducing a negative sign both

2023
03:17:55,279 --> 03:18:00,689
on the outside of our function outside our
function means we're affecting the y values.

2024
03:18:00,689 --> 03:18:06,120
So in multiplication means we're stretching
by a factor of three, the negative sign means

2025
03:18:06,120 --> 03:18:10,310
we reflect in the vertical direction,

2026
03:18:10,309 --> 03:18:15,168
here's stretching by a factor of three vertically,
before I apply the minus sign, and now the

2027
03:18:15,168 --> 03:18:19,020
minus sign reflects in the vertical direction.

2028
03:18:19,020 --> 03:18:26,029
Finally, in this last example, we're multiplying
by one quarter on the inside of our function,

2029
03:18:26,029 --> 03:18:31,040
we know that multiplication means stretch
or shrink. And since we're on the inside,

2030
03:18:31,040 --> 03:18:36,220
it's a horizontal motion, and it does the
opposite of what we expect. So instead of

2031
03:18:36,219 --> 03:18:41,648
shrinking by a factor of 1/4, horizontally,
it's actually going to stretch by the reciprocal,

2032
03:18:41,648 --> 03:18:44,139
a factor of four horizontally.

2033
03:18:44,139 --> 03:18:50,998
that'll look something like this. Notice that
stretching horizontally by a factor of four

2034
03:18:50,998 --> 03:18:55,689
looks kind of like shrinking vertically by
a factor of one half.

2035
03:18:55,689 --> 03:19:01,000
And that's actually borne out by the algebra,
because the square root of 1/4 x is the same

2036
03:19:01,000 --> 03:19:05,510
thing as the square root of 1/4 times the
square root of x, which is the same thing

2037
03:19:05,510 --> 03:19:12,119
as one half times the square root of x. And
so now we can see algebraically that vertical

2038
03:19:12,119 --> 03:19:18,520
shrink by a factor of one half is the same
as a horizontal stretch by a factor of four,

2039
03:19:18,520 --> 03:19:22,310
at least for this function, the square root
function.

2040
03:19:22,309 --> 03:19:27,988
This video gives some rules for transformations
of functions, which I'll repeat below. numbers

2041
03:19:27,988 --> 03:19:37,039
on the outside correspond to changes in the
y values, or vertical motions.

2042
03:19:37,040 --> 03:19:44,439
numbers on the inside of the function, affect
the x values, and result in horizontal motions.

2043
03:19:44,439 --> 03:19:47,760
Adding and subtracting

2044
03:19:47,760 --> 03:19:52,020
corresponds to translations or shifts.

2045
03:19:52,020 --> 03:19:57,988
multiplying and dividing by numbers corresponds
to stretches and shrinks

2046
03:19:57,988 --> 03:20:01,389
and putting in a negative sign.

2047
03:20:01,389 --> 03:20:04,269
corresponds to a reflection,

2048
03:20:04,270 --> 03:20:10,100
horizontal reflection, if the negative sign
is on the inside, and a vertical reflection,

2049
03:20:10,100 --> 03:20:13,010
if the negative sign is on the outside,

2050
03:20:13,010 --> 03:20:17,939
knowing these basic rules about transformations
empowers you to be able to sketch graphs of

2051
03:20:17,939 --> 03:20:24,040
much more complicated functions, like y equals
three times the square root of x plus two,

2052
03:20:24,040 --> 03:20:29,521
by simply considering the transformations,
one at a time.

2053
03:20:29,521 --> 03:20:35,770
a quadratic function is a function that can
be written in the form f of x equals A x squared

2054
03:20:35,770 --> 03:20:46,918
plus bx plus c, where a, b and c are real
numbers. And a is not equal to zero. The reason

2055
03:20:46,918 --> 03:20:53,449
we require that A is not equal to zero is
because if a were equal to zero, we'd have

2056
03:20:53,449 --> 03:21:01,449
f of x is equal to b x plus c, which is called
a linear function. So by making sure that

2057
03:21:01,449 --> 03:21:06,459
a is not zero, we make sure there's really
an x squared term which is the hallmark of

2058
03:21:06,459 --> 03:21:08,959
a quadratic function.

2059
03:21:08,959 --> 03:21:13,939
Please pause the video for a moment and decide
which of these equations represent quadratic

2060
03:21:13,939 --> 03:21:15,869
functions.

2061
03:21:15,869 --> 03:21:23,110
The first function can definitely be written
in the form g of x equals A x squared plus

2062
03:21:23,110 --> 03:21:30,840
bx plus c. fact it's already written in that
form, where a is negative five, B is 10, and

2063
03:21:30,840 --> 03:21:32,199
C is three.

2064
03:21:32,199 --> 03:21:39,859
The second equation is also a quadratic function,
because we can think of it as f of x equals

2065
03:21:39,859 --> 03:21:46,559
one times x squared plus zero times x plus
zero.

2066
03:21:46,559 --> 03:21:53,379
So it is in the right form, where A is one,
B is zero, and C is zero.

2067
03:21:53,379 --> 03:21:59,108
It's perfectly fine for the coefficient of
x and the constant term to be zero for a quadratic

2068
03:21:59,109 --> 03:22:03,590
function, we just need the coefficient of
x squared to be nonzero, so the x squared

2069
03:22:03,590 --> 03:22:06,380
term is preserved.

2070
03:22:06,379 --> 03:22:13,269
The third equation is not a quadratic function.
It's a linear function, because there's no

2071
03:22:13,270 --> 03:22:17,069
x squared term.

2072
03:22:17,068 --> 03:22:21,998
The fourth function might not look like a
quadratic function. But if we rewrite it by

2073
03:22:21,998 --> 03:22:28,708
expanding out the X minus three squared, let's
see what happens. We get y equals two times

2074
03:22:28,709 --> 03:22:36,760
x minus three times x minus three plus four.
So that's two times x squared minus 3x minus

2075
03:22:36,760 --> 03:22:45,228
3x, plus nine plus four, continuing, I get
2x squared,

2076
03:22:45,228 --> 03:22:53,260
minus 12 access, plus 18 plus four, in other
words, y equals 2x squared minus 12x plus

2077
03:22:53,260 --> 03:23:02,488
22. So in fact, our function can be written
in the right form, and it is a quadratic function.

2078
03:23:02,488 --> 03:23:08,488
A function that is already written in the
form y equals A x squared plus bx plus c is

2079
03:23:08,488 --> 03:23:12,520
said to be in standard form.

2080
03:23:12,520 --> 03:23:17,289
So our first example g of x is in standard
form

2081
03:23:17,289 --> 03:23:23,270
a function that's written in the format of
the last function that is in the form of y

2082
03:23:23,270 --> 03:23:33,529
equals a times x minus h squared plus k for
some numbers, a, h, and K. That said to be

2083
03:23:33,529 --> 03:23:35,760
in vertex form,

2084
03:23:35,760 --> 03:23:40,498
I'll talk more about standard form and vertex
form in another video.

2085
03:23:40,498 --> 03:23:45,869
In this video, we identified some quadratic
functions, and talked about the difference

2086
03:23:45,869 --> 03:23:49,529
between standard form

2087
03:23:49,529 --> 03:23:54,180
and vertex form.

2088
03:23:54,180 --> 03:23:59,529
This video is about graphing quadratic functions.
a quadratic function, which is typically written

2089
03:23:59,529 --> 03:24:08,069
in standard form, like this, or sometimes
in vertex form, like this, always has the

2090
03:24:08,068 --> 03:24:11,350
graph that looks like a parabola.

2091
03:24:11,350 --> 03:24:17,229
This video will show how to tell whether the
problem is pointing up or down. How to find

2092
03:24:17,228 --> 03:24:21,788
its x intercepts, and how to find its vertex.

2093
03:24:21,789 --> 03:24:30,890
The bare bones basic quadratic function is
f of x equals x squared, it goes to the origin,

2094
03:24:30,889 --> 03:24:38,568
since f of zero is zero squared, which is
zero, and it is a parabola pointing upwards

2095
03:24:38,568 --> 03:24:40,738
like this.

2096
03:24:40,738 --> 03:24:46,270
The vertex of a parabola is its lowest point
if it's pointing upwards, and its highest

2097
03:24:46,270 --> 03:24:53,908
point if it's pointing downwards. So in this
case, the vertex is at 00.

2098
03:24:53,908 --> 03:24:58,949
The x intercepts are where the graph crosses
the x axis. In other words, where y is zero

2099
03:24:58,949 --> 03:25:05,529
In this function, y equals zero means that
x squared is zero, which happens only when

2100
03:25:05,529 --> 03:25:11,350
x is zero. So the x intercept, there's only
one of them is also zero.

2101
03:25:11,350 --> 03:25:17,869
The second function, y equals negative 3x
squared also goes to the origin. Since the

2102
03:25:17,869 --> 03:25:25,140
functions value when x is zero, is y equals
zero. But in this case, the parabola is pointing

2103
03:25:25,139 --> 03:25:27,818
downwards.

2104
03:25:27,818 --> 03:25:32,100
That's because thinking about transformations
of functions, a negative sign on the outside

2105
03:25:32,100 --> 03:25:38,600
reflects the function vertically over the
x axis, making the problem instead of pointing

2106
03:25:38,600 --> 03:25:43,819
upwards, reflecting the point downwards.

2107
03:25:43,819 --> 03:25:49,010
The number three on the outside stretches
the graph vertically by a factor of three.

2108
03:25:49,010 --> 03:25:55,309
So it makes it kind of long and skinny like
this.

2109
03:25:55,309 --> 03:26:00,299
In general, a negative coefficient to the
x squared term means the problem will be pointing

2110
03:26:00,299 --> 03:26:05,248
down. Whereas a positive coefficient, like
here, the coefficients, one means the parabola

2111
03:26:05,248 --> 03:26:06,738
is pointing up.

2112
03:26:06,738 --> 03:26:09,648
Alright, that roll over here.

2113
03:26:09,648 --> 03:26:16,129
So if a is bigger than zero, the parabola
opens up.

2114
03:26:16,129 --> 03:26:22,159
And if the value of the coefficient a is less
than zero, then the parabola opens down.

2115
03:26:22,159 --> 03:26:27,728
In this second example, we can see again that
the vertex is at 00. And the x intercept is

2116
03:26:27,728 --> 03:26:30,408
x equals zero.

2117
03:26:30,408 --> 03:26:33,538
Let's look at this third example.

2118
03:26:33,539 --> 03:26:38,168
If we multiplied our expression out, we'd
see that the coefficient of x squared would

2119
03:26:38,168 --> 03:26:43,760
be to a positive number. So that means our
parabola is going to be opening up.

2120
03:26:43,760 --> 03:26:50,020
But the vertex of this parabola will no longer
be at the origin. In fact, we can find the

2121
03:26:50,020 --> 03:26:55,319
parabolas vertex by thinking about transformations
of functions.

2122
03:26:55,318 --> 03:27:02,318
Our function is related to the function y
equals to x squared, by moving it to the right

2123
03:27:02,318 --> 03:27:13,889
by three and up by four. Since y equals 2x
squared has a vertex at 00. If we move that

2124
03:27:13,889 --> 03:27:21,549
whole parabola including the vertex, right
by three, and up by four, the vertex will

2125
03:27:21,549 --> 03:27:26,099
end up at the point three, four.

2126
03:27:26,100 --> 03:27:30,170
So a parabola will look something like this.

2127
03:27:30,170 --> 03:27:36,680
Notice how easy it was to just read off the
vertex when our quadratic function is written

2128
03:27:36,680 --> 03:27:44,050
in this form. In fact, any parabola any quadratic
function written in the form a times x minus

2129
03:27:44,049 --> 03:27:53,269
h squared plus k has a vertex at h k. By the
same reasoning, we're moving the parabola

2130
03:27:53,270 --> 03:27:59,779
with a vertex at the origin to the right by
H, and by K.

2131
03:27:59,779 --> 03:28:06,300
That's why this form of a quadratic function
is called the vertex form.

2132
03:28:06,299 --> 03:28:11,879
Notice that this parabola has no x intercepts
because it does not cross the x axis.

2133
03:28:11,879 --> 03:28:19,118
For our final function, we have g of x equals
5x squared plus 10x plus three,

2134
03:28:19,119 --> 03:28:23,590
we know the graph of this function will be
a parabola pointing upwards, because the coefficient

2135
03:28:23,590 --> 03:28:26,139
of x squared is positive.

2136
03:28:26,139 --> 03:28:32,519
To find the x intercepts, we can set y equals
zero, since the x intercepts is where our

2137
03:28:32,520 --> 03:28:37,229
graph crosses the x axis, and that's where
the y value is zero.

2138
03:28:37,228 --> 03:28:43,858
So zero equals 5x squared plus 10x plus three,
I'm going to use the quadratic formula to

2139
03:28:43,859 --> 03:28:52,029
solve that. So x is negative 10 plus or minus
the square root of 10 squared minus four times

2140
03:28:52,029 --> 03:28:59,748
five times three, all over two times five.
That simplifies to x equals negative 10 plus

2141
03:28:59,748 --> 03:29:03,590
or minus the square root of 40 over 10,

2142
03:29:03,590 --> 03:29:08,139
which simplifies further to x equals negative
10 over 10, plus or minus the square root

2143
03:29:08,139 --> 03:29:16,478
of 40 over 10, which is negative one plus
or minus two square root of 10 over 10, or

2144
03:29:16,478 --> 03:29:21,288
negative one plus or minus square root of
10 over five.

2145
03:29:21,289 --> 03:29:27,409
Since the square root of 10 is just a little
bit bigger than three, this works out to approximately

2146
03:29:27,408 --> 03:29:34,359
about negative two fifths and negative eight
foot or so somewhere around right here. So

2147
03:29:34,359 --> 03:29:40,600
our parabola is going to look something like
this. Notice that it goes through crosses

2148
03:29:40,600 --> 03:29:48,168
the x axis at y equals three, that's because
when we plug in x equals zero, and two here

2149
03:29:48,168 --> 03:29:56,969
we get y equal three, so the y intercept is
at three. Finally, we can find the vertex.

2150
03:29:56,969 --> 03:30:00,299
Since this function is written in standard
form,

2151
03:30:00,299 --> 03:30:10,208
On the Y equals a x plus a squared plus bx
plus c form, and not in vertex form, we can't

2152
03:30:10,209 --> 03:30:15,120
just read off the vertex like we couldn't
before. But there's a trick called the vertex

2153
03:30:15,120 --> 03:30:20,279
formula, which says that whenever you have
a function in a quadratic function in standard

2154
03:30:20,279 --> 03:30:24,040
form, the vertex has an x coordinate

2155
03:30:24,040 --> 03:30:34,090
of negative B over two A. So in this case,
that's an x coordinate of negative 10 over

2156
03:30:34,090 --> 03:30:39,988
two times five or negative one, which is kind
of like where I put it on the graph. To find

2157
03:30:39,988 --> 03:30:47,578
the y coordinate of that vertex, I can just
plug in negative one into my equation for

2158
03:30:47,578 --> 03:30:54,799
x, which gives me at y equals five times negative
one squared, plus 10, times negative one plus

2159
03:30:54,799 --> 03:30:59,420
three, which is negative two.

2160
03:30:59,420 --> 03:31:04,398
So I think I better redraw my graph a little
bit to put that vertex down at a y coordinate

2161
03:31:04,398 --> 03:31:10,939
of negative two where it's supposed to be.

2162
03:31:10,939 --> 03:31:15,828
Let's summarize the steps we use to graph
these quadratic functions. First of all, the

2163
03:31:15,828 --> 03:31:19,680
graph of a quadratic function has the shape
of a parabola.

2164
03:31:19,680 --> 03:31:27,318
The parabola opens up, if the coefficient
of x squared, which we'll call a is greater

2165
03:31:27,318 --> 03:31:34,658
than zero and down if a is less than zero.
To find the x intercepts, we set y equal to

2166
03:31:34,658 --> 03:31:44,199
zero, or in other words, f of x equal to zero
and solve for x, the find the vertex, we can

2167
03:31:44,199 --> 03:31:53,310
either read it off as h k, if our function
is in vertex form,

2168
03:31:53,309 --> 03:31:57,189
or we can use the vertex formula

2169
03:31:57,189 --> 03:32:01,260
and get the x coordinate

2170
03:32:01,260 --> 03:32:10,930
of the vertex to be negative B over to a if
our function is in standard form.

2171
03:32:10,930 --> 03:32:19,520
To find the y coordinate of the vertex in
this case, we just plug in the x coordinate

2172
03:32:19,520 --> 03:32:21,789
and figure out what y is.

2173
03:32:21,789 --> 03:32:29,590
Finally, we can always find additional points
on the graph by plugging in values of x.

2174
03:32:29,590 --> 03:32:34,870
In this video, we learned some tricks for
graphing quadratic functions. In particular,

2175
03:32:34,870 --> 03:32:43,220
we saw that the vertex can be read off as
h k, if our function is written in vertex

2176
03:32:43,219 --> 03:32:50,478
form, and the x coordinate of the vertex can
be calculated

2177
03:32:50,478 --> 03:32:57,198
as negative B over two A, if our function
is in standard form. For an explanation of

2178
03:32:57,199 --> 03:33:02,959
why this vertex formula works, please see
the my other video.

2179
03:33:02,959 --> 03:33:09,920
a quadratic and standard form looks like y
equals A x squared plus bx plus c, where a,

2180
03:33:09,920 --> 03:33:16,568
b and c are real numbers, and a is not zero.
a quadratic function in vertex form looks

2181
03:33:16,568 --> 03:33:24,028
like y equals a times x minus h squared plus
k, where a h and k are real numbers, and a

2182
03:33:24,029 --> 03:33:30,328
is not zero. When a functions in vertex form,
it's easy to read off the vertex is just the

2183
03:33:30,328 --> 03:33:34,488
ordered pair, H, okay.

2184
03:33:34,488 --> 03:33:40,568
This video explains how to get from vertex
form two standard form and vice versa.

2185
03:33:40,568 --> 03:33:46,549
Let's start by converting this quadratic function
from vertex form to standard form. That's

2186
03:33:46,549 --> 03:33:50,568
pretty straightforward, we just have to distribute
out.

2187
03:33:50,568 --> 03:33:58,668
So if I multiply out the X minus three squared,

2188
03:33:58,668 --> 03:34:07,289
I get minus four times x squared minus 6x
plus nine plus one, distributing the negative

2189
03:34:07,289 --> 03:34:16,918
four, I get negative 4x squared plus 24x minus
36 plus one. So that works out to minus 4x

2190
03:34:16,918 --> 03:34:26,709
squared plus 24x minus 35. And I have my quadratic
function now in standard form.

2191
03:34:26,709 --> 03:34:31,618
Now let's go the other direction and convert
a quadratic function that's already in standard

2192
03:34:31,619 --> 03:34:40,510
form into vertex form. That is, we want to
put it in the form of g of x equals a times

2193
03:34:40,510 --> 03:34:51,090
x minus h squared plus k, where the vertex
is at h k. To do this, it's handy to use the

2194
03:34:51,090 --> 03:34:53,870
vertex formula.

2195
03:34:53,870 --> 03:35:00,380
The vertex formula says that the x coordinate
of the vertex is given by negative

2196
03:35:00,379 --> 03:35:08,368
over to a, where A is the coefficient of x
squared, and B is the coefficient of x. So

2197
03:35:08,369 --> 03:35:16,649
in this case, we get an x coordinate of negative
eight over two times two or negative two.

2198
03:35:16,648 --> 03:35:24,118
To find the y coordinate of the vertex, we
just plug in the x coordinate into our formula

2199
03:35:24,119 --> 03:35:30,918
for a g of x. So that's g of negative two,
which is two times negative two squared plus

2200
03:35:30,918 --> 03:35:37,360
eight times negative two plus six. And that
works out to be negative two by coincidence.

2201
03:35:37,360 --> 03:35:44,461
So the vertex for our quadratic function has
coordinates negative two, negative two. And

2202
03:35:44,460 --> 03:35:50,429
if I want to write g of x in vertex form,
it's going to be a times x minus negative

2203
03:35:50,430 --> 03:36:00,789
two squared plus minus two. That's because
remember, we subtract H and we add K. So that

2204
03:36:00,789 --> 03:36:08,699
simplifies to g of x equals a times x plus
two squared minus two. And finally, we just

2205
03:36:08,699 --> 03:36:14,359
need to figure out what this leading coefficient
as. But notice, if we were to multiply, distribute

2206
03:36:14,359 --> 03:36:23,029
this out, then the coefficient of x squared
would end up being a. So therefore, the coefficient

2207
03:36:23,029 --> 03:36:27,520
of x squared here, which is a has to be the
same as the coefficient of x squared here,

2208
03:36:27,520 --> 03:36:33,149
which we conveniently also called a, in other
words, are a down here needs to be two. So

2209
03:36:33,148 --> 03:36:39,778
I'm going to write that as g of x equals two
times x plus two squared minus two, lots of

2210
03:36:39,779 --> 03:36:46,289
twos in this problem. And that's our quadratic
function in vertex form. If I want to check

2211
03:36:46,289 --> 03:36:53,569
my answer, of course, I could just distribute
out again, I'd get two times x squared plus

2212
03:36:53,568 --> 03:37:03,000
4x, plus two, minus two, in other words, 2x
squared plus 8x plus six, which checks out

2213
03:37:03,000 --> 03:37:09,010
to exactly what I started with. This video
showed how to get from vertex form to standard

2214
03:37:09,010 --> 03:37:14,408
form by distributing out and how to get from
standard form to vertex form by finding the

2215
03:37:14,408 --> 03:37:17,639
vertex using the vertex formula.

2216
03:37:17,639 --> 03:37:26,269
Suppose you have a quadratic function in the
form y equals x squared plus bx plus c, and

2217
03:37:26,270 --> 03:37:31,760
you want to find where the vertex is, when
you graph it.

2218
03:37:31,760 --> 03:37:40,969
The vertex formula says that the x coordinate
of this vertex is that negative B over two

2219
03:37:40,969 --> 03:37:42,879
A,

2220
03:37:42,879 --> 03:37:47,608
this video gives a justification for where
that formula comes from.

2221
03:37:47,609 --> 03:37:53,390
Let's start with a specific example. Suppose
I wanted to find the x intercepts and the

2222
03:37:53,389 --> 03:38:00,840
vertex for this quadratic function. To find
the x intercepts, I would set y equal to zero

2223
03:38:00,840 --> 03:38:07,510
and solve for x. So that's zero equals 3x
squared plus 7x minus five. And to solve for

2224
03:38:07,510 --> 03:38:13,418
x, I use the quadratic formula. So x is going
to be negative seven plus or minus the square

2225
03:38:13,418 --> 03:38:22,328
root of seven squared minus four times three
times negative five, all over two times three.

2226
03:38:22,328 --> 03:38:29,228
That simplifies to negative seven plus or
minus a squared of 109 over six, we could,

2227
03:38:29,228 --> 03:38:35,039
I could also write this as negative seven,
six, plus the square root of 109 over six,

2228
03:38:35,040 --> 03:38:41,029
or x equals negative seven, six minus the
square root of 109 over six, since the square

2229
03:38:41,029 --> 03:38:47,488
root of 109 is just a little bit bigger than
10, this can be approximated by negative seven

2230
03:38:47,488 --> 03:38:54,340
six plus 10, six, and negative seven, six
minus 10, six.

2231
03:38:54,340 --> 03:39:00,709
So pretty close to, I guess about what half
over here and pretty close to about negative

2232
03:39:00,709 --> 03:39:06,609
three over here, I'm just going to estimate
so that I can draw a picture of the function.

2233
03:39:06,609 --> 03:39:11,629
Since the leading coefficient three is positive,
I know my parabola is going to be opening

2234
03:39:11,629 --> 03:39:20,270
up and the intercepts are somewhere around
here and here. So roughly speaking, it's gonna

2235
03:39:20,270 --> 03:39:25,488
look something like this.

2236
03:39:25,488 --> 03:39:30,439
Now the vertex is going to be somewhere in
between the 2x intercepts, in fact, it's going

2237
03:39:30,439 --> 03:39:37,010
to be by symmetry, it'll be exactly halfway
in between the 2x intercepts. Since the x

2238
03:39:37,010 --> 03:39:43,260
intercepts are negative seven, six plus and
minus 100 squared of 109 over six, the number

2239
03:39:43,260 --> 03:39:47,978
halfway in between those is going to be exactly
negative seven, six,

2240
03:39:47,978 --> 03:39:56,049
right, because on the one hand, I have negative
seven six plus something. And on the other

2241
03:39:56,049 --> 03:39:59,858
hand, I have negative seven six minus that
same thing.

2242
03:39:59,859 --> 03:40:08,998
So negative seven, six will be exactly in
the middle. So my x coordinate of my vertex

2243
03:40:08,998 --> 03:40:16,670
will be at negative seven sex. Notice that
I got that number from the quadratic formula.

2244
03:40:16,670 --> 03:40:24,818
More generally, if I want to find the x intercepts
for any quadratic function, I set y equal

2245
03:40:24,818 --> 03:40:27,469
to zero

2246
03:40:27,469 --> 03:40:33,198
and solve for x using the quadratic formula,
negative b plus or minus the square root of

2247
03:40:33,199 --> 03:40:40,890
b squared minus four AC Oliver to a, b, x
intercepts will be at these two values, but

2248
03:40:40,889 --> 03:40:47,189
the x coordinate of the vertex, which is exactly
halfway in between the 2x intercepts will

2249
03:40:47,189 --> 03:40:53,710
be at negative B over two A. That's where
the vertex formula comes from.

2250
03:40:53,709 --> 03:40:58,628
And it turns out that this formula works even
when there are no x intercepts, even when

2251
03:40:58,629 --> 03:41:04,010
the quadratic formula gives us no solutions.
That vertex still has the x coordinate, negative

2252
03:41:04,010 --> 03:41:06,658
B over two A.

2253
03:41:06,658 --> 03:41:11,498
And that's the justification of the vertex
formula.

2254
03:41:11,498 --> 03:41:14,898
This video is about polynomials and their
graphs.

2255
03:41:14,898 --> 03:41:20,379
We call that a polynomial is a function like
this one, for example.

2256
03:41:20,379 --> 03:41:28,708
His terms are numbers times powers of x. I'll
start with some definitions. The degree of

2257
03:41:28,709 --> 03:41:35,899
a polynomial is the largest exponent. For
example, for this polynomial, the degree is

2258
03:41:35,898 --> 03:41:37,849
for

2259
03:41:37,850 --> 03:41:44,129
the leading term is the term with the largest
exponent. In the same example, the leading

2260
03:41:44,129 --> 03:41:51,350
term is 5x to the fourth, it's conventional
to write the polynomial in descending order

2261
03:41:51,350 --> 03:41:56,418
of powers of x. So the leading term is first.
But the leading term doesn't have to be the

2262
03:41:56,418 --> 03:42:04,498
first term. If I wrote the same polynomial
as y equals, say, two minus 17x minus 21x

2263
03:42:04,498 --> 03:42:11,658
cubed plus 5x. Fourth, the leading term would
still be the 5x to the fourth, even though

2264
03:42:11,658 --> 03:42:13,260
it was last.

2265
03:42:13,260 --> 03:42:20,029
The leading coefficient is the number in the
leading term. In this example, the leading

2266
03:42:20,029 --> 03:42:21,459
coefficient is five.

2267
03:42:21,459 --> 03:42:28,140
Finally, the constant term is the term with
no x's in it. In our same example, the constant

2268
03:42:28,139 --> 03:42:29,139
term is to

2269
03:42:29,139 --> 03:42:34,760
please pause the video for a moment and take
a look at this next example of polynomial

2270
03:42:34,760 --> 03:42:40,639
figure out what's the degree the leading term,
the leading coefficient and the constant term.

2271
03:42:40,639 --> 03:42:48,760
The degree is again, four, since that's the
highest power we have, and the leading term

2272
03:42:48,760 --> 03:42:56,340
is negative 7x. to the fourth, the leading
coefficient is negative seven, and the constant

2273
03:42:56,340 --> 03:42:58,148
term is 18.

2274
03:42:58,148 --> 03:43:04,510
In the graph of the polynomial Shown here
are the three marks points are called turning

2275
03:43:04,510 --> 03:43:12,469
points, because the polynomial turns around
and changes direction at those three points,

2276
03:43:12,469 --> 03:43:18,868
those same points can also be called local
extreme points, or they can be called local

2277
03:43:18,869 --> 03:43:25,710
maximum and minimum points. For this polynomial,
the degree is four, and the number of turning

2278
03:43:25,709 --> 03:43:28,068
points is three.

2279
03:43:28,068 --> 03:43:33,129
Let's compare the degree and the number of
turning points for these next three examples.

2280
03:43:33,129 --> 03:43:39,698
For the first one, the degree is to and there's
one turning point.

2281
03:43:39,699 --> 03:43:50,090
For the second example, that agree, is three,
and there's two turning points.

2282
03:43:50,090 --> 03:43:56,920
And for this last example, the degree is four,
and there's one turning point.

2283
03:43:56,920 --> 03:44:01,699
For this first example, and the next two,
the number of turning points is exactly one

2284
03:44:01,699 --> 03:44:07,869
less than the degree. So you might conjecture
that this is always true. But in fact, this

2285
03:44:07,869 --> 03:44:13,689
is not always true. In this last example,
the degree is four, but the number of turning

2286
03:44:13,689 --> 03:44:16,909
points is one, not three.

2287
03:44:16,909 --> 03:44:21,510
In fact, it turns out that while the number
of turning points is not always equal to a

2288
03:44:21,510 --> 03:44:28,629
minus one, it is always less than or equal
to the degree minus one. That's a useful factor.

2289
03:44:28,629 --> 03:44:33,680
Remember, when you're sketching graphs are
recognizing graphs of polynomials.

2290
03:44:33,680 --> 03:44:39,659
The end behavior of a function is how the
ends of the function look, as x gets bigger

2291
03:44:39,659 --> 03:44:45,248
and bigger heads towards infinity, or x gets
goes through larger and larger negative numbers

2292
03:44:45,248 --> 03:44:47,859
towards negative infinity.

2293
03:44:47,859 --> 03:44:54,818
In this first example, the graph of the function
goes down as x gets towards infinity as x

2294
03:44:54,818 --> 03:44:59,959
goes towards negative infinity. I can draw
this with two little arrows pointing

2295
03:44:59,959 --> 03:45:05,958
Down on either side, or I can say in words,
that the function is falling as we had left

2296
03:45:05,959 --> 03:45:09,459
and falling also as we had right.

2297
03:45:09,459 --> 03:45:16,600
In the second example, the graph rises to
the left and rises to the right. In the third

2298
03:45:16,600 --> 03:45:23,379
example, the graph falls to the left, but
rises to the right. And in the fourth example,

2299
03:45:23,379 --> 03:45:29,489
it rises to the left and falls to the right.
If you study these examples, and others, you

2300
03:45:29,488 --> 03:45:36,189
might notice there's a relationship between
the degree of the polynomial the leading coefficients

2301
03:45:36,189 --> 03:45:44,090
of the polynomials and the end behavior. Specifically,
these four types of end behavior are determined

2302
03:45:44,090 --> 03:45:51,728
by whether the degree is even or odd. And
by whether the leading coefficient is positive

2303
03:45:51,728 --> 03:45:54,878
or negative.

2304
03:45:54,879 --> 03:46:00,729
When the degree is even, and the leading coefficient
is positive, like in this example, where the

2305
03:46:00,728 --> 03:46:08,458
leading coefficient is one, we have this sorts
of n behavior rising on both sides.

2306
03:46:08,459 --> 03:46:14,069
When the degree is even at the leading coefficient
is negative, like in this example, we have

2307
03:46:14,068 --> 03:46:18,378
the end behavior that's falling on both sides

2308
03:46:18,379 --> 03:46:23,810
when the degree is odd, and the leading coefficient
is positive, that's like this example, with

2309
03:46:23,809 --> 03:46:30,699
the degree three and the leading coefficient
three, then we have this sort of end behavior.

2310
03:46:30,700 --> 03:46:35,979
And finally, when the degree is odd, and the
leading coefficient is negative, like in this

2311
03:46:35,978 --> 03:46:40,628
example, we have this sort of NBA havior.

2312
03:46:40,629 --> 03:46:48,409
I like to remember this chart just by thinking
of the most simple examples, y equals x squared,

2313
03:46:48,408 --> 03:46:55,079
y equals negative x squared, y equals x cubed,
and y equals negative x cubed. Because I know

2314
03:46:55,079 --> 03:47:00,209
by heart what those four examples look like,

2315
03:47:00,209 --> 03:47:08,009
then I just have to remember that any polynomial
with a even degree and positive leading coefficient

2316
03:47:08,010 --> 03:47:11,950
has the same end behavior as x squared.

2317
03:47:11,950 --> 03:47:18,020
And similarly, any polynomial with even degree
and negative leading coefficient has the same

2318
03:47:18,020 --> 03:47:25,729
end behavior as negative x squared. And similar
statements for x cubed and negative x cubed.

2319
03:47:25,728 --> 03:47:30,738
We can use facts about turning points and
behavior, to say something about the equation

2320
03:47:30,738 --> 03:47:34,270
of a polynomial just by looking at this graph.

2321
03:47:34,270 --> 03:47:42,699
In this example, because of the end behavior,
we know that the degree is odd,

2322
03:47:42,699 --> 03:47:46,569
we know that the leading coefficient

2323
03:47:46,568 --> 03:47:49,389
must be positive.

2324
03:47:49,389 --> 03:47:58,099
And finally, since there are 1234, turning
points, we know that the degree is greater

2325
03:47:58,100 --> 03:48:01,300
than or equal to five.

2326
03:48:01,299 --> 03:48:06,288
That's because the number of turning points
is less than or equal to the degree minus

2327
03:48:06,289 --> 03:48:10,529
one. And in this case, the number of turning
points we said was four.

2328
03:48:10,529 --> 03:48:18,238
And so solving that inequality, we get the
degree is bigger than equal to five.

2329
03:48:18,238 --> 03:48:23,129
Put in some of that information together,
we see the degree could be

2330
03:48:23,129 --> 03:48:31,059
five, or seven, or nine, or any odd number
greater than or equal to five. But it couldn't

2331
03:48:31,059 --> 03:48:38,939
be for example, three or six. Because even
numbers and and also numbers less than five

2332
03:48:38,939 --> 03:48:41,129
are right out.

2333
03:48:41,129 --> 03:48:46,500
This video gave a lot of definitions, including
the definition of degree

2334
03:48:46,500 --> 03:48:49,168
leading coefficients,

2335
03:48:49,168 --> 03:48:51,840
turning points,

2336
03:48:51,840 --> 03:48:54,930
and and behavior.

2337
03:48:54,930 --> 03:48:59,908
We saw that knowing the degree and the leading
coefficient can help you make predictions

2338
03:48:59,908 --> 03:49:06,189
about the number of turning points and the
end behavior, as well as vice versa.

2339
03:49:06,189 --> 03:49:11,068
This video is about exponential functions
and their graphs.

2340
03:49:11,068 --> 03:49:16,908
an exponential function is a function that
can be written in the form f of x equals a

2341
03:49:16,908 --> 03:49:24,420
times b to the x, where a and b are any real
numbers. As long as a is not zero, and B is

2342
03:49:24,420 --> 03:49:26,549
positive.

2343
03:49:26,549 --> 03:49:33,170
It's important to notice that for an exponential
function, the variable x is in the exponent.

2344
03:49:33,170 --> 03:49:37,609
This is different from many other functions
we've seen, for example, quite a quadratic

2345
03:49:37,609 --> 03:49:44,309
function like f of x equals 3x squared has
the variable in the base, so it's not an exponential

2346
03:49:44,309 --> 03:49:45,500
function.

2347
03:49:45,500 --> 03:49:53,010
For exponential functions, f of x equals a
times b dx. We require that A is not equal

2348
03:49:53,010 --> 03:50:00,100
to zero, because otherwise, we would have
f of x equals zero times b dx.

2349
03:50:00,100 --> 03:50:04,850
Which just means that f of x equals zero.
And this is called a constant function, not

2350
03:50:04,850 --> 03:50:11,149
an exponential function. Because f of x is
always equal to zero

2351
03:50:11,148 --> 03:50:17,389
in an exponential function, but we require
that B is bigger than zero, because otherwise,

2352
03:50:17,389 --> 03:50:27,578
for example, if b is equal to negative one,
say, we'd have f of x equals a times negative

2353
03:50:27,578 --> 03:50:34,038
one to the x. Now, this would make sense for
a lot of values of x. But if we try to do

2354
03:50:34,039 --> 03:50:40,600
something like f of one half, with our Bs,
negative one, that would be the same thing

2355
03:50:40,600 --> 03:50:45,340
as a times the square root of negative one,
which is not a real number,

2356
03:50:45,340 --> 03:50:52,158
we'd get the same problem for other values
of b, if the values of b were negative. And

2357
03:50:52,158 --> 03:50:59,299
if we tried b equals zero, we'd get a kind
of ridiculous thing a times zero to the x,

2358
03:50:59,299 --> 03:51:03,828
which again is always zero. So that wouldn't
count as an exponential either. So we can't

2359
03:51:03,828 --> 03:51:09,728
use any negative basis, and we can't use zero
basis, if we want an exponential function.

2360
03:51:09,728 --> 03:51:18,448
The number A in the expression f of x equals
a times b to the x is called the initial value.

2361
03:51:18,449 --> 03:51:21,029
And the number B is called the base.

2362
03:51:21,029 --> 03:51:28,998
The phrase initial value comes from the fact
that if we plug in x equals zero, we get a

2363
03:51:28,998 --> 03:51:34,529
times b to the zero, well, anything to the
zero is just one. So this is equal to A. In

2364
03:51:34,529 --> 03:51:40,920
other words, f of zero equals a. So if we
think of starting out,

2365
03:51:40,920 --> 03:51:44,700
when x equals zero, we get the y value

2366
03:51:44,700 --> 03:51:51,629
of a, that's why it's called the initial value.

2367
03:51:51,629 --> 03:51:59,890
Let's start out with this example, where y
equals a times b to the x next special function,

2368
03:51:59,889 --> 03:52:06,049
and we've set a equal to one and B equals
a two. Notice that the y intercept, the value

2369
03:52:06,049 --> 03:52:09,340
when x is zero, is going to be one.

2370
03:52:09,340 --> 03:52:20,318
If I change my a value, my initial value,
the y intercept changes, the function is stretched

2371
03:52:20,318 --> 03:52:21,318
out.

2372
03:52:21,318 --> 03:52:27,379
If I make the value of a go to zero, and then
negative,

2373
03:52:27,379 --> 03:52:35,219
then my initial value becomes negative, and
my graph flips across the x axis. Let's go

2374
03:52:35,219 --> 03:52:41,778
back to an a value of say one, and see what
happens when we change B. Right now, the B

2375
03:52:41,779 --> 03:52:52,869
value the basis two, if I increase B, my y
intercept sticks at one, but my graph becomes

2376
03:52:52,869 --> 03:53:03,020
steeper and steeper. If I put B back down
close to one, my graph becomes more flat at

2377
03:53:03,020 --> 03:53:05,959
exactly one, my graph is just a constant.

2378
03:53:05,959 --> 03:53:14,779
As B gets into fractional territory, point
8.7 point six, my graph starts to slope the

2379
03:53:14,779 --> 03:53:20,119
other way, it's decreasing now instead of
increasing, but notice that the y intercept

2380
03:53:20,119 --> 03:53:26,199
still hasn't changed, I can get it more and
more steep as my B gets farther and farther

2381
03:53:26,199 --> 03:53:33,550
away from one of course, when B goes to negative
territory, my graph doesn't make any sense.

2382
03:53:33,549 --> 03:53:44,418
So a changes the y intercept, and B changes
the steepness of the graph. So that it's added

2383
03:53:44,418 --> 03:53:50,949
whether it's increasing for B values bigger
than one, and decreasing for values of the

2384
03:53:50,949 --> 03:53:55,078
less than one.

2385
03:53:55,078 --> 03:53:58,978
we'll summarize all these observations on
the next slide.

2386
03:53:58,978 --> 03:54:04,568
We've seen that for an exponential function,
y equals a times b to the x, the parameter

2387
03:54:04,568 --> 03:54:13,159
or number a gives the y intercept, the parameter
B tells us how the graph is increasing or

2388
03:54:13,159 --> 03:54:19,889
decreasing. Specifically, if b is greater
than one, the graph is increasing.

2389
03:54:19,889 --> 03:54:23,868
And if b is less than one, the graph is decreasing.

2390
03:54:23,869 --> 03:54:29,459
The closer B is to the number one, the flatter
the graph.

2391
03:54:29,459 --> 03:54:37,600
So for example, if I were to graph y equals
point two five to the x and y equals point

2392
03:54:37,600 --> 03:54:44,379
four to the x, they would both be decreasing
graphs, since the base for both of them is

2393
03:54:44,379 --> 03:54:52,209
less than one. But point two five is farther
away from one and point four is closer to

2394
03:54:52,209 --> 03:54:58,560
one. So point four is going to be flatter.
And point two five is going to be more steep.

2395
03:54:58,559 --> 03:55:03,418
So in this picture,
This red graph would correspond to point two

2396
03:55:03,418 --> 03:55:11,728
five to the x, and the blue graph would correspond
to point four to the x. For all these exponential

2397
03:55:11,728 --> 03:55:17,458
functions, whether the graphs are decreasing
or increasing, they all have a horizontal

2398
03:55:17,459 --> 03:55:25,460
asymptote along the x axis. In other words,
at the line at y equals zero, the domain is

2399
03:55:25,459 --> 03:55:34,799
always from negative infinity to infinity,
and the range is always from zero to infinity,

2400
03:55:34,799 --> 03:55:43,309
because the range is always positive y values.
Actually, that's true. If a is greater than

2401
03:55:43,309 --> 03:55:52,109
zero, if a is less than zero, then our graph
flips over the x axis, our domain stays the

2402
03:55:52,109 --> 03:55:59,890
same, but our range becomes negative infinity
to zero. The most famous exponential function

2403
03:55:59,889 --> 03:56:04,868
is f of x equals e to the x. This function
is also sometimes written as f of x equals

2404
03:56:04,869 --> 03:56:13,420
x of x. The number E is Oilers number as approximately
2.7. This function has important applications

2405
03:56:13,420 --> 03:56:21,100
to calculus and to some compound interest
problems. In this video, we looked at exponential

2406
03:56:21,100 --> 03:56:28,879
functions, functions of the form a times b
to the x, where the variable is in the exponent,

2407
03:56:28,879 --> 03:56:37,720
we saw that they all have the same general
shape, either increasing like this, or decreasing

2408
03:56:37,719 --> 03:56:45,929
like this, unless a is negative, in which
case they flip over the x axis. They all have

2409
03:56:45,930 --> 03:56:55,859
a horizontal asymptote at y equals zero, the
x axis. In this video, we'll use exponential

2410
03:56:55,859 --> 03:57:02,748
functions to model real world examples. Let's
suppose you're hired for a job. The starting

2411
03:57:02,748 --> 03:57:10,039
salary is $40,000. With a guaranteed annual
raise of 3% each year, how much will your

2412
03:57:10,039 --> 03:57:16,899
salary be after one year, two years, five
years. And in general after two years, let

2413
03:57:16,898 --> 03:57:23,349
me chart out the information. The left column
will be the number of years since you're hired.

2414
03:57:23,350 --> 03:57:30,930
And the right column will be your salary.
When you start work, at zero years after you're

2415
03:57:30,930 --> 03:57:40,970
hired, your salary will be $40,000. After
one year, you'll have gotten a 3% raise. So

2416
03:57:40,969 --> 03:57:52,459
your salary will be the original 40,000 plus
3% of 40,000.03 times 40,000. I can think

2417
03:57:52,459 --> 03:58:00,528
of this first number as one at times 40,000.
And I can factor out the 40,000 from both

2418
03:58:00,529 --> 03:58:15,659
terms, to get 40,000 times one plus 0.03.
we rewrite this as 40,000 times 1.03. This

2419
03:58:15,659 --> 03:58:26,119
is your original salary multiplied by a growth
factor of 1.03. After two years, you'll get

2420
03:58:26,119 --> 03:58:33,418
a 3% raise from your previous year salary,
your previous year salary was 40,000 times

2421
03:58:33,418 --> 03:58:34,418
1.03.

2422
03:58:34,418 --> 03:58:36,668
But you'll add

2423
03:58:36,668 --> 03:58:45,559
3% of that, again, I can think of the first
number is one times 40,000 times 1.03. And

2424
03:58:45,559 --> 03:58:53,828
I can factor out the common factor of 40,000
times 1.03 from both terms, to get 40,000

2425
03:58:53,828 --> 03:59:08,920
times 1.03 times one plus 0.03. Let me rewrite
that as 40,000 times 1.03 times 1.03 or 40,000

2426
03:59:08,920 --> 03:59:19,068
times 1.3 squared. We can think of this as
your last year salary multiplied by the same

2427
03:59:19,068 --> 03:59:26,799
growth factor of 1.03. After three years,
a similar computation will give you that your

2428
03:59:26,799 --> 03:59:37,019
new salary is your previous year salary times
that growth factor of 1.03 that can be written

2429
03:59:37,020 --> 03:59:45,720
as 40,000 times 1.03 cubed. And in general,
if you're noticing the pattern, after two

2430
03:59:45,719 --> 03:59:53,760
years, your salary should be 40,000 times
1.03 to the t power. In other words, your

2431
03:59:53,760 --> 04:00:06,260
salary after two years is your original salary
multiplied By the growth factor of 1.03, taken

2432
04:00:06,260 --> 04:00:15,818
to the t power, let me write this as a formula
s of t, where S of t is your salary is equal

2433
04:00:15,818 --> 04:00:24,969
to 40,000 times 1.03 to the T. This is an
exponential function, that is a function of

2434
04:00:24,969 --> 04:00:38,760
the form a times b to the T, where your initial
value a is 40,000 and your base b is 1.03.

2435
04:00:38,760 --> 04:00:45,050
Notice that your base is the amount that your
salary gets multiplied by each year. Given

2436
04:00:45,050 --> 04:00:49,868
this formula, we can easily figure out what
your salary will be after, for example, five

2437
04:00:49,869 --> 04:01:01,279
years by plugging in five for T. I worked
that out on my calculator to be $46,370.96

2438
04:01:01,279 --> 04:01:08,459
to the nearest cent. exponential functions
are also useful in modeling population growth.

2439
04:01:08,459 --> 04:01:15,209
The United Nations estimated that the world
population in 2010 was 6.7 9 billion growing

2440
04:01:15,209 --> 04:01:20,739
at a rate of 1.1% per year. Assuming that
the growth rate stayed the same and will continue

2441
04:01:20,738 --> 04:01:26,309
to stay the same, we'll write an equation
for the population at t years after the year

2442
04:01:26,309 --> 04:01:39,109
2010 1.1%, written as a decimal is 0.011.
So if we work out a chart as before, we see

2443
04:01:39,109 --> 04:01:46,998
that after zero years since 2010, we have
our initial population 6.7 9 billion, after

2444
04:01:46,998 --> 04:01:59,680
one year, we'll take that 6.7 9 billion and
add 1.1% of it. That is point 011 times 6.79.

2445
04:01:59,680 --> 04:02:15,898
This works out to 6.79 times one plus point
011 or 6.79 times 1.011. Here we have our

2446
04:02:15,898 --> 04:02:27,599
initial population of 6.7 9 billion, and our
growth factor of 1.011. That's how much the

2447
04:02:27,600 --> 04:02:35,640
population got multiplied by in one year.
As before, we can work out that after two

2448
04:02:35,639 --> 04:02:45,010
years, our population becomes 6.79 times 1.011
squared, since they got multiplied by 1.011,

2449
04:02:45,010 --> 04:02:56,488
twice, and after two years, it'll be 6.79
times 1.011 to the t power. So our function

2450
04:02:56,488 --> 04:03:09,520
that models population is going to be 6.79
times 1.011 to the T. Here, t represents time

2451
04:03:09,520 --> 04:03:21,220
in years, since 2010. Just for fun, I'll plug
in t equals 40. That's 40 years since 2010.

2452
04:03:21,219 --> 04:03:31,408
So that's the year 2050. And I get 6.79 times
1.011 to the 40th power, which works out to

2453
04:03:31,408 --> 04:03:40,368
10 point 5 billion. That's the prediction
based on this exponential model.

2454
04:03:40,369 --> 04:03:45,629
The previous two examples were examples of
exponential growth. This last example is an

2455
04:03:45,629 --> 04:03:51,869
example of exponential decay. The drugs Seroquel
is metabolized and eliminated from the body

2456
04:03:51,869 --> 04:03:58,959
at a rate of 11% per hour. If 400 milligrams
are given how much remains in the body. 24

2457
04:03:58,959 --> 04:04:09,090
hours later, I'll chart out my information
where the left column will be time in hours

2458
04:04:09,090 --> 04:04:15,998
since the dose was given, and the right column
will be the number of milligrams of Seroquel

2459
04:04:15,998 --> 04:04:22,329
still on the body. zero hours after the dose
is given, we have the full 400 milligrams

2460
04:04:22,329 --> 04:04:30,668
in the body. One hour later, we have the formula
100 milligrams minus 11% of it, that's minus

2461
04:04:30,668 --> 04:04:40,078
point one one times 400. If I factor out the
400 from both terms, I get 400 times one minus

2462
04:04:40,078 --> 04:04:53,219
0.11 or 400 times point eight nine. The 400
represents the initial amount. The point eight

2463
04:04:53,219 --> 04:04:58,608
nine I'll call the growth factor, even though
in this case the quantity is decreasing, not

2464
04:04:58,609 --> 04:05:08,440
growing. So really it's kind of a shrink factor.
Let's see what happens after two hours. Now

2465
04:05:08,440 --> 04:05:15,550
I'll have 400 times 0.89 my previous amount,
it'll again get multiplied by point eight,

2466
04:05:15,549 --> 04:05:22,469
nine, so that's going to be 400 times 0.89
squared. And in general, after t hours, I'll

2467
04:05:22,469 --> 04:05:29,528
have 400 multiplied by this growth or shrinkage
factor of point eight nine, raised to the

2468
04:05:29,529 --> 04:05:36,270
t power. Since each hour, the amount of Seroquel
gets multiplied by point eight, nine, that

2469
04:05:36,270 --> 04:05:44,289
number less than one. All right, my exponential
decay function as f of t equals 400 times

2470
04:05:44,289 --> 04:05:54,560
0.89 to the T, where f of t represents the
number of milligrams of Seroquel in the body.

2471
04:05:54,559 --> 04:06:01,958
And t represents the number of hours since
the dose was given. To find out how much is

2472
04:06:01,959 --> 04:06:13,239
in the body after 24 hours, I just plug in
24 for t. This works out to about 24.4 milligrams,

2473
04:06:13,238 --> 04:06:17,788
I hope you notice the common form for the
functions used to model all three of these

2474
04:06:17,789 --> 04:06:26,418
examples. The functions are always in the
form f of t equals a times b to the T, where

2475
04:06:26,418 --> 04:06:36,449
a represented the initial amount, and B represented
the growth factor. To find the growth factor

2476
04:06:36,449 --> 04:06:45,449
B, we started with the percent increase or
decrease. We wrote it as a decimal. And then

2477
04:06:45,449 --> 04:06:50,800
we either added or subtracted it from one,
depending on whether the quantity was increasing

2478
04:06:50,799 --> 04:06:58,738
or decreasing. Let me show you that as a couple
examples. In the first example, we had a percent

2479
04:06:58,738 --> 04:07:09,010
increase of 3% on the race as a decimal, I'll
call this r, this was point O three. And to

2480
04:07:09,010 --> 04:07:16,619
get the growth factor, we added that to one
to get 1.03. In the world population example,

2481
04:07:16,619 --> 04:07:29,640
we had a 1.1% increase, we wrote this as point
011 and added it to one to get 1.011. In the

2482
04:07:29,639 --> 04:07:42,349
drug example, we had a decrease of 11%. We
wrote that as a decimal. And we subtracted

2483
04:07:42,350 --> 04:07:50,988
the point one one from one to get 0.89. In
fact, you can always write the growth factor

2484
04:07:50,988 --> 04:07:58,838
B as one plus the percent change written as
a decimal. If you're careful to make that

2485
04:07:58,838 --> 04:08:04,539
percent change, negative when the quantity
is decreasing and positive when the quantity

2486
04:08:04,539 --> 04:08:09,189
is increasing. Since here one plus negative
point 11

2487
04:08:09,189 --> 04:08:14,210
gives us the correct growth factor of point
eight nine which is less than one as a good

2488
04:08:14,209 --> 04:08:23,599
check. Remember that if your quantity is increasing,
the base b should be bigger than one. And

2489
04:08:23,600 --> 04:08:30,529
if the quantity is decreasing, then B should
be less than one.

2490
04:08:30,529 --> 04:08:35,129
exponential functions can also be used to
model bank accounts and loans with compound

2491
04:08:35,129 --> 04:08:42,229
interest, as we'll see in another video. This
video is about interpreting exponential functions.

2492
04:08:42,228 --> 04:08:49,728
An antique car is worth $50,000 now, and its
value increases by 7%. Each year, let's write

2493
04:08:49,728 --> 04:09:00,448
an equation to model its value x years from
now. After one year, it's value is the 50,000

2494
04:09:00,449 --> 04:09:10,459
plus 0.07 times the 50,000. That's because
its value has grown by 7% or point oh seven

2495
04:09:10,459 --> 04:09:20,199
times 50,000. this can be written as 50,000
times one plus point O seven. Notice that

2496
04:09:20,199 --> 04:09:28,359
adding 7% to the original value is the same
as multiplying the original value by one plus

2497
04:09:28,359 --> 04:09:41,770
point oh seven or by 1.07. After two years,
the value will be 50,000 times one plus 0.07

2498
04:09:41,770 --> 04:09:51,748
squared, or 50,000 times 1.07 squared. That's
because the previous year's value is multiplied

2499
04:09:51,748 --> 04:10:06,408
again by 1.7. In general, after x years We
have x, the antique cars value will be 50,000

2500
04:10:06,408 --> 04:10:16,728
times 1.07 to the x. That's because the original
value of 50,000 gets multiplied by 1.07x times

2501
04:10:16,728 --> 04:10:24,019
one time for each year. If we dissect this
equation, we see that the number of 50,000

2502
04:10:24,020 --> 04:10:29,709
comes from the original value of the car.
The one point is seven, which I call the growth

2503
04:10:29,709 --> 04:10:41,949
factor comes from one plus point 707. The
point O seven being the, the percent increase

2504
04:10:41,949 --> 04:10:50,010
written as a decimal. So the form of this
equation is an exponential equation, V of

2505
04:10:50,010 --> 04:10:57,738
x equals a times b to the x, where A is the
initial value, and B is the growth factor.

2506
04:10:57,738 --> 04:11:06,478
But we could also write this as a times one
plus r to the x, where A is still the initial

2507
04:11:06,478 --> 04:11:07,478
value,

2508
04:11:07,478 --> 04:11:08,478
but

2509
04:11:08,478 --> 04:11:15,709
R is the percent increase written as a decimal.
This same equation will come up in the next

2510
04:11:15,709 --> 04:11:22,559
example, here, my Toyota Prius is worth only
3000. Now, and its value is decreasing by

2511
04:11:22,559 --> 04:11:34,889
5% each year. So after one year, its value
will be 3000 minus 0.05 times 3000. That is

2512
04:11:34,889 --> 04:11:48,738
3000 times one minus 0.05. I can also write
that as 3000 times 0.95. Decreasing the value

2513
04:11:48,738 --> 04:12:00,020
by 5% is like multiplying the value by one
minus point oh five, or by point nine, five.

2514
04:12:00,020 --> 04:12:08,658
After two years, the value will be multiplied
by point nine five again. So the value will

2515
04:12:08,658 --> 04:12:17,908
be 3000 times point nine, five squared. And
after x years, the value will be 3000 times

2516
04:12:17,908 --> 04:12:26,998
point nine five to the x. So my equation for
the value is 3000 times point nine five to

2517
04:12:26,998 --> 04:12:36,408
the x. This is again, an equation of the form
V of x equals a times b to the x, where here,

2518
04:12:36,408 --> 04:12:47,588
a is 3000, the initial value, and B is point
nine, five, I'll still call that the growth

2519
04:12:47,588 --> 04:12:53,738
factor, even though we're actually declining
value not growing. Now remember where this

2520
04:12:53,738 --> 04:13:01,788
point nine five came from, it came from taking
one and subtracting point 05, because of the

2521
04:13:01,789 --> 04:13:09,129
5%, decrease in value, so I can again, write
my equation in the form a times well, this

2522
04:13:09,129 --> 04:13:19,829
time times one minus r to the x, where R is
our point 05. That's our percent decrease

2523
04:13:19,828 --> 04:13:26,920
written as a decimal. Please take a moment
to study this equation and the previous one.

2524
04:13:26,920 --> 04:13:35,408
They say that when you have an exponential
function, the number here is the initial value.

2525
04:13:35,408 --> 04:13:42,088
If it's written in this form, B is your growth
factor. But you can think of B as being either

2526
04:13:42,088 --> 04:13:50,510
one minus r, where r is the percent decrease,
or one plus r, where r is the percent increase.

2527
04:13:50,510 --> 04:13:55,898
In this example, we're given a function f
of x to model the number of bacteria in a

2528
04:13:55,898 --> 04:14:03,510
petri dish x hours after 12 o'clock noon,
we want to know what was the number of bacteria

2529
04:14:03,510 --> 04:14:09,168
at noon, and by what percent, the number of
bacteria is increasing every hour, we can

2530
04:14:09,168 --> 04:14:14,408
see from the equation that the number of bacteria
is increasing and not decreasing, because

2531
04:14:14,408 --> 04:14:21,458
the base of the exponential function 1.45
is bigger than one. Notice that our equation

2532
04:14:21,459 --> 04:14:30,189
f of x equals 12 times 1.45 to the X has the
form of a times b to the x, or we can think

2533
04:14:30,189 --> 04:14:43,389
of it as a times one plus r to the x. Here
a is 12, b is 1.45 and r is 0.45. Based on

2534
04:14:43,389 --> 04:14:49,158
this familiar form, we can recognize that
the initial amount of bacteria is going to

2535
04:14:49,158 --> 04:15:00,949
be 12 12,000. Since those are our units and
this value 1.45 is our growth factor. What

2536
04:15:00,949 --> 04:15:08,859
the number of bacteria is multiplied by each
hour? Well are the point four five is the

2537
04:15:08,859 --> 04:15:18,529
the rate of increase, in other words, a 45%
increase each hour. So our answers to the

2538
04:15:18,529 --> 04:15:29,359
questions are 12,045%. In this example, the
population of salamanders is modeled by this

2539
04:15:29,359 --> 04:15:37,529
exponential function, where x is the number
of years since 2015. Notice that the number

2540
04:15:37,529 --> 04:15:42,850
of salamanders is decreasing, because the
base of our exponential function point seven,

2541
04:15:42,850 --> 04:15:50,100
eight is less than one. So if we recognize
the form of our exponential function, a times

2542
04:15:50,100 --> 04:15:59,579
b to the x, or we can think of this as a times
one minus r to the x, where A is our initial

2543
04:15:59,579 --> 04:16:08,998
value, and r is our percent decrease written
as a decimal.

2544
04:16:08,998 --> 04:16:18,779
Our initial value is 3000. So that's the number
of salamanders zero years after 2015. Our

2545
04:16:18,779 --> 04:16:27,529
growth factor B is 0.78. But if I write that
as one minus r, I see that R has to be one

2546
04:16:27,529 --> 04:16:40,720
minus 0.78, or 0.22. In other words, our population
is decreasing by 22% each year. In this video,

2547
04:16:40,719 --> 04:16:47,028
we saw that exponential functions can be written
in the form f of x equals a times b to the

2548
04:16:47,029 --> 04:16:58,050
x, where A is the initial value. And B is
the growth factor. We also saw that they can

2549
04:16:58,049 --> 04:17:05,878
be written in the form a times one plus r
to the x when the amount is increasing. And

2550
04:17:05,879 --> 04:17:16,609
as a times one minus r to the x when the amount
is decreasing. In this format, R is the percent

2551
04:17:16,609 --> 04:17:28,350
increase or the percent decrease written as
a decimal. So 15% increase will be an R value

2552
04:17:28,350 --> 04:17:43,690
of 0.15 and a growth factor B of 1.15. Whereas
a 12% decrease will be an R value of point

2553
04:17:43,690 --> 04:17:55,779
one, two, and a B value of one minus point
one, two, or 0.7. Sorry, 0.88. These observations

2554
04:17:55,779 --> 04:18:03,439
help us quickly interpret exponential functions.
For example, here, we have an initial value

2555
04:18:03,439 --> 04:18:17,889
of 100 and a 15% increase. And here, we have
an initial value of 50 and a 40%. decrease.

2556
04:18:17,889 --> 04:18:23,118
exponential functions can be used to model
compound interest for loans and bank accounts.

2557
04:18:23,119 --> 04:18:30,291
Suppose you invest $200 in a bank account
that earns 3% interest every year. If you

2558
04:18:30,290 --> 04:18:37,418
make no deposits or withdrawals, how much
money will you have accumulated after 10 years,

2559
04:18:37,418 --> 04:18:41,939
because 3% of the money that's in the bank
is getting added each year, the money in the

2560
04:18:41,939 --> 04:18:51,359
bank gets multiplied by 1.03 each year. So
after one year, the amount will be 200 times

2561
04:18:51,359 --> 04:19:02,180
1.3, after two years 200 times one point O
three squared, and and after t years 200 times

2562
04:19:02,180 --> 04:19:08,479
one point O three to the t power. So the function
modeling the amount of money, I'll call it

2563
04:19:08,478 --> 04:19:20,418
P of t is given by 200 times 1.03 to the T.
More generally, if you invest a dollars

2564
04:19:20,418 --> 04:19:32,628
at an annual interest rate of our for t years.
In the end, you'll have P of t equals a times

2565
04:19:32,629 --> 04:19:37,079
one plus r to the T

2566
04:19:37,078 --> 04:19:45,039
here r needs to be written as a decimal, so
0.03 In our example, for the 3% annual interest

2567
04:19:45,040 --> 04:19:51,909
rate. Going back to our specific example,
After 10 years, the amount of money is going

2568
04:19:51,908 --> 04:20:03,038
to be P of 10 which is 200 times 1.03 to the
10 which works out to 206 $68.78 to the nearest

2569
04:20:03,039 --> 04:20:10,239
cent. In this problem, we've assumed that
the interest accumulates once per year. But

2570
04:20:10,238 --> 04:20:14,510
in the next few examples, we'll see what happens
when the interest rate accumulates more frequently,

2571
04:20:14,510 --> 04:20:23,079
twice a year, or every month. For example,
let's deposit $300 in an account that earns

2572
04:20:23,079 --> 04:20:31,059
4.5% annual interest compounded semi annually,
this means two times a year or every six months.

2573
04:20:31,059 --> 04:20:40,748
A 4.5% annual interest rate compounded two
times a year means that we're actually getting

2574
04:20:40,748 --> 04:20:53,039
4.5 over 2% interest, every time the interest
is compounded. That is every six months. That's

2575
04:20:53,040 --> 04:21:05,890
2.25% interest every half a year. Note that
2.25% is the same as 0.02 to five as a decimal.

2576
04:21:05,889 --> 04:21:17,128
So every time we earn interest, our money
gets multiplied by 1.02 to five. Let me make

2577
04:21:17,129 --> 04:21:25,399
a chart of what happens. After zero years,
which is also zero half years, we have our

2578
04:21:25,398 --> 04:21:33,760
original $300. for half a year, that's one
half year, our money gets earns interest one

2579
04:21:33,760 --> 04:21:42,078
time, so we multiply the 300 by 1.02 to five,
after one year, that's two half years, our

2580
04:21:42,078 --> 04:21:52,770
money earns interest two times. So we multiply
300 by 1.0225 squared. continuing this way,

2581
04:21:52,770 --> 04:22:01,680
after 1.5 years, that's three half years,
we have 300 times 1.02 to five cubed. And

2582
04:22:01,680 --> 04:22:09,750
after two years or four half years, we have
300 times 1.02 to five to the fourth. In general,

2583
04:22:09,750 --> 04:22:20,158
after t years, which is to T half years, our
money will grow to 300 times 1.02 to five

2584
04:22:20,158 --> 04:22:28,299
to the two t power. Because we've compounded
interest to tee times our formula for our

2585
04:22:28,299 --> 04:22:35,328
amount of money is P of t equals 300 times
one point O two to five to the two t where

2586
04:22:35,328 --> 04:22:45,639
t is the number of years. To finish the problem,
after seven years, we'll have P of $7, which

2587
04:22:45,639 --> 04:22:59,849
is 300 times 1.02 to five to the two times
seven or 14 power. And that works out to $409.65

2588
04:22:59,850 --> 04:23:06,470
to the nearest cent. In this next example,
we're going to take out a loan for $1,200

2589
04:23:06,469 --> 04:23:14,719
in annual interest rate of 6% compounded monthly.
Although loans and bank accounts might feel

2590
04:23:14,719 --> 04:23:19,130
different, they're mathematically the same.
It's like from the bank's point of view, they're

2591
04:23:19,130 --> 04:23:25,278
investing money in you and getting interest
on their money from you. So we can work them

2592
04:23:25,279 --> 04:23:36,050
out with the same kind of math 6% annual interest
rate compounded monthly means you're compounding

2593
04:23:36,049 --> 04:23:42,250
12 times a year. So each time you compound
interest, you're just going to get six over

2594
04:23:42,250 --> 04:23:56,148
12% interest. That's point 5% interest. And
as a decimal, that's 0.005. Let's try it out

2595
04:23:56,148 --> 04:24:05,689
again, what happens. Time is zero, of course,
you'll have the original loan amount of 1200.

2596
04:24:05,689 --> 04:24:15,809
After one year, that's 12 months, your loan
has had interest added to it 12 times so it

2597
04:24:15,809 --> 04:24:22,510
gets multiplied by 1.05 to the 12th.

2598
04:24:22,510 --> 04:24:30,648
After two years, that's 24 months, it's had
interest added to it 24 times, so it gets

2599
04:24:30,648 --> 04:24:39,108
multiplied by 1.05 to the 24th power. Similarly,
after three years or 36 months, your loan

2600
04:24:39,109 --> 04:24:49,890
amount will be 1200 times 1.05 to the 36th
power. And in general, after two years, that's

2601
04:24:49,889 --> 04:24:59,939
12 t months. So the interest will be compounded
12 t times and so we have to raise the 1.05

2602
04:24:59,939 --> 04:25:08,120
To the 12 t power. This gives us the general
formula for the money owed is P of t equals

2603
04:25:08,120 --> 04:25:20,399
1200 times 1.05 to the 12 T, where T is the
number of years. In particular, after three

2604
04:25:20,398 --> 04:25:31,398
years, we'll have to pay back a total of 1200
times 1.05 to the 12 times three or 36 power,

2605
04:25:31,398 --> 04:25:42,799
which works out to $1,436.02 to the nearest
cent. These last two problems follow a general

2606
04:25:42,799 --> 04:25:51,938
pattern. If A is the initial amount of the
loan or bank account, and r is the annual

2607
04:25:51,939 --> 04:26:03,529
interest rate, compounded n times per year,
then our formula for the amount of money is

2608
04:26:03,529 --> 04:26:13,620
going to be a times one plus r over n to the
n t. This formula exactly matches what we

2609
04:26:13,620 --> 04:26:20,680
did in this problem. First, we took the interest
rate our which was 6%, and divided it by the

2610
04:26:20,680 --> 04:26:25,838
number of compounding periods each year 12.
We wrote that as a decimal and added one to

2611
04:26:25,838 --> 04:26:35,059
it. That's where we got the 1.05 from. We
raised this to not to the number of years,

2612
04:26:35,059 --> 04:26:39,818
but to 12 times the number of years. That's
the number of compounding periods per year,

2613
04:26:39,818 --> 04:26:46,628
times the number of years. And we multiplied
all that by the initial amount of money, which

2614
04:26:46,629 --> 04:26:52,979
was 1200. This formula for compound interest
is a good one to memorize. But it's also important

2615
04:26:52,978 --> 04:26:57,408
to be able to reason your way through it,
like we did in this chart. There's one more

2616
04:26:57,408 --> 04:27:02,908
type of compound interest. And that's interest
compounded continuously. You can think of

2617
04:27:02,908 --> 04:27:09,950
continuous compounding as the limit of compounding
more and more frequently 10 times a year,

2618
04:27:09,950 --> 04:27:14,969
100 times a year 1000 times a year a million
times a year. In the limit, you get continuous

2619
04:27:14,969 --> 04:27:24,809
compounding. The formula for continuous compounding
is P of t equals a times e to the RT, where

2620
04:27:24,809 --> 04:27:37,269
p of t is the amount of money, t is the time
in years. A is the initial amount of money.

2621
04:27:37,270 --> 04:27:49,120
And R is the annual interest rate written
as a decimal. So 0.025 in this problem from

2622
04:27:49,120 --> 04:27:57,000
the 2.5% annual interest rate. He represents
the famous constant Oilers constant, which

2623
04:27:57,000 --> 04:28:12,059
is about 2.718. So in this problem, we have
P of t is 4000 times e to the 0.025 T. And

2624
04:28:12,059 --> 04:28:21,549
after five years, we'll have P of five, which
is 4000 times e to the 0.0 to five times five,

2625
04:28:21,549 --> 04:28:32,228
which works out to $4,532.59 to the nearest
cent. To summarize, if R represents the annual

2626
04:28:32,228 --> 04:28:43,398
interest rate written as a decimal, that is
2% would be 0.02. And t represents the number

2627
04:28:43,398 --> 04:28:50,719
of years and a represents the initial amount
of money, then for just simple annual interest

2628
04:28:50,719 --> 04:29:00,608
compounded once a year. Our formula is P of
t is a times one plus r to the T for compound

2629
04:29:00,609 --> 04:29:08,309
interest compounded n times per year. Our
formula is P of t is a times one plus r over

2630
04:29:08,309 --> 04:29:16,930
n to the n T. And for compound interest compounded
continuously, we get P of t is a times e to

2631
04:29:16,930 --> 04:29:18,430
the RT.

2632
04:29:18,430 --> 04:29:24,670
In this video, we looked at three kinds of
compound interest problems, simple annual

2633
04:29:24,670 --> 04:29:33,908
interest, interest compounded and times per
year, and continuously compounded interest.

2634
04:29:33,908 --> 04:29:42,118
This video introduces logarithms. logarithms
are a way of writing exponents. The expression

2635
04:29:42,119 --> 04:29:52,949
log base a of B equals c means that a to the
C equals b. In other words, log base a of

2636
04:29:52,949 --> 04:30:02,119
B is the exponent that you raise a to to get
BE THE NUMBER A It is called the base of the

2637
04:30:02,119 --> 04:30:07,489
logarithm. It's also called the base when
we write it in this exponential form. Some

2638
04:30:07,488 --> 04:30:14,139
students find it helpful to remember this
relationship. log base a of B equals c means

2639
04:30:14,139 --> 04:30:17,778
a to the C equals b, by drawing arrows,

2640
04:30:17,779 --> 04:30:20,510
a to the C equals b.

2641
04:30:20,510 --> 04:30:30,779
Other students like to think of it in terms
of asking a question, log base a of B, asks,

2642
04:30:30,779 --> 04:30:41,439
What power do you raise a to in order to get
b? Let's look at some examples. log base two

2643
04:30:41,439 --> 04:30:50,408
of eight is three, because two to the three
equals eight. In general, log base two of

2644
04:30:50,408 --> 04:30:59,359
y is asking you the question, What power do
you have to raise to to to get y? So for example,

2645
04:30:59,359 --> 04:31:08,130
log base two of 16 is four, because it's asking
you the question to what power equals 16?

2646
04:31:08,129 --> 04:31:15,128
And the answer is four. Please pause the video
and try some of these other examples. log

2647
04:31:15,129 --> 04:31:22,079
base two of two is asking, What power do you
raise to two to get two? And the answer is

2648
04:31:22,079 --> 04:31:34,369
one. Two to the one equals two. log base two
of one half is asking two to what power gives

2649
04:31:34,369 --> 04:31:40,790
you one half? Well, to get one half, you need
to raise two to a negative power. So that

2650
04:31:40,790 --> 04:31:49,699
would be two to the negative one. So the answer
is negative one. log base two of 1/8 means

2651
04:31:49,699 --> 04:31:58,658
what power do we raise to two in order to
get 1/8. Since 1/8, is one over two cubed,

2652
04:31:58,658 --> 04:32:06,418
we have to raise two to the negative three
power to get one over two cubed. So our exponent

2653
04:32:06,418 --> 04:32:13,689
is negative three. And that's our answer to
our log expression. Finally, log base two

2654
04:32:13,689 --> 04:32:21,979
of one is asking to what power equals one.
Well, anything raised to the zero power gives

2655
04:32:21,978 --> 04:32:28,918
us one, so this log expression evaluates to
zero. Notice that we can get positive negative

2656
04:32:28,918 --> 04:32:36,299
and zero answers for our logarithm expressions.
Please pause the video and figure out what

2657
04:32:36,299 --> 04:32:46,929
these logs evaluate to. to work out log base
10 of a million. Notice that a million is

2658
04:32:46,930 --> 04:32:54,078
10 to the sixth power. Now we're asking the
question, What power do we raise tend to to

2659
04:32:54,078 --> 04:32:59,748
get a million? So that is what power do we
raise 10 to to get 10 to the six? Well, of

2660
04:32:59,748 --> 04:33:11,030
course, the answer is going to be six. Similarly,
since point O one is 10 to the minus three,

2661
04:33:11,030 --> 04:33:16,271
this log expression is the same thing as asking,
what's the log base 10 of 10 to the minus

2662
04:33:16,271 --> 04:33:22,859
three? Well, what power do you have to raise
10? to to get 10 to the minus three? Of course,

2663
04:33:22,859 --> 04:33:31,020
the answer is negative three. Log base 10
of zero is asking, What power do we raise

2664
04:33:31,020 --> 04:33:38,189
10 to to get zero. If you think about it,
there's no way to raise 10 to an exponent

2665
04:33:38,188 --> 04:33:43,468
get zero. Raising 10 to a positive exponent
gets us really big positive numbers. Raising

2666
04:33:43,469 --> 04:33:49,840
10 to a negative exponent is like one over
10 to a power that's giving us tiny fractions,

2667
04:33:49,840 --> 04:33:55,000
but they're still positive numbers, we're
never going to get zero. Even if we raise

2668
04:33:55,000 --> 04:33:59,919
10 to the zero power, we'll just get one.
So there's no way to get zero and the log

2669
04:33:59,919 --> 04:34:05,599
base 10 of zero does not exist. If you try
it on your calculator using the log base 10

2670
04:34:05,599 --> 04:34:11,159
button, you'll get an error message. Same
thing happens when we do log base 10 of negative

2671
04:34:11,159 --> 04:34:17,109
100. We're asking 10 to what power equals
negative 100. And there's no exponent that

2672
04:34:17,109 --> 04:34:23,170
will work. And more generally, it's possible
to take the log of numbers that are greater

2673
04:34:23,169 --> 04:34:29,929
than zero, but not for numbers that are less
than or equal to zero. In other words, the

2674
04:34:29,930 --> 04:34:36,309
domain of the function log base a of x, no
matter what base you're using for a, the domain

2675
04:34:36,309 --> 04:34:44,131
is going to be all positive numbers. A few
notes on notation. When you see ln of x, that's

2676
04:34:44,131 --> 04:34:50,729
called natural log, and it means the log base
e of x where he is that famous number that's

2677
04:34:50,729 --> 04:34:59,041
about 2.718. When you see log of x with no
base at all, by convention, that means log

2678
04:34:59,041 --> 04:35:05,770
base 10 of x And it's called the common log.
Most scientific calculators have buttons for

2679
04:35:05,770 --> 04:35:13,449
natural log, and for common log. Let's practice
rewriting expressions with logs in them. log

2680
04:35:13,449 --> 04:35:20,949
base three of one nine is negative two, can
be rewritten as the expression three to the

2681
04:35:20,949 --> 04:35:26,100
negative two equals 1/9.

2682
04:35:26,099 --> 04:35:34,750
Log of 13 is shorthand for log base 10 of
13. So that can be rewritten as 10 to the

2683
04:35:34,750 --> 04:35:45,760
1.11394 equals 13. Finally, in this last expression,
ln means natural log, or log base e, so I

2684
04:35:45,760 --> 04:35:52,878
can rewrite this equation as log base e of
whenever E equals negative one. Well, that

2685
04:35:52,879 --> 04:36:00,898
means the same thing as e to the negative
one equals one over e, which is true. Now

2686
04:36:00,898 --> 04:36:05,559
let's go the opposite direction. We'll start
with exponential equations and rewrite them

2687
04:36:05,559 --> 04:36:14,600
as logs. Remember that log base a of B equals
c means the same thing as a to the C equals

2688
04:36:14,599 --> 04:36:23,979
b, the base stays the same in both expressions.
So for this example, the base of three in

2689
04:36:23,979 --> 04:36:29,317
the exponential equation, that's going to
be the same as the base in our log. Now I

2690
04:36:29,317 --> 04:36:33,438
just have to figure out what's in the argument
of the log. And what goes on the other side

2691
04:36:33,438 --> 04:36:40,519
of the equal sign. Remember that the answer
to a log is an exponent. So the thing that

2692
04:36:40,520 --> 04:36:48,067
goes in this box should be my exponent for
my exponential equation. In other words, you.

2693
04:36:48,067 --> 04:36:57,919
And I'll put the 9.78 as the argument of my
log. This works, because log base three of

2694
04:36:57,919 --> 04:37:06,108
9.78 equals view means the same thing as three
to the U equals 9.78, which is just what we

2695
04:37:06,109 --> 04:37:13,831
started with. In the second example, the base
of my exponential equation is E. So the base

2696
04:37:13,830 --> 04:37:22,051
of my log is going to be the answer to my
log is an exponent. In this case, the exponent

2697
04:37:22,051 --> 04:37:32,430
3x plus seven. And the other expression, the
four minus y becomes my argument of my log.

2698
04:37:32,430 --> 04:37:41,638
Let me check, log base e of four minus y equals
3x plus seven means e to the 3x plus seven

2699
04:37:41,638 --> 04:37:47,159
equals four minus y, which is just what I
started with. I can also rewrite log base

2700
04:37:47,159 --> 04:37:56,409
e as natural log. This video introduced the
idea of logs, and the fact that log base a

2701
04:37:56,409 --> 04:38:07,618
of B equal c means the same thing as a to
the C equals b. So log base a of B is asking

2702
04:38:07,618 --> 04:38:16,669
you the question, What power exponent Do you
raise a to in order to get b. In this video,

2703
04:38:16,669 --> 04:38:22,968
we'll work out the graph, so some log functions
and also talk about their domains. For this

2704
04:38:22,969 --> 04:38:29,240
first example, let's graph a log function
by hand by plotting some points. The function

2705
04:38:29,240 --> 04:38:36,648
we're working with is y equals log base two
of x, I'll make a chart of x and y values.

2706
04:38:36,648 --> 04:38:41,229
Since we're working this out by hand, I want
to pick x values for which it's easy to compute

2707
04:38:41,229 --> 04:38:47,879
log base two of x. So I'll start out with
the x value of one because log base two of

2708
04:38:47,879 --> 04:38:56,789
one is zero, log base anything of one is 02
is another x value that's easy to compute

2709
04:38:56,789 --> 04:39:03,809
log base two of two, that's asking, What power
do I raise to two to get to one? And the answer

2710
04:39:03,809 --> 04:39:11,969
is one. Power other powers of two are easy
to work with. So for example, log base two

2711
04:39:11,969 --> 04:39:18,340
of four that saying what power do I raise
to two to get four, so the answer is two.

2712
04:39:18,340 --> 04:39:26,349
Similarly, log base two of eight is three
and log base two of 16 is four. Let me also

2713
04:39:26,349 --> 04:39:33,919
work with some fractional values for X. If
x is one half, then log base two of one half

2714
04:39:33,919 --> 04:39:39,329
that saying what power do I raise to two to
get one half? Well, that needs a power of

2715
04:39:39,330 --> 04:39:48,430
negative one. It's also easy to compute by
hand, the log base two of 1/4 and 1/8. log

2716
04:39:48,430 --> 04:39:57,740
base two of 1/4 is negative two since two
to the negative two is 1/4. And similarly,

2717
04:39:57,740 --> 04:40:04,850
log base two of one eight is negative f3 I'll
put some tick marks on my x and y axes. Please

2718
04:40:04,849 --> 04:40:10,817
pause the video and take a moment to plot
these points. Let's see, I have the point,

2719
04:40:10,817 --> 04:40:17,579
one zero, that's here to one that's here,
for two,

2720
04:40:17,580 --> 04:40:26,190
that is here, and then eight, three, which
is here. And the fractional x values, one

2721
04:40:26,189 --> 04:40:33,457
half goes with negative one, and 1/4 with
negative two 1/8 with negative three. And

2722
04:40:33,457 --> 04:40:39,739
if I connect the dots, I get a graph that
looks something like this. If I had smaller

2723
04:40:39,740 --> 04:40:44,700
and smaller fractions, I would keep getting
more and more negative answers when I took

2724
04:40:44,700 --> 04:40:51,200
log base two of them, so my graph is getting
more and more negative, my y values are getting

2725
04:40:51,200 --> 04:40:56,920
more and more negative, as x is getting close
to zero. Now I didn't draw any parts of the

2726
04:40:56,919 --> 04:41:02,189
graph over here with negative X values, I
didn't put any negative X values in my chart,

2727
04:41:02,189 --> 04:41:08,599
that omission is no accident. Because if you
try to take the log base two or base anything

2728
04:41:08,599 --> 04:41:15,829
of a negative number, like say negative four
or something, there's no answer. This doesn't

2729
04:41:15,830 --> 04:41:24,951
exist because there's no power that you can
raise to to to get a negative number. So there

2730
04:41:24,951 --> 04:41:29,360
are no points on the graph for negative X
values. And similarly, there are no points

2731
04:41:29,360 --> 04:41:35,549
on the graph where x is zero, because you
can't take log base two of zero, there's no

2732
04:41:35,549 --> 04:41:42,378
power you can raise to two to get zero. I
want to observe some key features of this

2733
04:41:42,378 --> 04:41:51,297
graph. First of all, the domain is x values
greater than zero. In interval notation, I

2734
04:41:51,297 --> 04:41:58,029
can write that as a round bracket because
I don't want to include zero to infinity,

2735
04:41:58,029 --> 04:42:05,039
the range is going to be the y values, while
they go all the way down into the far reaches

2736
04:42:05,040 --> 04:42:11,069
of the negative numbers. And the graph gradually
increases y value is getting bigger and bigger.

2737
04:42:11,069 --> 04:42:17,279
So the range is actually all real numbers
are an interval notation negative infinity

2738
04:42:17,279 --> 04:42:25,590
to infinity. Finally, I want to point out
that this graph has a vertical asymptote at

2739
04:42:25,590 --> 04:42:35,887
the y axis, that is at the line x equals zero.
I'll draw that on my graph with a dotted line.

2740
04:42:35,887 --> 04:42:42,569
A vertical asymptote is a line that our functions
graph gets closer and closer to. So this is

2741
04:42:42,569 --> 04:42:50,189
the graph of y equals log base two of x. But
if I wanted to graph say, y equals log base

2742
04:42:50,189 --> 04:42:57,250
10 of x, it would look very similar, it would
still have a domain of X values greater than

2743
04:42:57,250 --> 04:43:02,878
zero, a range of all real numbers and a vertical
asymptote at the y axis, it will still go

2744
04:43:02,878 --> 04:43:10,360
through the point one zero, but it would go
through the point 10 one instead, because

2745
04:43:10,360 --> 04:43:20,159
log base 10 of 10 is one, it would look pretty
much the same, just a lot flatter over here.

2746
04:43:20,159 --> 04:43:24,930
But even though it doesn't look like it with
the way I've drawn it, it's still gradually

2747
04:43:24,930 --> 04:43:34,950
goes up to n towards infinity. In fact, the
graph of y equals log base neaa of X for a

2748
04:43:34,950 --> 04:43:42,387
bigger than one looks pretty much the same,
and has the same three properties. Now that

2749
04:43:42,387 --> 04:43:46,969
we know what the basic log graph looks like,
we can plot at least rough graphs of other

2750
04:43:46,970 --> 04:43:52,968
log functions without plotting points. Here
we have the graph of natural log of X plus

2751
04:43:52,968 --> 04:43:57,750
five. And again, I'm just going to draw a
rough graph. If I did want to do a more accurate

2752
04:43:57,750 --> 04:44:02,477
graph, I probably would want to plot some
points. But I know that roughly a log graph,

2753
04:44:02,477 --> 04:44:09,930
if it was just like y equals ln of x, that
would look something like this, and it would

2754
04:44:09,930 --> 04:44:17,271
go through the point one zero, with a vertical
asymptote along the y axis. Now if I want

2755
04:44:17,271 --> 04:44:23,389
a graph, ln of x plus five, that just shifts
our graph by five units, it'll still have

2756
04:44:23,389 --> 04:44:27,450
the same vertical asymptote. Since the vertical
line shifted up by five units, there's still

2757
04:44:27,450 --> 04:44:33,119
a vertical line, but instead of going through
one zero, it'll go through the point one,

2758
04:44:33,119 --> 04:44:42,680
five. So I'll draw a rough sketch here. Let's
compare our starting function y equals ln

2759
04:44:42,680 --> 04:44:51,080
x and the transformed version y equals ln
x plus five in terms of the domain, the range

2760
04:44:51,080 --> 04:44:59,750
and the vertical asymptote. Our original function
y equals ln x had a domain of zero to infinity

2761
04:44:59,750 --> 04:45:08,659
Since adding five on the outside affects the
y values, and the domain is the x values,

2762
04:45:08,659 --> 04:45:16,340
this transformation doesn't change the domain.
So the domain is still from zero to infinity.

2763
04:45:16,340 --> 04:45:22,529
Now the range of our original y equals ln
x was from negative infinity to infinity.

2764
04:45:22,529 --> 04:45:28,680
Shifting up by five does affect the y values,
and the range is talking about the y values.

2765
04:45:28,680 --> 04:45:33,260
But since the original range was all real
numbers, if you add five to all set of all

2766
04:45:33,259 --> 04:45:38,147
real numbers, you still get the set of all
real numbers. So in this case, the range doesn't

2767
04:45:38,148 --> 04:45:43,958
change either. And finally, we already saw
that the original vertical asymptote of the

2768
04:45:43,957 --> 04:45:49,647
y axis x equals zero, when we shift that up
by five units, it's still the vertical line

2769
04:45:49,648 --> 04:45:57,490
x equals zero. In this next example, we're
starting with a log base 10 function. And

2770
04:45:57,490 --> 04:46:02,978
since the plus two is on the inside, that
means we shift that graph left by two. So

2771
04:46:02,977 --> 04:46:09,718
I'll draw our basic log function. Here's our
basic log function. So I'll think of that

2772
04:46:09,718 --> 04:46:15,317
as y equals log of x going through the point
one, zero,

2773
04:46:15,317 --> 04:46:23,000
here's its vertical asymptote. Now I need
to shift everything left by two. So my vertical

2774
04:46:23,000 --> 04:46:28,610
asymptote shifts left, and now it's at the
line x equals negative two, instead of at

2775
04:46:28,610 --> 04:46:37,090
x equals zero, and my graph, let's see my
point, one zero gets shifted to, let's see

2776
04:46:37,090 --> 04:46:45,270
negative one zero, since I'm subtracting two
from the axis, and here's a rough sketch of

2777
04:46:45,270 --> 04:46:54,069
the resulting graph. Let's compare the features
of the two graphs drawn here. We're talking

2778
04:46:54,069 --> 04:47:01,207
about domains, the original had a domain of
from zero to infinity. But now I've shifted

2779
04:47:01,207 --> 04:47:07,119
that left. So I've subtracted two from all
my x values. And here's my new domain, which

2780
04:47:07,119 --> 04:47:15,227
I can also verify just by looking at the picture.
My range was originally from negative infinity

2781
04:47:15,227 --> 04:47:20,218
to infinity, well, shifting left only affects
the x value. So it doesn't even affect the

2782
04:47:20,218 --> 04:47:26,690
range. So my range is still negative infinity
to infinity. My vertical asymptote was originally

2783
04:47:26,689 --> 04:47:32,967
at x equals zero. And since I subtract two
from all x values, that shifts it to x equals

2784
04:47:32,968 --> 04:47:38,190
negative two. In this last problem, I'm not
going to worry about drawing this graph. I'll

2785
04:47:38,189 --> 04:47:44,567
just use algebra to compute its domain. So
let's think about what's the issue, when you're

2786
04:47:44,567 --> 04:47:50,079
taking the logs of things? Well, you can't
take the log of a negative number is zero.

2787
04:47:50,080 --> 04:47:56,010
So whatever's inside the argument of the log
function, whatever is being fed into log had

2788
04:47:56,009 --> 04:48:03,269
better be greater than zero. So I'll write
that down, we need to minus 3x to be greater

2789
04:48:03,270 --> 04:48:08,797
than zero. Now it's a matter of solving an
inequality to it's got to be greater than

2790
04:48:08,797 --> 04:48:15,759
3x. So two thirds is greater than x. In other
words, x has to be less than two thirds. So

2791
04:48:15,759 --> 04:48:23,340
our domain is all the x values from negative
infinity to two thirds, not including two

2792
04:48:23,340 --> 04:48:31,420
thirds. It's a good idea to memorize the basic
shape of the graph of a log function. It looks

2793
04:48:31,419 --> 04:48:36,759
something like this, go through the point
one zero, and has a vertical asymptote on

2794
04:48:36,759 --> 04:48:44,798
the y axis. Also, if you remember that you
can't take the log of a negative number, or

2795
04:48:44,798 --> 04:48:52,878
zero, then that helps you quickly compute
domains for log functions. Whatever's inside

2796
04:48:52,878 --> 04:49:01,779
the log function, you set that greater than
zero, and solve. This video is about combining

2797
04:49:01,779 --> 04:49:07,500
logs and exponents. Please pause the video
and take a moment to use your calculator to

2798
04:49:07,500 --> 04:49:15,707
evaluate the following four expressions. Remember,
that log base 10 on your calculator is the

2799
04:49:15,707 --> 04:49:24,817
log button. While log base e on your calculator
is the natural log button, you should find

2800
04:49:24,817 --> 04:49:34,409
that the log base 10 of 10 cubed is three.
The log base e of e to the 4.2 is 4.2 10 to

2801
04:49:34,409 --> 04:49:45,387
the log base 10 of 1000 is 1000. And eat the
log base e of 9.6 is 9.6. In each case, the

2802
04:49:45,387 --> 04:49:51,718
log and the exponential function with the
same base undo each other and we're left with

2803
04:49:51,718 --> 04:49:59,420
the exponent. In fact, it's true that for
any base a the log base a of a to the x is

2804
04:49:59,419 --> 04:50:05,289
equal to x, the same sort of cancellation
happens if we do the exponential function

2805
04:50:05,290 --> 04:50:10,770
in the log function with the same base in
the opposite order. For example, we take 10

2806
04:50:10,770 --> 04:50:15,869
to the power of log base 10 of 1000, the 10
to the power and the log base 10 undo each

2807
04:50:15,869 --> 04:50:21,770
other, and we're left with the 1000s. This
happens for any base, say, a data log base

2808
04:50:21,770 --> 04:50:31,067
a of x is equal to x. We can describe this
by saying that an exponential function and

2809
04:50:31,067 --> 04:50:40,169
a log function with the same base undo each
other. If you're familiar with the language

2810
04:50:40,169 --> 04:50:46,869
of inverse functions, the exponential function
and log function are inverses. Let's see why

2811
04:50:46,869 --> 04:50:55,289
these roles hold for the first log role. log
base a of a dx is asking the question, What

2812
04:50:55,290 --> 04:51:03,270
power do we raise a two in order to get a
to the x? In other words, a two what power

2813
04:51:03,270 --> 04:51:11,430
is a dx? Well, the answer is clearly x. And
that's why log base a of a to the x equals

2814
04:51:11,430 --> 04:51:15,290
x. For the second log rule,

2815
04:51:15,290 --> 04:51:25,120
notice that the log base a of x means the
power we raise a two to get x. But this expression

2816
04:51:25,119 --> 04:51:30,137
is saying that we're supposed to raise a to
that power. If we raise a to the power, we

2817
04:51:30,137 --> 04:51:37,128
need to raise a two to get x, then we'll certainly
get x. Now let's use these two rules. In some

2818
04:51:37,128 --> 04:51:44,610
examples. If we want to find three to the
log base three of 1.43 to the power and log

2819
04:51:44,610 --> 04:51:51,860
base three undo each other, so we're left
with 1.4. If we want to find ln of e iliacs.

2820
04:51:51,860 --> 04:51:59,060
Remember that ln means log base e. So we're
taking log base e of e to the x, well, those

2821
04:51:59,060 --> 04:52:05,398
functions undo each other, and we're left
with x. If we want to take 10 to the log of

2822
04:52:05,398 --> 04:52:12,600
three z, remember that log without a base
written implies that the base is 10. So really,

2823
04:52:12,599 --> 04:52:18,750
we want to take 10 to the log base 10 of three
z will tend to a power and log base 10 undo

2824
04:52:18,750 --> 04:52:26,409
each other. So we're left with a three z.
Finally, this last statement hold is ln of

2825
04:52:26,409 --> 04:52:36,419
10 to the x equal to x, well, ln means log
base e. So we're taking log base e of 10 to

2826
04:52:36,419 --> 04:52:41,159
the x, notice that the base of the log and
the base of the exponential function are not

2827
04:52:41,159 --> 04:52:47,939
the same. So they don't undo each other. And
in fact, log base e of 10 to the x is not

2828
04:52:47,939 --> 04:52:53,919
usually equal to x, we can check with one
example. Say if x equals one, then log base

2829
04:52:53,919 --> 04:52:59,629
e of 10 to the one, that's log base e of 10.
And we can check on the calculator that's

2830
04:52:59,630 --> 04:53:06,558
equal to 2.3. And some more decimals, which
is not the same thing as one. So this statement

2831
04:53:06,558 --> 04:53:13,900
is false, it does not hold. We need the basis
to be the same for logs and exponents to undo

2832
04:53:13,900 --> 04:53:23,000
each other. In this video, we saw that logs
and exponents with the same base undo each

2833
04:53:23,000 --> 04:53:26,250
other. Specifically,

2834
04:53:26,250 --> 04:53:33,779
log base a of a to the x is equal to x and
a to the log base a of x is also equal to

2835
04:53:33,779 --> 04:53:34,779
x

2836
04:53:34,779 --> 04:53:42,860
for any values of x and any base a. This video
is about rules or properties of logs. The

2837
04:53:42,860 --> 04:53:46,779
log rules are closely related to the exponent
rules. So let's start by reviewing some of

2838
04:53:46,779 --> 04:53:51,378
the exponent rules. To keep things simple,
we'll write everything down with a base of

2839
04:53:51,378 --> 04:53:57,297
two. Even though the exponent rules hold for
any base. We know that if we raise two to

2840
04:53:57,297 --> 04:54:04,039
the zero power, we get one, we have a product
rule for exponents, which says that two to

2841
04:54:04,040 --> 04:54:12,120
the M times two to the n is equal to two to
the m plus n. In other words, if we multiply

2842
04:54:12,119 --> 04:54:18,590
two numbers, then we add the exponents. We
also have a quotient rule that says that two

2843
04:54:18,590 --> 04:54:26,590
to the M divided by n to the n is equal to
two to the m minus n. In words, that says

2844
04:54:26,590 --> 04:54:34,520
that if we divide two numbers, then we subtract
the exponents. Finally, we have a power rule

2845
04:54:34,520 --> 04:54:43,317
that says if we take a power to a power, then
we multiply the exponents. Each of these exponent

2846
04:54:43,317 --> 04:54:50,270
rules can be rewritten as a log rule. The
first rule, two to the zero equals one can

2847
04:54:50,270 --> 04:54:57,990
be rewritten in terms of logs as log base
two of one equals zero. That's because log

2848
04:54:57,990 --> 04:55:05,090
base two of one equals zero mean To the zero
equals one. The second rule, the product rule

2849
04:55:05,090 --> 04:55:14,990
can be rewritten in terms of logs by saying
log of x times y equals log of x plus log

2850
04:55:14,990 --> 04:55:21,990
of y. I'll make these base two to agree with
my base that I'm using for my exponent rules.

2851
04:55:21,990 --> 04:55:28,860
In words, that says the log of the product
is the sum of the logs. Since logs really

2852
04:55:28,860 --> 04:55:34,637
represent exponent, this is saying that when
you multiply two numbers together, you add

2853
04:55:34,637 --> 04:55:41,329
their exponents, which is just what we said
for the exponent version. The quotient rule

2854
04:55:41,330 --> 04:55:48,830
for exponents can be rewritten in terms of
logs by saying the log of x divided by y is

2855
04:55:48,830 --> 04:55:58,160
equal to the log of x minus the log of y.
In words, we can say that the log of the quotient

2856
04:55:58,159 --> 04:56:04,968
is equal to the difference of the logs. Since
logs are really exponents, another way of

2857
04:56:04,968 --> 04:56:11,770
saying the same thing is that when you divide
two numbers, you subtract their exponents.

2858
04:56:11,770 --> 04:56:18,600
That's how we described the exponent rule
above. Finally, the power rule for exponents

2859
04:56:18,599 --> 04:56:27,317
can be rewritten in terms of logs by saying
the log of x to the n is equal to n times

2860
04:56:27,317 --> 04:56:34,090
log of x. Sometimes people describe this rule
by saying when you take the log of an expression

2861
04:56:34,090 --> 04:56:40,148
with an exponent, you can bring down the exponent
and multiply. If we think of x as being some

2862
04:56:40,148 --> 04:56:45,409
power of two, this is really saying when we
take a power to a power, we multiply their

2863
04:56:45,409 --> 04:56:50,340
exponents. That's exactly how we described
the power rule above. It doesn't really matter

2864
04:56:50,340 --> 04:56:56,819
if you multiply this exponent on the left
side, or on the right side. But it's more

2865
04:56:56,819 --> 04:57:03,189
traditional to multiply it on the left side.
I've given these rules with the base of two,

2866
04:57:03,189 --> 04:57:07,887
but they actually work for any base. To help
you remember them, please take a moment to

2867
04:57:07,887 --> 04:57:17,509
write out the log roles using a base of a
you should get the following chart. Let's

2868
04:57:17,509 --> 04:57:23,137
use the log rules to rewrite the following
expressions as a sum or difference of logs.

2869
04:57:23,137 --> 04:57:30,369
In the first expression, we have a log base
10 of a quotient. So we can rewrite the log

2870
04:57:30,369 --> 04:57:39,307
of the quotient as the difference of the logs.
Now we still have the log of a product, I

2871
04:57:39,308 --> 04:57:47,727
can rewrite the log of a product as the sum
of the logs. So that is log of y plus log

2872
04:57:47,727 --> 04:57:55,099
of z. When I put things together, I have to
be careful because here I'm subtracting the

2873
04:57:55,099 --> 04:58:02,529
entire log expression. So I need to subtract
both terms of this song on the make sure I

2874
04:58:02,529 --> 04:58:08,270
do that by putting them in parentheses. Now
I can simplify a little bit by distributing

2875
04:58:08,270 --> 04:58:16,340
the negative sign. And here's my final answer.
In my next expression, I have the log of a

2876
04:58:16,340 --> 04:58:21,909
product. So I can rewrite that as the sum
of two logs.

2877
04:58:21,909 --> 04:58:31,042
I can also use my power rule to bring down
the exponent T and multiply it in the front.

2878
04:58:31,042 --> 04:58:38,909
That gives me the final expression log of
five plus t times log of to one common mistake

2879
04:58:38,909 --> 04:58:47,968
on this problem is to rewrite this expression
as t times log of five times two. In fact,

2880
04:58:47,968 --> 04:58:54,218
those two expressions are not equal. Because
the T only applies to the two, not to the

2881
04:58:54,218 --> 04:58:59,729
whole five times two, we can't just bring
it down in front using the power wall. After

2882
04:58:59,729 --> 04:59:08,349
all, the power rule only applies to a single
expression raised to an exponent, and not

2883
04:59:08,349 --> 04:59:14,457
to a product like this. And these next examples,
we're going to go the other direction. Here

2884
04:59:14,457 --> 04:59:18,729
we're given sums and differences of logs.
And we want to wrap them up into a single

2885
04:59:18,729 --> 04:59:24,989
log expression. By look at the first two pieces,
that's a difference of logs. So I know I can

2886
04:59:24,990 --> 04:59:34,978
rewrite it as the log of a quotient. Now I
have the sum of two logs. So I can rewrite

2887
04:59:34,977 --> 04:59:43,889
that as the log of a product. I'll clean that
up a little bit and rewrite it as log base

2888
04:59:43,889 --> 04:59:54,610
five of a times c over B. In my second example,
I can rewrite the sum of my logs as the log

2889
04:59:54,610 --> 05:00:03,369
of a product now, I will Like to rewrite this
difference of logs as the log of a quotient.

2890
05:00:03,369 --> 05:00:09,817
But I can't do it yet, because of that factor
of two multiplied in front. But I can use

2891
05:00:09,817 --> 05:00:15,669
the power rule backwards to put that two back
up in the exponent. So I'll do that first.

2892
05:00:15,669 --> 05:00:23,909
So I will copy down the ln of x plus one times
x minus one, and rewrite this second term

2893
05:00:23,909 --> 05:00:31,968
as ln of x squared minus one squared. Now
I have a straightforward difference of two

2894
05:00:31,968 --> 05:00:40,990
logs, which I can rewrite as the log of a
quotient. I can actually simplify this some

2895
05:00:40,990 --> 05:00:50,030
more. Since x plus one times x minus one is
the same thing as x squared minus one. I can

2896
05:00:50,029 --> 05:01:00,119
cancel factors to get ln of one over x squared
minus one. In this video, we saw four rules

2897
05:01:00,119 --> 05:01:07,329
for logs that are related to exponent rules.
First, we saw that the log with any base of

2898
05:01:07,330 --> 05:01:15,330
one is equal to zero. Second, we saw the product
rule, the log of a product is equal to the

2899
05:01:15,330 --> 05:01:23,968
sum of the logs. We saw the quotient rule,
the log of a quotient is the difference of

2900
05:01:23,968 --> 05:01:30,639
the logs. And we saw the power rule. When
you take a log of an expression with an exponent

2901
05:01:30,639 --> 05:01:37,200
in it, you can bring down the exponent and
multiply it. It's worth noticing that there's

2902
05:01:37,200 --> 05:01:44,137
no log rule that helps you split up the log
of a psalm. In particular, the log of a psalm

2903
05:01:44,137 --> 05:01:52,489
is not equal to the sum of the logs. If you
think about logs and exponent rules going

2904
05:01:52,490 --> 05:02:00,740
together, this kind of makes sense, because
there's also no rule for rewriting the sum

2905
05:02:00,740 --> 05:02:04,450
of two exponential expressions.

2906
05:02:04,450 --> 05:02:10,308
Log rules will be super handy, as we start
to solve equations using logs. Anytime you

2907
05:02:10,308 --> 05:02:17,760
have an equation like this one that has variables
in the exponent logs are the tool of choice

2908
05:02:17,759 --> 05:02:23,387
for getting those variables down where you
can solve for them. In this video, I'll do

2909
05:02:23,387 --> 05:02:29,797
a few examples of solving equations with variables
in the exponent. For our first example, let's

2910
05:02:29,797 --> 05:02:37,250
solve for x the equation five times two to
the x plus one equals 17. As my first step,

2911
05:02:37,250 --> 05:02:42,707
I'm going to isolate the difficult spot, the
part that has the variables and the exponent.

2912
05:02:42,707 --> 05:02:47,889
In this example, I can do that by dividing
both sides by five. That gives me two to the

2913
05:02:47,889 --> 05:02:56,250
x plus one equals 17 over five. Next, I'm
going to take the log of both sides, it's

2914
05:02:56,250 --> 05:03:00,919
possible to take the log with any base, but
I prefer to take the log base 10, or the log

2915
05:03:00,919 --> 05:03:06,149
base e for the simple reason that my calculator
has buttons for those logs. So in this example,

2916
05:03:06,150 --> 05:03:12,610
I'll just take the log base 10. So I can omit
the base because it's a base 10 is implied

2917
05:03:12,610 --> 05:03:19,500
here. And that gives me this expression. As
my next step, I'm going to use log roles to

2918
05:03:19,500 --> 05:03:24,547
bring down my exponent and multiply it on
the front. It's important to use parentheses

2919
05:03:24,547 --> 05:03:31,217
here because the entire x plus one needs to
get multiplied by log two. So that was my

2920
05:03:31,218 --> 05:03:37,409
third step using the log roles. Now all my
variables are down from the exponent where

2921
05:03:37,409 --> 05:03:42,040
I can work with them, but I still need to
solve for x. Right now x is trapped in the

2922
05:03:42,040 --> 05:03:47,100
parentheses. So I'm going to free it from
the parentheses by distributing. So I get

2923
05:03:47,099 --> 05:03:55,887
x log two plus log two equals log of 17 fifths.
Now I will try to isolate x by moving all

2924
05:03:55,887 --> 05:04:01,520
my terms with x's in them to one side, and
all my terms without x's in them to the other

2925
05:04:01,520 --> 05:04:08,850
side. Finally, I factor out my x. Well, it's
already kind of factored out, and I divide

2926
05:04:08,849 --> 05:04:17,797
to isolate it. So I read out what I did. So
far I distributed. I moved all the x terms

2927
05:04:17,797 --> 05:04:24,739
on one side, and the terms without x's on
the other side. And then I isolated the x

2928
05:04:24,740 --> 05:04:31,830
by factoring out and dividing. We have an
exact solution for x. This is correct, but

2929
05:04:31,830 --> 05:04:36,830
maybe not so useful if you want a decimal
answer. At this point, you can type in everything

2930
05:04:36,830 --> 05:04:46,870
into your calculator using parentheses liberally
to get a decimal answer about 0.765. It's

2931
05:04:46,869 --> 05:04:51,829
always a good idea to check your work by typing
in this decimal answer and seeing if the equation

2932
05:04:51,830 --> 05:04:57,540
checks out. This next example is trickier
because there are variables in the exponent

2933
05:04:57,540 --> 05:05:05,180
in two places with two different bases. First
step is normally to clean things up and simplify

2934
05:05:05,180 --> 05:05:10,207
by isolating the tricky stuff. But in this
example, there's nothing really to clean up

2935
05:05:10,207 --> 05:05:15,590
or simplifier, or no way to isolate anything
more than it already is. So we'll go ahead

2936
05:05:15,590 --> 05:05:20,400
to the next step and take the log of both
sides. Again, I'll go ahead and use log base

2937
05:05:20,400 --> 05:05:27,548
10. But we couldn't use log base e instead.
Next, we can use log roles to bring down the

2938
05:05:27,547 --> 05:05:35,789
exponents. This gives me 2x minus three in
parentheses, log two equals x minus two, log

2939
05:05:35,790 --> 05:05:43,138
five. Now I'm going to distribute things out
to free the excess from the parentheses. That

2940
05:05:43,137 --> 05:05:52,950
gives me 2x log two minus three log two equals
x log five minus two log five. Now I need

2941
05:05:52,950 --> 05:05:59,350
to group the x terms on one side, and the
other terms that don't involve x on the other

2942
05:05:59,349 --> 05:06:08,099
side. So I'll keep the 2x log two on the left,
put the minus log x log five on the left,

2943
05:06:08,099 --> 05:06:13,759
and that gives me the minus two log five that
stays on the right and a plus three log two

2944
05:06:13,759 --> 05:06:22,799
on the right. Finally, I need to isolate x
by factoring and dividing by factoring I just

2945
05:06:22,799 --> 05:06:27,009
mean I factor out the x from all the terms
on the left, so that gives me x times the

2946
05:06:27,009 --> 05:06:37,989
quantity to log two minus log five. And that
equals this mess on the right. Now I can divide

2947
05:06:37,990 --> 05:06:48,540
the right side by the quantity on the left
side. When you type this into your calculator,

2948
05:06:48,540 --> 05:06:53,850
I encourage you to type the whole thing in
rather than type bits and pieces in because

2949
05:06:53,849 --> 05:06:58,159
if you round off, you'll get a less accurate
answer than if you type the whole thing in

2950
05:06:58,159 --> 05:07:06,648
at once. In this example, when I type it in,
I get a final answer of about 5.106.

2951
05:07:06,648 --> 05:07:12,670
In this equation, we have the variable t in
the exponent in two places. The letter E is

2952
05:07:12,669 --> 05:07:19,500
not a variable, it represents the number e
whose decimal approximation is about 2.7.

2953
05:07:19,500 --> 05:07:24,148
Because there's already an E and the expression,
it's going to be handy to use natural log

2954
05:07:24,148 --> 05:07:29,468
in this problem instead of log base 10. But
before we take the log of both sides, let's

2955
05:07:29,468 --> 05:07:35,468
clean things up. by isolating the tricky parts,
we could at least divide both sides by five.

2956
05:07:35,468 --> 05:07:42,409
And that will give us either the negative
0.05 t is equal to three fifths it is 0.2

2957
05:07:42,409 --> 05:07:47,799
T. One way to proceed would now be to clean
things out further by dividing both sides

2958
05:07:47,799 --> 05:07:51,898
by either the point two t, but I'm going to
take a different approach and go ahead and

2959
05:07:51,898 --> 05:08:01,290
take the natural log of both sides. That gives
me ln of e to the minus 0.05 t is equal to

2960
05:08:01,290 --> 05:08:08,479
ln of three fifths e to the 0.2 t. Now on
the left side, I can immediately use my log

2961
05:08:08,479 --> 05:08:18,099
rule to bring down my exponent and get minus
0.05 t, L and E. But on the right side, I

2962
05:08:18,099 --> 05:08:23,519
can't bring down the exponent yet because
this e to the point two t is multiplied by

2963
05:08:23,520 --> 05:08:28,797
three fifths. So before I can bring down the
exponent, I need to split up this product

2964
05:08:28,797 --> 05:08:36,110
using the product rule. So I can rewrite this
as ln of three fifths plus ln e to the 0.2

2965
05:08:36,110 --> 05:08:47,887
T. And now I can bring down the exponent.
So that third step was using log roles to

2966
05:08:47,887 --> 05:08:56,128
ultimately bring down my exponents. Now ln
of E is a really nice expression, ln IV means

2967
05:08:56,128 --> 05:09:02,431
log base e of E. So that's asking what power
do I raise E to in order to get e? And the

2968
05:09:02,431 --> 05:09:10,570
answer is one. So anytime I have ln of E,
I can just replace that with one. That's why

2969
05:09:10,570 --> 05:09:14,990
using natural log is a little bit handier
here than using log base 10. You can make

2970
05:09:14,990 --> 05:09:23,770
that simplification. Next, I'm ready to solve
for t. So I don't need to distribute here.

2971
05:09:23,770 --> 05:09:29,610
But I do need to bring my T terms to one side
of my terms without t to the other side. So

2972
05:09:29,610 --> 05:09:37,779
let's see. I'll put my T terms on the left
and my team turns without T on the right.

2973
05:09:37,779 --> 05:09:45,619
And finally I'm going to isolate t by factoring
and dividing. So by factoring I mean I factor

2974
05:09:45,619 --> 05:09:57,147
out my T 
and now I can divide. Using my calculator,

2975
05:09:57,148 --> 05:10:04,718
I can get a decimal answer of 2.04 Three,
three. This video gave some examples of solving

2976
05:10:04,718 --> 05:10:10,477
equations with variables in the exponent.
And the key idea was to take the log of both

2977
05:10:10,477 --> 05:10:21,270
sides and use the log properties to bring
the exponents down. This video, give some

2978
05:10:21,270 --> 05:10:26,718
examples of equations with logs in them like
this one. In order to solve the equations

2979
05:10:26,718 --> 05:10:32,540
like this one, we have to free the variable
from the log. And we'll do that using exponential

2980
05:10:32,540 --> 05:10:38,830
functions. My first step in solving pretty
much any kind of equation is to simplify it

2981
05:10:38,830 --> 05:10:44,000
and isolate the tricky part. In this case,
the tricky part is the part with the log in

2982
05:10:44,000 --> 05:10:51,599
it. So I can isolate it by first adding three
to both sides. That gives me two ln 2x plus

2983
05:10:51,599 --> 05:11:01,180
five equals four, and then I can divide both
sides by two. Now that I've isolated the tricky

2984
05:11:01,180 --> 05:11:07,520
part, I still need to solve for x, but x is
trapped inside the log function. to free it,

2985
05:11:07,520 --> 05:11:13,200
I need to somehow undo the log function. Well
log functions and exponential functions undo

2986
05:11:13,200 --> 05:11:18,727
each other. And since this is a log base e,
I need to use the exponential function base

2987
05:11:18,727 --> 05:11:26,270
e also. So I'm going to take e to the power
of both sides. In other words, I'll take e

2988
05:11:26,270 --> 05:11:33,420
to the ln 2x plus five, and that's going to
equal e to the power of two.

2989
05:11:33,419 --> 05:11:40,189
Now e to the ln of anything, I'll just write
a for anything. That means e to the log base

2990
05:11:40,189 --> 05:11:46,859
e of a, that you the power and log base e
undo each other. So we just get a, I'm going

2991
05:11:46,860 --> 05:11:53,200
to use that principle over here, e to the
ln of 2x plus five, the E and the log base

2992
05:11:53,200 --> 05:11:59,030
e undo each other. And we're left with 2x
plus five on the left side. So 2x plus five

2993
05:11:59,029 --> 05:12:04,360
is equal to E squared. And from there, it's
easy to finish the problem and solve for x

2994
05:12:04,360 --> 05:12:11,047
by subtracting five from both sides and then
dividing by two. So I'll just write the third

2995
05:12:11,047 --> 05:12:17,967
step here is to just say finish solving for
x. There's one last step we need to do when

2996
05:12:17,968 --> 05:12:25,909
solving equations with logs in them. And that's
to check our answers. Because we may get extraneous

2997
05:12:25,909 --> 05:12:31,529
solutions. an extraneous solution is a solution
that comes out of the solving process, but

2998
05:12:31,529 --> 05:12:36,807
doesn't actually satisfy the original equation.
And those can happen for equations with logs

2999
05:12:36,808 --> 05:12:42,317
in them, because we might get a solution that
makes the argument of the log negative or

3000
05:12:42,317 --> 05:12:52,898
zero, and we can't take the log of a negative
numbers zero. So let's check and plug in the

3001
05:12:52,898 --> 05:13:01,020
solution of E squared minus five over two.
We'll plug that in for x in our original equation

3002
05:13:01,020 --> 05:13:09,840
and see if that works. So let's see the twos
cancel here. So I get two ln e squared minus

3003
05:13:09,840 --> 05:13:17,049
five plus five, minus three, I want that to
equal one. And now my fives cancel. And so

3004
05:13:17,049 --> 05:13:23,648
I have two ln e squared minus three that I
want to equal one. Well, I think this is going

3005
05:13:23,649 --> 05:13:34,900
to work out because let's see, ln is log base
e. and log base e of E squared that's asking

3006
05:13:34,900 --> 05:13:38,890
the question, What power do I raise e two
to get e squared? Well, I have to raise it

3007
05:13:38,889 --> 05:13:46,750
to the power of two to get e squared. So this
becomes two times two minus three. And does

3008
05:13:46,750 --> 05:13:53,080
that equal one, four minus three does equal
one. So that all checks out. And we didn't

3009
05:13:53,080 --> 05:13:58,780
have any problem with taking the log of negative
numbers zero, we didn't get any extraneous

3010
05:13:58,779 --> 05:14:04,619
solutions. So this is our solution. The second
equation is a little bit trickier, because

3011
05:14:04,619 --> 05:14:10,637
there's a log into places. Now notice that
this is a log where there's no base written.

3012
05:14:10,637 --> 05:14:16,340
So a base 10 is implied. So I'm already thinking
in order to undo a log base 10, I'm going

3013
05:14:16,340 --> 05:14:23,080
to want to take a 10 to the power of both
sides. Of course, it's still a good idea to

3014
05:14:23,080 --> 05:14:29,060
isolate the tricky part, but there's nothing
really to isolate here. So I'm going to say,

3015
05:14:29,060 --> 05:14:37,298
can't do it here. So we'll just jump right
to straight step two, and take 10 to the power

3016
05:14:37,297 --> 05:14:46,887
of both sides. Okay, so that's going to give
me 10 to the whole thing, log x plus three

3017
05:14:46,887 --> 05:14:52,389
plus log x, that whole thing is in the exponent
equals 10 to the One Power. Well, I know what

3018
05:14:52,389 --> 05:14:58,750
to do with the right side 10 to the one is
just 10. But what do I do with the 10 to these

3019
05:14:58,750 --> 05:14:59,840
two things added

3020
05:14:59,840 --> 05:15:00,930
up Up,

3021
05:15:00,930 --> 05:15:06,869
while remembering my exponent rules, I know
that when you add up the exponent, that's

3022
05:15:06,869 --> 05:15:12,419
what happens when you multiply two things.
So this is the same thing as 10 to the log

3023
05:15:12,419 --> 05:15:18,369
x plus three times 10 to the log x, right,
because when you multiply two things, you

3024
05:15:18,369 --> 05:15:24,439
add the exponent, so these are the same. Okay,
now we're in business, because 10 to the log

3025
05:15:24,439 --> 05:15:31,797
base 10, those undo each other. And so this
whole expression here simplifies to x plus

3026
05:15:31,797 --> 05:15:38,967
three. Similarly, 10 to the log base, 10 of
x is just x. So I'm multiplying x plus three

3027
05:15:38,968 --> 05:15:46,030
by x, that's equal to 10. Now I have an equation
I can deal with, it's a quadratic, so I'm

3028
05:15:46,029 --> 05:15:52,397
going to first multiply out to make it look
more like a quadratic, get everything to one

3029
05:15:52,398 --> 05:15:59,409
side. So is equal to zero. And, and now I
can either factor or use the quadratic formula.

3030
05:15:59,409 --> 05:16:06,500
I think this one factors, it looks like X
plus five times x minus two, so I'm going

3031
05:16:06,500 --> 05:16:14,060
to get x is negative five, or x is two. So
that was all the third step to finish solving

3032
05:16:14,060 --> 05:16:20,860
for x. Finally, we need to check our solutions
to make sure we haven't gotten some extraneous

3033
05:16:20,860 --> 05:16:27,378
ones. So let's see if x equals negative five.
If I plug that into my original equation,

3034
05:16:27,378 --> 05:16:36,047
that says, I'm checking that log of negative
five plus three, plus log of negative five,

3035
05:16:36,047 --> 05:16:40,879
checking that's equal to one. Well, this is
giving me a queasy feeling. And I hope it's

3036
05:16:40,880 --> 05:16:48,540
giving you a queasy feeling too, because log
of negative two does not exist, right, you

3037
05:16:48,540 --> 05:16:53,750
can't take the log of a negative number. Same
thing with log of negative five. So x equals

3038
05:16:53,750 --> 05:17:00,297
negative five is an extraneous solution, it
doesn't actually solve our original equation.

3039
05:17:00,297 --> 05:17:08,029
Let's check the other solution, x equals two.
So now we're checking to see if log of two

3040
05:17:08,029 --> 05:17:15,360
plus three plus log of two is equal to one.
Since there's no problem with taking logs

3041
05:17:15,360 --> 05:17:20,670
of negative numbers, or zero here, this should
work out fine. And just to finish checking,

3042
05:17:20,669 --> 05:17:27,423
we can see let's see this is log of five plus
log of two, we want that equal one. But using

3043
05:17:27,423 --> 05:17:34,540
my log rules, the sum of two logs is the log
of the product. So log of five times two,

3044
05:17:34,540 --> 05:17:41,148
we want that to equal one. And that's just
log base 10 of 10. And that definitely equals

3045
05:17:41,148 --> 05:17:46,909
one because log base 10 of 10 says, What power
do I raise tend to to get 10. And that power

3046
05:17:46,909 --> 05:17:56,079
is one. So the second solution x equals two
does check out. And that's our final answer.

3047
05:17:56,080 --> 05:18:00,350
Before I leave this problem, I do want to
mention that some people have an alternative

3048
05:18:00,349 --> 05:18:05,659
approach. Some people like to start with the
original equation, and then use log roles

3049
05:18:05,659 --> 05:18:12,790
to combine everything into one log expression.
So since we have the sum of two logs, we know

3050
05:18:12,790 --> 05:18:18,388
that's the same as the log of a product, right,
so we can rewrite the left side as log of

3051
05:18:18,387 --> 05:18:25,849
x plus three times x, that equals one, then
we do the same trick of taking 10 to the power

3052
05:18:25,849 --> 05:18:32,717
of both sides. And as before, the 10 to the
power and the log base 10 undo each other,

3053
05:18:32,718 --> 05:18:40,690
and we get x plus three times x equals 10,
just like we did before. In the first solution,

3054
05:18:40,689 --> 05:18:45,639
we ended up using exponent rules to rewrite
things. In the second alternative method,

3055
05:18:45,639 --> 05:18:50,689
we use log rules to rewrite things. So the
methods are really pretty similar, pretty

3056
05:18:50,689 --> 05:18:54,849
equivalent, and they certainly will get us
to the same answer. So we've seen a couple

3057
05:18:54,849 --> 05:19:02,137
examples of equations with logs in them and
how to solve them. And the key step is to

3058
05:19:02,137 --> 05:19:13,279
use exponential functions to undo the log.
In other words, take e to the power of both

3059
05:19:13,279 --> 05:19:25,090
sides to undo natural log, and take 10 to
the power of both sides. To undo log base

3060
05:19:25,090 --> 05:19:27,560
10.

3061
05:19:27,560 --> 05:19:33,030
In this video, we'll use exponential equations
in some real life applications, like population

3062
05:19:33,029 --> 05:19:40,217
growth and radioactive decay. I'll also introduce
the ideas of half life and doubling time.

3063
05:19:40,218 --> 05:19:47,350
In this first example, let's suppose we invest
$1,600 in a bank account that earns 6.5% annual

3064
05:19:47,349 --> 05:19:52,889
interest compounded once a year. How many
years will it take until the account has $2,000

3065
05:19:52,889 --> 05:19:59,450
in it if you don't make any further deposits
or withdrawals since our money is earning

3066
05:19:59,450 --> 05:20:05,540
six point 5% interest each year, that means
that every year the money gets multiplied

3067
05:20:05,540 --> 05:20:18,817
by 1.065. So after t years, my 1600 gets multiplied
by 1.065 to the t power. I'll write this in

3068
05:20:18,817 --> 05:20:30,308
function notation as f of t equals 1600 times
1.065 to the T, where f of t is the amount

3069
05:20:30,308 --> 05:20:41,080
of money after t years. Now we're trying to
figure out how long it will take to get $2,000

3070
05:20:41,080 --> 05:20:48,580
$2,000 is a amount of money. So that's an
amount for f of t. And we're trying to solve

3071
05:20:48,580 --> 05:20:55,290
for T the amount of time. So let me write
out my equation and make a note that I'm solving

3072
05:20:55,290 --> 05:21:04,840
for t. Now to solve for t, I want to first
isolate the tricky part. So I'm going to the

3073
05:21:04,840 --> 05:21:08,790
tricky part is the part with the exponential
in it. So I'm going to divide both sides by

3074
05:21:08,790 --> 05:21:18,190
1600. That gives me 2000 over 1600 equals
1.065. To the tee, I can simplify this a little

3075
05:21:18,189 --> 05:21:24,317
bit further as five fourths. Now that I've
isolated the tricky part, my next step is

3076
05:21:24,317 --> 05:21:29,727
going to be to take the log of both sides.
That's because I have a variable in the exponent.

3077
05:21:29,727 --> 05:21:33,829
And I know that if I log take the log of both
sides, I can use log roles to bring that exponent

3078
05:21:33,830 --> 05:21:40,850
down where I can solve for it. I think I'll
use log base e this time. So I have ln fi

3079
05:21:40,849 --> 05:21:48,759
force equals ln of 1.065 to the T. Now by
the power rule for logs, on the right side,

3080
05:21:48,759 --> 05:21:56,919
I can bring that exponent t down and multiply
it in the front. Now it's easy to isolate

3081
05:21:56,919 --> 05:22:07,329
t just by dividing both sides by ln of 1.065.
Typing that into my calculator, I get that

3082
05:22:07,330 --> 05:22:17,308
t is approximately 3.54 years. And the next
example, we have a population of bacteria

3083
05:22:17,308 --> 05:22:24,290
that initially contains 1.5 million bacteria,
and it's growing by 12% per day, we want to

3084
05:22:24,290 --> 05:22:34,659
find the doubling time, the doubling time
means the amount of time it takes for a quantity

3085
05:22:34,659 --> 05:22:41,110
to double in size. For example, the amount
of time it takes to get from the initial 1.5

3086
05:22:41,110 --> 05:22:47,540
million bacteria to 3 million bacteria would
be the doubling time. Let's start by writing

3087
05:22:47,540 --> 05:22:56,558
an equation for the amount of bacteria. So
if say, P of t represents the number of bacteria

3088
05:22:56,558 --> 05:23:08,968
in millions, then my equation and T represents
time in days, then P of t is going to be given

3089
05:23:08,968 --> 05:23:19,887
by the initial amount of bacteria times the
growth factor 1.12 to the T. That's because

3090
05:23:19,887 --> 05:23:25,360
my population of bacteria is growing by 12%
per day. That means the number of bacteria

3091
05:23:25,360 --> 05:23:30,760
gets multiplied by one point 12. Since we're
looking for the doubling time, we're looking

3092
05:23:30,759 --> 05:23:39,309
for the t value when P of D will be twice
as big. So I can set P of t to be three and

3093
05:23:39,310 --> 05:23:56,298
solve for t. As before, I'll start by isolating
the tricky part, taking the log

3094
05:23:56,297 --> 05:24:08,237
bringing the T down. And finally solving for
T. Let's see three over 1.5 is two so I can

3095
05:24:08,238 --> 05:24:19,450
write this as ln two over ln 1.12. Using my
calculator, that's about 6.12 days. It's an

3096
05:24:19,450 --> 05:24:32,387
interesting fact that doubling time only depends
on the growth rate, that 12% growth, not the

3097
05:24:32,387 --> 05:24:37,137
initial population. In fact, I could have
figured out the doubling time without even

3098
05:24:37,137 --> 05:24:41,647
knowing how many bacteria were in my initial
population. Let me show you how that would

3099
05:24:41,648 --> 05:24:48,100
work. If I didn't know how many I started
with, I could still write P of t equals a

3100
05:24:48,099 --> 05:24:55,989
times 1.12 to the t where a is our initial
population that I don't know what it is.

3101
05:24:55,990 --> 05:25:01,121
Then if I want to figure out how long it takes
for my population, to double Well, if I start

3102
05:25:01,120 --> 05:25:08,869
with a and double that I get to a. So I'll
set my population equal to two a. And I'll

3103
05:25:08,869 --> 05:25:19,297
solve for t. Notice that my A's cancel. And
so when I take the log of both sides and bring

3104
05:25:19,297 --> 05:25:32,099
the T down and solve for t, I get the exact
same thing as before, it didn't matter what

3105
05:25:32,099 --> 05:25:38,009
the initial population was, I didn't even
have to know what it was. In this next example,

3106
05:25:38,009 --> 05:25:42,217
we're told the initial population, and we're
told the doubling time, we're not told by

3107
05:25:42,218 --> 05:25:49,218
what percent the population increases each
minute, or by what number, we're forced to

3108
05:25:49,218 --> 05:25:53,440
multiply the population by each minute. So
we're gonna have to solve for that, I do know

3109
05:25:53,439 --> 05:25:59,340
that I want to use an equation of the form
y equals a times b to the T, where T is going

3110
05:25:59,340 --> 05:26:04,920
to be the number of minutes, and y is going
to be the number of bacteria. And I know that

3111
05:26:04,919 --> 05:26:11,459
my initial amount a is 350. So I can really
write y equals 350 times b to the T. Now the

3112
05:26:11,459 --> 05:26:19,119
doubling time tells me that when 15 minutes
have elapsed, my population is going to be

3113
05:26:19,119 --> 05:26:28,619
twice as big, or 700. plugging that into my
equation, I have 700 equals 350 times b to

3114
05:26:28,619 --> 05:26:34,789
the 15. Now I need to solve for b. Let me
clean this up a little bit by dividing both

3115
05:26:34,790 --> 05:26:43,798
sides by 350. That gives me 700 over 350 equals
b to the 15th. In other words, two equals

3116
05:26:43,797 --> 05:26:51,479
b to the 15th. To solve for B, I don't have
to actually use logs here, because my variables

3117
05:26:51,479 --> 05:26:57,270
in the base not in the exponent, so I don't
need to bring that exponent down. Instead,

3118
05:26:57,270 --> 05:27:03,439
the easiest way to solve this is just by taking
the 15th root of both sides or equivalently.

3119
05:27:03,439 --> 05:27:12,467
The 1/15 power. That's because if I take B
to the 15th to the 1/15, I multiply by exponents,

3120
05:27:12,468 --> 05:27:19,690
that gives me B to the one is equal to two
to the 1/15. In other words, B is two to the

3121
05:27:19,689 --> 05:27:29,789
1/15, which as a decimal is approximately
1.047294. I like to use a lot of decimals

3122
05:27:29,790 --> 05:27:34,170
if I'm doing a decimal approximation and these
kind of problems to increase accuracy. But

3123
05:27:34,169 --> 05:27:38,467
of course, the most accurate thing is just
to leave B as it is. And I'll do that and

3124
05:27:38,468 --> 05:27:50,190
rewrite my equation is y equals 350 times
two to the 1/15 to the T. Now I'd like to

3125
05:27:50,189 --> 05:27:56,079
work this problem one more time. And this
time, I'm going to use the form of the equation

3126
05:27:56,080 --> 05:28:04,370
y equals a times e to the RT. This is called
a continuous growth model. It looks different,

3127
05:28:04,369 --> 05:28:09,750
but it's actually an equivalent form to this
growth model over here. And I'll say more

3128
05:28:09,750 --> 05:28:14,450
about why these two forms are equivalent at
the end, I can use the same general ideas

3129
05:28:14,450 --> 05:28:21,950
to solve in this form. So I know that my initial
amount is 350. And I know again that when

3130
05:28:21,950 --> 05:28:33,530
t is 15, my Y is 700. So I plug in 700 here
350 e to the r times 15. And I solve for r.

3131
05:28:33,529 --> 05:28:44,869
Again, I'm going to simplify things by dividing
both sides by 350. That gives me two equals

3132
05:28:44,869 --> 05:28:51,840
e to the r times 15. This time my variable
does end up being in the exponent. So I do

3133
05:28:51,840 --> 05:28:56,830
want to take the log of both sides, I'm going
to use natural log, because I already have

3134
05:28:56,830 --> 05:29:01,818
an E and my problem. So natural log and E
are kind of more harmonious the other than

3135
05:29:01,817 --> 05:29:09,259
then common log with base 10 and E. So I take
the log of both sides. Now I can pull the

3136
05:29:09,259 --> 05:29:14,789
exponent down. So that's our times 15 times
ln of E. Well Elena V is just one, right?

3137
05:29:14,790 --> 05:29:19,540
Because Elena V is asking what power do I
raise E to to get e that's just one. And so

3138
05:29:19,540 --> 05:29:27,218
I get 15 r equals ln two. So r is equal to
ln two divided by 15. Let me plug that back

3139
05:29:27,218 --> 05:29:35,250
in to my original equation as e to the ln
two over 15 times t. Now I claimed that these

3140
05:29:35,250 --> 05:29:40,279
two equations were actually the same thing
just looking different. And the way to see

3141
05:29:40,279 --> 05:29:49,099
that is if I start with this equation, and
I rewrite as e to the ln two over 15 like

3142
05:29:49,099 --> 05:29:50,099
that

3143
05:29:50,099 --> 05:29:58,090
to the tee. Well, I claim that this quantity
right here is the same thing as my two to

3144
05:29:58,090 --> 05:30:06,727
the 1/15. And in fact One way to see that
is e to the ln two already to the 115. The

3145
05:30:06,727 --> 05:30:10,509
Right, that's the same, because every time
I take a power to a power, I multiply exponents.

3146
05:30:10,509 --> 05:30:18,840
But what's Ed Oh, and two, E and ln undo each
other, so that's just 350 times two to the

3147
05:30:18,840 --> 05:30:24,887
1/15 to the T TA, the equations are really
the same. And these, you'll always be able

3148
05:30:24,887 --> 05:30:30,897
to find two different versions of an exponential
equation, sort of the standard one, a times

3149
05:30:30,898 --> 05:30:38,738
b to the T, or the continuous growth one,
a times e to the RT. In this last example,

3150
05:30:38,738 --> 05:30:43,569
we're going to work with half life, half life
is pretty much like doubling time, it just

3151
05:30:43,569 --> 05:30:55,099
means the amount of time that it takes for
a quantity to decrease to half as much as

3152
05:30:55,099 --> 05:31:02,509
we originally started with, we're told that
the half life of radioactive carbon 14 is

3153
05:31:02,509 --> 05:31:11,009
5750 years. So that means it takes that long
for a quantity of radioactive carbon 14 to

3154
05:31:11,009 --> 05:31:16,807
decay, so that you just have half as much
left and the rest is nonradioactive form.

3155
05:31:16,808 --> 05:31:21,718
So we're told a sample of bone that originally
contained 200 grams of radioactive carbon

3156
05:31:21,718 --> 05:31:29,670
14 now contains only 40 grams, we're supposed
to find out how old the sample is, is called

3157
05:31:29,669 --> 05:31:36,957
carbon dating. Let's use the continuous growth
model this time. So our final amount, so this

3158
05:31:36,957 --> 05:31:49,099
is our amount of radioactive c 14 is going
to be the initial amount

3159
05:31:49,099 --> 05:31:54,250
times e to the RT, we could have used the
other model to we could have used f of t equals

3160
05:31:54,250 --> 05:31:59,468
a times b to the T, but I just want to use
the continuous model for practice. So we know

3161
05:31:59,468 --> 05:32:13,659
that our half life is 5750. So what that means
is when t is 5750, our amount is going to

3162
05:32:13,659 --> 05:32:19,290
be one half of what we started with. Let me
see if I can plug that into my equation and

3163
05:32:19,290 --> 05:32:26,968
figure out use that to figure out what r is.
That's ours called the continuous growth rate.

3164
05:32:26,968 --> 05:32:35,350
So I plug in one half a for the final amount,
a is still the initial amount, e to the r

3165
05:32:35,349 --> 05:32:46,279
and I have 5750 I can cancel my A's. And now
I want to solve for r, r is in my exponent.

3166
05:32:46,279 --> 05:32:50,987
So I do need to take the log of both sides
to solve for it. I'm going to use log base

3167
05:32:50,988 --> 05:32:57,120
e since I already have an E and my problem,
log base e is more compatible with E then

3168
05:32:57,119 --> 05:33:04,227
log base 10 is okay. Now, on the left side,
I still have log of one half and natural log

3169
05:33:04,227 --> 05:33:09,887
of one half on the right side, ln n e to a
power those undo each other. So I'm left with

3170
05:33:09,887 --> 05:33:20,649
R times 5750. Now I can solve for r, it's
ln one half over 5750 could work that as a

3171
05:33:20,650 --> 05:33:25,909
decimal, but it's actually more accurate just
to keep it in exact form. So now I can rewrite

3172
05:33:25,909 --> 05:33:40,707
my equation, I have f of t equals a times
e to the ln one half over five 750 t. Now

3173
05:33:40,707 --> 05:33:46,750
I can use that to figure out my problem. And
my problem the bone originally contained 200

3174
05:33:46,750 --> 05:33:53,468
grams, that's my a, I want to figure out when
it's going to contain only 40 grams, that's

3175
05:33:53,468 --> 05:34:00,030
my final amount. And so I need to solve for
t. I'll clean things up by dividing both sides

3176
05:34:00,029 --> 05:34:10,501
by 200. Let's say 40 over 200 is 1/5. Now
I'm going to take the ln of both sides. And

3177
05:34:10,501 --> 05:34:19,029
ln and e to the power undo each other. So
I'm left with Elena 1/5 equals ln of one half

3178
05:34:19,029 --> 05:34:27,889
divided by five 750 T. And finally I can solve
for t but super messy but careful use of my

3179
05:34:27,889 --> 05:34:35,509
calculator gives me an answer of 13,351 years
approximately.

3180
05:34:35,509 --> 05:34:41,397
That kind of makes sense in terms of the half
life because to get from 200 to 40 you have

3181
05:34:41,398 --> 05:34:45,638
to decrease by half a little more than two
times right or increasing by half once would

3182
05:34:45,637 --> 05:34:50,659
get you to 100 decreasing to half again would
get you to 50 a little more than 40 and two

3183
05:34:50,659 --> 05:34:57,349
half lives is getting pretty close to 13,000
years. This video introduced a lot of new

3184
05:34:57,349 --> 05:35:04,797
things it introduced continue Less growth
model, which is another equivalent way of

3185
05:35:04,797 --> 05:35:16,939
writing an exponential function. The relationship
is that the B in this example, is the same

3186
05:35:16,939 --> 05:35:27,619
thing as EDR. In that version, it also introduced
the ideas of doubling time. And HalfLife the

3187
05:35:27,619 --> 05:35:32,409
amount of time it takes a quantity to double
or decreased to half in an exponential growth

3188
05:35:32,409 --> 05:35:42,701
model. Recall that a linear equation is equation
like for example, 2x minus y equals one. It's

3189
05:35:42,701 --> 05:35:47,990
an equation without any x squared or y squared
in it, something that could be rewritten in

3190
05:35:47,990 --> 05:35:55,120
the form y equals mx plus b, the equation
for a line a system of linear equations is

3191
05:35:55,119 --> 05:36:03,669
a collection of two or more linear equations.
For example, I could have these two equations.

3192
05:36:03,669 --> 05:36:10,089
A solution to a system of equations is that
an x value and a y value that satisfy both

3193
05:36:10,090 --> 05:36:18,590
of the equations. For example, the ordered
pair two three, that means x equals two, y

3194
05:36:18,590 --> 05:36:26,950
equals three is a solution to this system.
Because if I plug in x equals two, and y equals

3195
05:36:26,950 --> 05:36:33,968
three into the first equation, it checks out
since two times two minus three is equal to

3196
05:36:33,968 --> 05:36:40,170
one. And if I plug in x equals two and y equals
three into the second equation, it also checks

3197
05:36:40,169 --> 05:36:49,279
out two plus three equals five. However, the
ordered pair one for that is x equals one,

3198
05:36:49,279 --> 05:36:55,849
y equals four is not a solution to the system.
Even though this X and Y value work in the

3199
05:36:55,849 --> 05:37:02,707
second equation, since one plus four does
equal five, it doesn't work in the first equation,

3200
05:37:02,707 --> 05:37:10,567
because two times one minus four is not equal
to one. In this video, we'll use systematic

3201
05:37:10,567 --> 05:37:17,229
methods to find the solutions to systems of
linear equations. In this first example, we

3202
05:37:17,229 --> 05:37:22,181
want to solve this system of equations, there
are two main methods we could use, we could

3203
05:37:22,181 --> 05:37:29,900
use the method of substitution, or we could
use the method of elimination. If we use the

3204
05:37:29,900 --> 05:37:38,128
method of substitution, the main idea is to
isolate one variable in one equation, and

3205
05:37:38,128 --> 05:37:44,440
then substitute it in to the other equation.
For example, we can start with the first equation

3206
05:37:44,439 --> 05:37:53,807
3x minus two y equals four, and isolate the
x by adding two y to both sides, and then

3207
05:37:53,808 --> 05:38:01,080
dividing both sides by three. Think I'll rewrite
that a little bit by breaking up the fraction

3208
05:38:01,080 --> 05:38:07,900
into two fractions four thirds plus two thirds
y. Now, I'm going to copy down the second

3209
05:38:07,900 --> 05:38:16,530
equation 5x plus six y equals two. And I'm
going to substitute in my expression for x.

3210
05:38:16,529 --> 05:38:24,887
That gives me five times four thirds plus
two thirds y plus six y equals two. And now

3211
05:38:24,887 --> 05:38:30,750
I've got an equation with only one variable
in it y. So I can solve for y as a number.

3212
05:38:30,750 --> 05:38:38,830
First, I'm going to distribute the five so
that gives me 20 thirds plus 10 thirds y plus

3213
05:38:38,830 --> 05:38:45,548
six y equals two. And now I'm going to keep
all my y terms, my terms with y's in them

3214
05:38:45,547 --> 05:38:51,547
on the left side, but I'll move all my terms
without y's in them to the right side. At

3215
05:38:51,547 --> 05:38:56,110
this point, I could just add up all my fractions
and solve for y. But since I don't really

3216
05:38:56,110 --> 05:39:00,360
like working with fractions, I think I'll
do the trick of clearing the denominators

3217
05:39:00,360 --> 05:39:05,968
here. So I'm going to actually multiply both
sides by my common denominator of three just

3218
05:39:05,968 --> 05:39:10,637
to get rid of the denominators and not have
to work with fractions. So let me write that

3219
05:39:10,637 --> 05:39:20,237
down. Distributing the three, I get 10 y plus
eight t and y equals six minus 20. Now add

3220
05:39:20,238 --> 05:39:28,290
things together. So that's 28 y equals negative
14. So that means that y is going to be negative

3221
05:39:28,290 --> 05:39:32,110
14 over 28, which is negative one half.

3222
05:39:32,110 --> 05:39:38,680
So I've solved for y. And now I can go back
and plug y into either of my equations to

3223
05:39:38,680 --> 05:39:47,977
solve for x, I plug it into my first equation.
So I'm plugging in negative one half for y.

3224
05:39:47,977 --> 05:39:55,058
That gives me 3x plus one equals four. So
3x equals three, which means that x is equal

3225
05:39:55,058 --> 05:40:01,370
to one. I've solved my system of equations
and gotten x equals one On y equals minus

3226
05:40:01,369 --> 05:40:09,680
one half, I can also write that as an ordered
pair, one, negative one half for my solution.

3227
05:40:09,680 --> 05:40:14,297
Now let's go back and solve the same system,
but use a different method, the method of

3228
05:40:14,297 --> 05:40:23,750
elimination, the key idea to the method of
elimination is to multiply each equation by

3229
05:40:23,750 --> 05:40:34,718
a constant to make the coefficients of one
variable match. Let me start by copying down

3230
05:40:34,718 --> 05:40:42,080
my two equations. Say I'm trying to make the
coefficient of x match. One way to do that

3231
05:40:42,080 --> 05:40:48,730
is to multiply the first equation by five,
and the second equation by three. That way,

3232
05:40:48,729 --> 05:40:55,739
the coefficient of x will be 15 for both equations,
so let me do that. So for the first equation,

3233
05:40:55,740 --> 05:41:00,950
I'm going to multiply both sides by five.
And for the second equation, I'm going to

3234
05:41:00,950 --> 05:41:09,979
multiply both sides by three. That gives me
for the first equation 15x minus 10, y equals

3235
05:41:09,979 --> 05:41:19,110
20. And for the second equation, 15x plus
18, y is equal to six. Notice that the equate

3236
05:41:19,110 --> 05:41:25,378
the coefficients of x match. So if I subtract
the second equation from the first, the x

3237
05:41:25,378 --> 05:41:32,779
term will completely go away, it'll be zero
times x, and I'll be left with let's see negative

3238
05:41:32,779 --> 05:41:44,259
10 y minus 18, y is going to give me minus
28 y. And if I do 20 minus six, that's going

3239
05:41:44,259 --> 05:41:52,237
to give me 14. solving for y, I get y is 14
over minus 28, which is minus one half just

3240
05:41:52,238 --> 05:41:59,228
like before. Now we can continue, like we
did in the previous solution, and substitute

3241
05:41:59,227 --> 05:42:06,290
that value of y into either one of the equations.
I'll put it again in here. And my solution

3242
05:42:06,290 --> 05:42:13,280
proceeds as before. So once again, I get the
solution that x equals one and y equals minus

3243
05:42:13,279 --> 05:42:20,599
one half. Before we go on to the next problem,
let me show you graphically what this means.

3244
05:42:20,599 --> 05:42:29,147
Here I've graphed the equations 3x minus two,
y equals four, and 5x plus six y equals two.

3245
05:42:29,148 --> 05:42:36,488
And we can see that these two lines intersect
in the point with coordinates one negative

3246
05:42:36,488 --> 05:42:42,388
one half, just like we predicted by solving
equations algebraically. Let's take a look

3247
05:42:42,387 --> 05:42:49,297
at another system of equations. I'm going
to rewrite the first equation, so the x term

3248
05:42:49,297 --> 05:42:55,329
is on the left side with the y term, and the
constant term stays on the right. And I'll

3249
05:42:55,330 --> 05:43:02,580
rewrite or copy down the second equation.
Since the coefficient of x in the first equation

3250
05:43:02,580 --> 05:43:08,270
is minus four, and the second equation is
three, I'm going to try using the method of

3251
05:43:08,270 --> 05:43:14,119
elimination and multiply the first equation
by three and the second equation by four.

3252
05:43:14,119 --> 05:43:19,009
That'll give me a coefficient of x of negative
12. And the first equation, and 12. And the

3253
05:43:19,009 --> 05:43:23,799
second equation, those are equal and opposite,
right, so I'll be able to add together my

3254
05:43:23,799 --> 05:43:31,619
equations to cancel out access. So let's do
that. My first equation becomes negative 12x

3255
05:43:31,619 --> 05:43:39,009
plus 24, y equals three, and I'll put everything
by three. And my second equation, I'll multiply

3256
05:43:39,009 --> 05:43:46,859
everything by four. So that's 12x minus 24,
y equals eight. Now something kind of funny

3257
05:43:46,860 --> 05:43:53,369
has happened here, not only do the x coefficients
match has in with opposite signs, but the

3258
05:43:53,369 --> 05:44:00,169
Y coefficients do also. So if I add together
my two equations, in order to cancel out the

3259
05:44:00,169 --> 05:44:06,457
x term, I'm also going to cancel out the y
term, and I'll just get zero plus zero is

3260
05:44:06,457 --> 05:44:12,369
equal to three plus eight is 11. Well, that's
a contradiction, we can't have zero equal

3261
05:44:12,369 --> 05:44:19,817
to 11. And that shows that these two equations
actually have no solution.

3262
05:44:19,817 --> 05:44:25,628
Let's look at this situation graphically.
If we graph our two equations, we see that

3263
05:44:25,628 --> 05:44:31,440
they're parallel lines with the same slope.
This might be more clear, if I isolate y in

3264
05:44:31,439 --> 05:44:36,727
each equation, the first equation, I get y
equals, let's see, dividing by eight that's

3265
05:44:36,727 --> 05:44:43,329
the same thing as four eighths or one half
x plus one eight. And the second equation,

3266
05:44:43,330 --> 05:44:50,760
if I isolate y, let's say minus six y equals
minus 3x plus two divided by minus six, that's

3267
05:44:50,759 --> 05:44:59,487
y equals one half x minus 1/3. So indeed,
they have the same slope. And so they're parallel

3268
05:44:59,488 --> 05:45:04,409
with different In intercepts, and so they
can have no intersection. And so it makes

3269
05:45:04,409 --> 05:45:10,729
sense that we have no solution to our system
of linear equations. This kind of system that

3270
05:45:10,729 --> 05:45:18,298
has no solution is called an inconsistent
system. In this third example, yet a third

3271
05:45:18,298 --> 05:45:24,797
behavior happens. This time, I think I'm going
to solve by substitution because I already

3272
05:45:24,797 --> 05:45:30,579
have X with a coefficient of one. So it's
really easy to just isolate X in the first

3273
05:45:30,580 --> 05:45:39,228
equation, and then plug in to the second equation
to get three times six minus five y plus 15,

3274
05:45:39,227 --> 05:45:46,647
y equals 18. If I distribute out, I get the
strange phenomenon that the 15 y's cancel

3275
05:45:46,648 --> 05:45:54,488
and I just get 18 equals 18, which is always
true. This is what's called a dependent system

3276
05:45:54,488 --> 05:46:00,850
of linear equations. If you look more closely,
you can see that the second equation is really

3277
05:46:00,849 --> 05:46:06,169
just a constant multiple, the first equation
is just three times every term is three times

3278
05:46:06,169 --> 05:46:10,939
as big as the corresponding term and the first
equation. So there's no new information in

3279
05:46:10,939 --> 05:46:16,539
the second equation, anything, any x and y
values that satisfy the first one will satisfy

3280
05:46:16,540 --> 05:46:23,250
the second one. So this system of equations
has infinitely many solutions. Any ordered

3281
05:46:23,250 --> 05:46:32,580
pair x y, where X plus five y equals six,
or in other words, X's minus five y plus six

3282
05:46:32,580 --> 05:46:40,318
will satisfy this system of equations. That
would include a y value of zero corresponding

3283
05:46:40,317 --> 05:46:49,477
x value of six or a y value of one corresponding
to an x value of one, or a y value of 1/3.

3284
05:46:49,477 --> 05:46:55,259
Corresponding to an x value of 13 thirds just
by plugging into this equation will work.

3285
05:46:55,259 --> 05:47:00,679
Graphically, if I graph both of these equations,
the lines will just be on top of each other,

3286
05:47:00,680 --> 05:47:06,099
so I'll just see one line. In this video,
we've solved systems of linear equations,

3287
05:47:06,099 --> 05:47:13,199
using the method of substitution and the method
of elimination. We've seen that systems of

3288
05:47:13,200 --> 05:47:19,317
linear equations can have one solution. When
the lines that the equations represent intersect

3289
05:47:19,317 --> 05:47:26,797
in one point, they can be inconsistent, and
have no solutions that corresponds to parallel

3290
05:47:26,797 --> 05:47:33,029
lines, or they can be dependent and have infinitely
many solutions that corresponds to the lines

3291
05:47:33,029 --> 05:47:39,647
lying on top of each other. In this video,
I'll work through a problem involving distance

3292
05:47:39,648 --> 05:47:46,317
rate and time. The key relationship to keep
in mind is that the rate of travel is the

3293
05:47:46,317 --> 05:47:51,559
distance traveled divided by the time it takes
to travel again. For example, if you're driving

3294
05:47:51,560 --> 05:47:57,888
60 miles an hour, that's your rate. And that's
because you're going a distance of 60 miles

3295
05:47:57,887 --> 05:48:02,750
in one hour. Sometimes it's handy to rewrite
that relationship by multiplying both sides

3296
05:48:02,750 --> 05:48:10,020
by T time. And that gives us that R times
T is equal to D. In other words, distance

3297
05:48:10,020 --> 05:48:15,727
is equal to rate times time. There's one more
important principle to keep in mind. And that's

3298
05:48:15,727 --> 05:48:23,590
the idea that rates add. For example, if you
normally walk at three miles per hour, but

3299
05:48:23,590 --> 05:48:29,709
you're walking on a moving sidewalk, that's
going at a rate of two miles per hour, then

3300
05:48:29,709 --> 05:48:34,919
your total speed of travel with respect to
you know something stationary is going to

3301
05:48:34,919 --> 05:48:43,179
be three plus two, or five miles per hour.
All right, that is a formula as our one or

3302
05:48:43,180 --> 05:48:51,207
the first rate per second rate is equal to
the total rate. Let's use those two key ideas.

3303
05:48:51,207 --> 05:48:59,000
distance equals rate times time, and rates
add in the following problem. else's boat

3304
05:48:59,000 --> 05:49:04,119
has a top speed of six miles per hour and
still water. While traveling on a river at

3305
05:49:04,119 --> 05:49:10,319
top speed. She went 10 miles upstream in the
same amount of time she went 30 miles downstream,

3306
05:49:10,319 --> 05:49:15,579
we're supposed to find the rate of the river
currents. I'm going to organize the information

3307
05:49:15,580 --> 05:49:17,568
in this problem into a chart.

3308
05:49:17,567 --> 05:49:23,110
During the course of Elsa stay, there were
two situations we need to keep in mind. For

3309
05:49:23,110 --> 05:49:27,779
one period of time she was going upstream.
And for another period of time she was going

3310
05:49:27,779 --> 05:49:35,439
downstream. For each of those, I'm going to
chart out the distance you traveled. The rate

3311
05:49:35,439 --> 05:49:42,109
she went at and the time it took when she
was going upstream. She went a total distance

3312
05:49:42,110 --> 05:49:49,547
of 10 miles. When she was going downstream
she went a longer distance of 30 miles. But

3313
05:49:49,547 --> 05:49:55,207
the times to travel those two distances were
the same. Since I don't know what that time

3314
05:49:55,207 --> 05:50:02,930
was, I'll just give it a variable I'll call
it T now Think about her rate of travel, the

3315
05:50:02,930 --> 05:50:07,189
rate she traveled upstream was slower because
she was going against the current and faster

3316
05:50:07,189 --> 05:50:11,889
when she was going downstream with the current.
We don't know what the speed of the current

3317
05:50:11,889 --> 05:50:17,628
is, that's what we're trying to figure out.
So maybe I'll give it a variable R. But we

3318
05:50:17,628 --> 05:50:23,488
do know that in still water also can go six
miles per hour. And when she's going downstream,

3319
05:50:23,488 --> 05:50:29,260
since she's going with the direction of the
current rates should add, and her rate downstream

3320
05:50:29,259 --> 05:50:40,279
should be six plus R, that's her rate, and
still water plus the rate of the current.

3321
05:50:40,279 --> 05:50:46,707
On the other hand, when she's going upstream,
then she's going against the current, so her

3322
05:50:46,707 --> 05:50:53,149
rate of six miles per hour, we need to subtract
the rate of the current from that. Now that

3323
05:50:53,150 --> 05:50:57,250
we've charted out our information, we can
turn it into equations using the fact that

3324
05:50:57,250 --> 05:51:04,137
distance equals rate times time, we actually
have two equations 10 equals six minus R times

3325
05:51:04,137 --> 05:51:12,047
T, and 30 is equal to six plus R times T.
Now that we've converted our situation into

3326
05:51:12,047 --> 05:51:17,149
a system of equations, our next job is to
solve the system of equations. In this example,

3327
05:51:17,150 --> 05:51:23,580
I think the easiest way to proceed is to isolate
t in each of the two equations. So in the

3328
05:51:23,580 --> 05:51:28,200
first equation, I'll divide both sides by
six minus r, and a second equation R divided

3329
05:51:28,200 --> 05:51:36,760
by six plus R. That gives me 10 over six minus
r equals T, and 30 over six plus r equals

3330
05:51:36,759 --> 05:51:45,477
t. Now if I set my T variables equal to each
other, I get, I get 10 over six minus r is

3331
05:51:45,477 --> 05:51:52,989
equal to 30 over six plus R. I'm making progress
because now I have a single equation, the

3332
05:51:52,990 --> 05:51:58,670
single variable that I need to solve. Since
the variable R is trapped in the denominator,

3333
05:51:58,669 --> 05:52:03,887
I'm going to proceed by clearing the denominator.
So I'm going to multiply both sides by the

3334
05:52:03,887 --> 05:52:14,529
least common denominator, that is six minus
r times six plus R. Once I cancel things out,

3335
05:52:14,529 --> 05:52:23,689
I get that the six plus r times 10 is equal
to 30 times six minus r, if I distribute,

3336
05:52:23,689 --> 05:52:36,637
I'm going to get 60 plus xR equals 180 minus
30 ar, which I can now solve, let's see, that's

3337
05:52:36,637 --> 05:52:43,909
going to be 40 r is equal to 120. So our,
the speed of my current is going to be three

3338
05:52:43,909 --> 05:52:51,099
miles per hour. This is all that the problem
asked for the speed of the current. If I also

3339
05:52:51,099 --> 05:52:57,579
wanted to solve for the other unknown time,
I could do so by plugging in R into one of

3340
05:52:57,580 --> 05:53:04,270
my equations and solving for T. In this video,
I saw the distance rate and time problem by

3341
05:53:04,270 --> 05:53:12,279
charting out my information for the two situations
and my problem using the fact that rates add

3342
05:53:12,279 --> 05:53:17,930
to fill in some of my boxes, and then using
the formula distance equals rate times time

3343
05:53:17,930 --> 05:53:24,080
to build a system of equations. In this video,
I'll do a standard mixture problem in which

3344
05:53:24,080 --> 05:53:30,340
we have to figure out what quantity of two
solutions to mix together. household bleach

3345
05:53:30,340 --> 05:53:39,387
contains 6% sodium hypochlorite. The other
94% is water. How much household bleach should

3346
05:53:39,387 --> 05:53:50,957
be combined with 70 liters of a weaker 1%
hypochlorite solution in order to form a solution,

3347
05:53:50,957 --> 05:53:59,349
that's 2.5% sodium hypochlorite. I want to
turn this problem into a system of equations.

3348
05:53:59,349 --> 05:54:05,977
So I'm asking myself what quantities are going
to be equal to each other? Well, the total

3349
05:54:05,977 --> 05:54:14,829
amount of sodium hypochlorite that has symbol
and a co o before mixing,

3350
05:54:14,830 --> 05:54:23,270
it should equal the total amount of sodium
hypochlorite after mixing. Also, the total

3351
05:54:23,270 --> 05:54:30,378
amount of water before mixing should equal
the total amount of water after. And finally,

3352
05:54:30,378 --> 05:54:37,790
there's just the total amount of solution.
In my two jugs, sodium hypochlorite together

3353
05:54:37,790 --> 05:54:47,290
with water should equal the total amount of
solution after that gives me a hint for what

3354
05:54:47,290 --> 05:54:51,909
I'm looking for. But before I start reading
out equations, I find it very helpful to chart

3355
05:54:51,909 --> 05:55:05,387
out my quantities. So I've got the 6% solution.
The household leech, I've got the 1% solution.

3356
05:55:05,387 --> 05:55:16,128
And I've got my Desired Ending 2.5% solution.
Now on each of those solutions, I've got a

3357
05:55:16,128 --> 05:55:26,580
certain volume of sodium hypochlorite. I've
also got a volume of water. And I've got a

3358
05:55:26,580 --> 05:55:37,200
total volume of solution. Let me see which
of these boxes I can actually fill in, I know

3359
05:55:37,200 --> 05:55:45,640
that I'm adding 70 liters of the 1% solution.
So I can put a 70 in the total volume of solution

3360
05:55:45,640 --> 05:55:53,509
here. I don't know what volume of the household
bleach, I want to add, that's what I'm trying

3361
05:55:53,509 --> 05:56:02,949
to find out. So I'm going to just call that
volume x. Now since my 2.5% solution is made

3362
05:56:02,950 --> 05:56:08,409
by combining my other two solutions, I know
its volume is going to be the sum of these

3363
05:56:08,409 --> 05:56:16,680
two volumes, so I'll write 70 plus x in this
box. Now, the 6% solution means that whatever

3364
05:56:16,680 --> 05:56:22,580
the volume of solution is, 6% of that is the
sodium hypochlorite. So the volume of the

3365
05:56:22,580 --> 05:56:31,070
sodium hypochlorite is going to be 0.06 times
x, the volume of water and that solution is

3366
05:56:31,069 --> 05:56:40,000
whatever's left, so that's going to be x minus
0.06x, or point nine four times X with the

3367
05:56:40,000 --> 05:56:46,988
following the same reasoning for the 1% solution
1% of the 70 liters is a sodium hypochlorite.

3368
05:56:46,988 --> 05:56:56,530
So that's going to be 0.01 times 70. Or point
seven, the volume of water and that solution

3369
05:56:56,529 --> 05:57:05,699
is going to be 99% or point nine, nine times
seven day. That works out to 69.3. Finally,

3370
05:57:05,700 --> 05:57:15,317
for the 2.5% solution, the volume of the sodium
hypochlorite is going to be 0.0 to five times

3371
05:57:15,317 --> 05:57:22,599
the volume of solution 70 plus x and the volume
of water is going to be the remainder. So

3372
05:57:22,599 --> 05:57:32,919
that's 0.975 times 70 plus x. Now I've already
used the fact that the volume of solutions

3373
05:57:32,919 --> 05:57:39,359
before added up is the volume of solution
after in writing a 70 plus x in this box.

3374
05:57:39,360 --> 05:57:43,830
But I haven't yet used the fact that the volume
of the sodium hypochlorite is preserved before

3375
05:57:43,830 --> 05:57:53,040
and after. So I can write that down as an
equation. So that means 0.06x plus 0.7 is

3376
05:57:53,040 --> 05:58:01,270
equal to 0.025 times 70 plus x. Now I've got
an equation with a variable. I'll try to solve

3377
05:58:01,270 --> 05:58:06,850
it. Since I don't like all these decimals,
I'm going to multiply both sides of my equation

3378
05:58:06,849 --> 05:58:15,057
by let's see, 1000 should get rid of all the
decimals. After distributing, I get 60x plus

3379
05:58:15,058 --> 05:58:27,090
700 equals 25 times 70 plus x. Distributing
some more, I get 60x plus 700 is equal to

3380
05:58:27,090 --> 05:58:42,409
1750 plus 25x. So let's see 60 minus 25 is
35x is equal to 1050. Which gives me that

3381
05:58:42,409 --> 05:58:49,139
x equals 30 liters of the household bleach.

3382
05:58:49,139 --> 05:58:54,610
Notice that I never actually had to use the
fact that the water quantity of water before

3383
05:58:54,610 --> 05:59:00,659
mixing is equal to the quantity of water after
that is I never use the information in this

3384
05:59:00,659 --> 05:59:08,450
column. In fact, that information is redundant.
Once I know that the quantities of sodium

3385
05:59:08,450 --> 05:59:15,520
hypochlorite add up, and the total volume
of solutions add up. The fact that that volume

3386
05:59:15,520 --> 05:59:21,689
of waters add up is just redundant information.
The techniques to use to solve this equation

3387
05:59:21,689 --> 05:59:28,039
involving solutions can be used to solve many
many other equations involving mixtures of

3388
05:59:28,040 --> 05:59:36,977
items. My favorite method is to first make
a chart involving the types of mixtures and

3389
05:59:36,977 --> 05:59:43,930
the types of items in your mixture. Fill in
as many boxes size I can and then use the

3390
05:59:43,930 --> 05:59:53,159
fact that the quantities add. This video is
about rational functions and their graphs.

3391
05:59:53,159 --> 05:59:59,000
Recall that a rational function is a function
that can be written as a ratio or quotient

3392
05:59:59,000 --> 06:00:07,308
of two power. No Here's an example. The simpler
function, f of x equals one over x is also

3393
06:00:07,308 --> 06:00:14,520
considered a rational function, you can think
of one and x as very simple polynomials. The

3394
06:00:14,520 --> 06:00:20,110
graph of this rational function is shown here.
This graph looks different from the graph

3395
06:00:20,110 --> 06:00:27,290
of a polynomial. For one thing, its end behavior
is different. The end behavior of a function

3396
06:00:27,290 --> 06:00:33,030
is the way the graph looks when x goes through
really large positive, or really large negative

3397
06:00:33,029 --> 06:00:38,840
numbers, we've seen that the end behavior
of a polynomial always looks like one of these

3398
06:00:38,840 --> 06:00:44,700
cases. That is why marches off to infinity
or maybe negative infinity, as x gets really

3399
06:00:44,700 --> 06:00:50,200
big or really negative. But this rational
function has a different type of end behavior.

3400
06:00:50,200 --> 06:00:54,600
Notice, as x gets really big, the y values
are leveling off

3401
06:00:54,599 --> 06:01:00,779
at about a y value of three. And similarly,
as x values get really negative, our graph

3402
06:01:00,779 --> 06:01:07,849
is leveling off near the line y equals three,
I'll draw that line, y equals three on my

3403
06:01:07,849 --> 06:01:16,119
graph, that line is called a horizontal asymptote.
A horizontal asymptote is a horizontal line

3404
06:01:16,119 --> 06:01:21,529
that our graph gets closer and closer to as
x goes to infinity, or as X goes to negative

3405
06:01:21,529 --> 06:01:26,750
infinity, or both. There's something else
that's different about this graph from a polynomial

3406
06:01:26,750 --> 06:01:32,599
graph, look at what happens as x gets close
to negative five. As we approach negative

3407
06:01:32,599 --> 06:01:37,717
five with x values on the right, our Y values
are going down towards negative infinity.

3408
06:01:37,718 --> 06:01:42,690
And as we approach the x value of negative
five from the left, our Y values are going

3409
06:01:42,689 --> 06:01:49,340
up towards positive infinity. We say that
this graph has a vertical asymptote at x equals

3410
06:01:49,340 --> 06:01:56,439
negative five. A vertical asymptote is a vertical
line that the graph gets closer and closer

3411
06:01:56,439 --> 06:02:02,567
to. Finally, there's something really weird
going on at x equals two, there's a little

3412
06:02:02,567 --> 06:02:10,000
open circle there, like the value at x equals
two is dug out. That's called a hole. A hole

3413
06:02:10,000 --> 06:02:16,718
is a place along the curve of the graph where
the function doesn't exist. Now that we've

3414
06:02:16,718 --> 06:02:19,600
identified some of the features of our rational
functions graph,

3415
06:02:19,599 --> 06:02:24,627
I want to look back at the equation and see
how we could have predicted those features

3416
06:02:24,628 --> 06:02:31,040
just by looking at the equation. To find horizontal
asymptotes, we need to look at what our function

3417
06:02:31,040 --> 06:02:36,580
is doing when x goes through really big positive
or really big negative numbers. Looking at

3418
06:02:36,580 --> 06:02:42,990
our equation for our function, the numerator
is going to be dominated by the 3x squared

3419
06:02:42,990 --> 06:02:47,659
term when x is really big, right, because
three times x squared is going to be absolutely

3420
06:02:47,659 --> 06:02:53,619
enormous compared to this negative 12. If
x is a big positive or negative number, in

3421
06:02:53,619 --> 06:02:59,610
the denominator, the denominator will be dominated
by the x squared term. Again, if x is a really

3422
06:02:59,610 --> 06:03:04,920
big positive or negative number, like a million,
a million squared will be much, much bigger

3423
06:03:04,919 --> 06:03:10,797
than three times a million or negative 10.
For that reason, to find the end behavior,

3424
06:03:10,797 --> 06:03:18,387
or the horizontal asymptote, for our function,
we just need to look at the term on the numerator

3425
06:03:18,387 --> 06:03:22,637
and the term on the denominator that have
the highest exponent, those are the ones that

3426
06:03:22,637 --> 06:03:29,468
dominate the expression in size. So as x gets
really big, our functions y values are going

3427
06:03:29,468 --> 06:03:37,220
to be approximately 3x squared over x squared,
which is three. That's why we have a horizontal

3428
06:03:37,220 --> 06:03:45,430
asymptote at y equals three. Now our vertical
asymptotes, those tend to occur where our

3429
06:03:45,430 --> 06:03:51,308
denominator of our function is zero. That's
because the function doesn't exist when our

3430
06:03:51,308 --> 06:03:56,150
denominator is zero. And when we get close
to that place where our denominator is zero,

3431
06:03:56,150 --> 06:04:01,680
we're going to be dividing by tiny, tiny numbers,
which will make our Y values really big in

3432
06:04:01,680 --> 06:04:06,920
magnitude. So to check where our denominators
zero, let's factor our function. In fact,

3433
06:04:06,919 --> 06:04:11,909
I'm going to go ahead and factor the numerator
and the denominator. So the numerator factors,

3434
06:04:11,909 --> 06:04:18,468
let's see, pull out the three, I get x squared
minus four, factor in the denominator, that

3435
06:04:18,468 --> 06:04:25,040
factors into X plus five times x minus two,
I can factor a little the numerator a little

3436
06:04:25,040 --> 06:04:34,950
further, that's three times x minus two times
x plus two over x plus 5x minus two. Now,

3437
06:04:34,950 --> 06:04:41,319
when x is equal to negative five, my denominator
will be zero, but my numerator will not be

3438
06:04:41,319 --> 06:04:49,398
zero. That's what gives me the vertical asymptote
at x equals negative five. Notice that when

3439
06:04:49,398 --> 06:04:57,292
x equals two, the denominator is zero, but
the numerator is also zero. In fact, if I

3440
06:04:57,292 --> 06:05:01,909
cancelled the x minus two factor from the
numerator, and in nominator, I get a simplified

3441
06:05:01,909 --> 06:05:11,279
form for my function that agrees with my original
function as long as x is not equal to two.

3442
06:05:11,279 --> 06:05:16,760
That's because when x equals two, the simplified
function exists, but the original function

3443
06:05:16,760 --> 06:05:23,637
does not, it's zero over zero, it's undefined.
But for every other x value, including x values

3444
06:05:23,637 --> 06:05:29,829
near x equals to our original functions just
the same as this function. And that's why

3445
06:05:29,830 --> 06:05:36,468
our function only has a vertical asymptote
at x equals negative five, not one at x equals

3446
06:05:36,468 --> 06:05:41,069
two, because the x minus two factor is no
longer in the function after simplifying,

3447
06:05:41,069 --> 06:05:45,779
it does have a hole at x equals two, because
the original function is not defined there,

3448
06:05:45,779 --> 06:05:52,189
even though the simplified version is if we
want to find the y value of our hole, we can

3449
06:05:52,189 --> 06:05:59,599
just plug in x equals two into our simplified
version of our function, that gives a y value

3450
06:05:59,599 --> 06:06:06,769
of three times two plus two over two plus
seven, or 12 ninths, which simplifies to four

3451
06:06:06,770 --> 06:06:15,477
thirds. So our whole is that to four thirds.
Now that we've been through one example in

3452
06:06:15,477 --> 06:06:22,520
detail, let's summarize our findings. We find
the vertical asymptotes and the holes by looking

3453
06:06:22,520 --> 06:06:29,477
where the denominator is zero. The holes happen
where the denominator and numerator are both

3454
06:06:29,477 --> 06:06:35,930
zero and those factors cancel out. The vertical
asymptotes are all other x values where the

3455
06:06:35,930 --> 06:06:42,010
denominator is zero, we find the horizontal
asymptotes. By considering the highest power

3456
06:06:42,009 --> 06:06:47,477
term on the numerator and the denominator,
I'll explain this process in more detail in

3457
06:06:47,477 --> 06:06:55,567
three examples. In the first example, if we
circle the highest power terms, that simplifies

3458
06:06:55,567 --> 06:07:03,648
to 5x over 3x squared, which is five over
3x. As x gets really big, the denominator

3459
06:07:03,648 --> 06:07:09,958
is going to be huge. So I'm going to be dividing
five by a huge, huge number, that's going

3460
06:07:09,957 --> 06:07:15,147
to be going very close to zero. And therefore
we have a horizontal asymptote

3461
06:07:15,148 --> 06:07:24,398
at y equals zero. In the second example, the
highest power terms, 2x cubed, over 3x cubed

3462
06:07:24,398 --> 06:07:30,200
simplifies to two thirds. So as x gets really
big, we're going to be heading towards two

3463
06:07:30,200 --> 06:07:36,950
thirds, and we have a horizontal asymptote
at y equals two thirds. In the third example,

3464
06:07:36,950 --> 06:07:45,270
the highest power terms, x squared over 2x
simplifies to x over two. As x gets really

3465
06:07:45,270 --> 06:07:51,887
big, x over two is getting really big. And
therefore, we don't have a horizontal asymptote

3466
06:07:51,887 --> 06:07:59,279
at all. This is going to infinity, when x
gets through go through big positive numbers,

3467
06:07:59,279 --> 06:08:07,237
and is going to negative infinity when x goes
through a big negative numbers. So in this

3468
06:08:07,238 --> 06:08:12,840
case, the end behavior is kind of like that
of a polynomial, and there's no horizontal

3469
06:08:12,840 --> 06:08:18,887
asymptote. In general, when the degree of
the numerator is smaller than the degree of

3470
06:08:18,887 --> 06:08:23,520
the denominator, we're in this first case
where the denominator gets really big compared

3471
06:08:23,520 --> 06:08:29,250
to the numerator and we go to zero. In the
second case, where the degree of the numerator

3472
06:08:29,250 --> 06:08:34,297
and the degree of the dominant are equal,
things cancel out, and so we get a horizontal

3473
06:08:34,297 --> 06:08:42,750
asymptote at the y value, that's equal to
the ratio of the leading coefficients. Finally,

3474
06:08:42,750 --> 06:08:46,968
in the third case, when the degree of the
numerator is bigger than the degree of the

3475
06:08:46,968 --> 06:08:52,290
denominator, then the numerator is getting
really big compared to the denominator, so

3476
06:08:52,290 --> 06:08:57,840
we end up with no horizontal asymptote. Final
Finally, let's apply all these observations

3477
06:08:57,840 --> 06:09:03,379
to one more example. Please pause the video
and take a moment to find the vertical asymptotes,

3478
06:09:03,379 --> 06:09:09,817
horizontal asymptotes and holes for this rational
function. To find the vertical asymptotes

3479
06:09:09,817 --> 06:09:15,547
and holes, we need to look at where the denominator
is zero. In fact, it's going to be handy to

3480
06:09:15,547 --> 06:09:20,289
factor both the numerator and the denominator.
Since there if there are any common factors,

3481
06:09:20,290 --> 06:09:25,128
we might have a whole instead of a vertical
asymptote. The numerator is pretty easy to

3482
06:09:25,128 --> 06:09:31,887
factor. Let's see that's 3x times x plus one
for the denominator or first factor out an

3483
06:09:31,887 --> 06:09:40,029
x. And then I'll factor some more using a
guess and check method. I know that I'll need

3484
06:09:40,029 --> 06:09:47,967
a 2x and an X to multiply together to the
2x squared and I'll need a three and a minus

3485
06:09:47,968 --> 06:09:56,030
one or alpha minus three and a one. Let's
see if that works. If I multiply out 2x minus

3486
06:09:56,029 --> 06:10:02,860
one times x plus three that does get me back
to x squared plus 5x minus three, so that

3487
06:10:02,861 --> 06:10:08,590
checks out. Now I noticed that I have a common
factor of x in both the numerator and the

3488
06:10:08,590 --> 06:10:14,270
denominator. So that's telling me I'm going
to have a hole at x equals zero. In fact,

3489
06:10:14,270 --> 06:10:21,889
I could rewrite my rational function by cancelling
out that common factor. And that's equivalent,

3490
06:10:21,889 --> 06:10:28,290
as long as x is not equal to zero. So the
y value of my whole is what I get when I plug

3491
06:10:28,290 --> 06:10:35,628
zero into my simplified version, that would
be three times zero plus one over two times

3492
06:10:35,628 --> 06:10:42,797
zero minus one times zero plus three, which
is three over negative three or minus one.

3493
06:10:42,797 --> 06:10:49,619
So my whole is at zero minus one. Now all
the remaining places in my denominator that

3494
06:10:49,619 --> 06:10:55,239
make my denominator zero will get me vertical
asymptotes. So I'll have a vertical asymptote,

3495
06:10:55,240 --> 06:11:04,780
when 2x minus one times x plus three equals
zero, that is, when 2x minus one is zero,

3496
06:11:04,779 --> 06:11:13,840
or x plus three is zero. In other words, when
x is one half, or x equals negative three.

3497
06:11:13,840 --> 06:11:20,299
Finally, to find my horizontal asymptotes,
I just need to consider the highest power

3498
06:11:20,299 --> 06:11:28,759
term in the numerator and the denominator.
That simplifies to three over 2x, which is

3499
06:11:28,759 --> 06:11:35,359
bottom heavy, right? When x gets really big,
this expression is going to zero. And that

3500
06:11:35,360 --> 06:11:41,718
means that we have a horizontal asymptote
at y equals zero. So we found the major features

3501
06:11:41,718 --> 06:11:48,150
of our graph, the whole, the vertical asymptotes
and the horizontal asymptotes. Together, this

3502
06:11:48,150 --> 06:11:54,458
would give us a framework for what the graph
of our function looks like. horizontal asymptote

3503
06:11:54,457 --> 06:12:02,599
at y equals zero, vertical asymptotes at x
equals one half, and x equals minus three

3504
06:12:02,599 --> 06:12:09,909
at a hole at the point zero minus one. plotting
a few more points, or using a graphing calculator

3505
06:12:09,909 --> 06:12:17,950
of graphing program, we can see that our actual
function will look something like this.

3506
06:12:17,950 --> 06:12:24,159
Notice that the x intercept when x is negative
one, corresponds to where the numerator of

3507
06:12:24,159 --> 06:12:29,529
our rational function or reduced rational
function is equal to zero. That's because

3508
06:12:29,529 --> 06:12:33,949
a zero on the numerator that doesn't make
the denominator zero makes the whole function

3509
06:12:33,950 --> 06:12:40,000
zero. And an X intercept is where the y value
of the whole function is zero. In this video,

3510
06:12:40,000 --> 06:12:44,797
we learned how to find horizontal asymptotes
have rational functions. By looking at the

3511
06:12:44,797 --> 06:12:50,779
highest power terms, we learned to find the
vertical asymptotes and holes. By looking

3512
06:12:50,779 --> 06:12:56,307
at the factored version of the functions.
The holes correspond to the x values that

3513
06:12:56,308 --> 06:13:03,317
make the numerator and denominator zero, his
corresponding factors cancel. The vertical

3514
06:13:03,317 --> 06:13:09,099
asymptotes correspond to the x values that
make the denominator zero, even after factoring

3515
06:13:09,099 --> 06:13:13,957
any any common and in common factors in the
numerator denominator.

3516
06:13:13,957 --> 06:13:19,750
This video is about combining functions by
adding them subtracting them multiplying and

3517
06:13:19,750 --> 06:13:26,270
dividing them. Suppose we have two functions,
f of x equals x plus one and g of x equals

3518
06:13:26,270 --> 06:13:34,218
x squared. One way to combine them is by adding
them together. This notation, f plus g of

3519
06:13:34,218 --> 06:13:43,909
x means the function defined by taking f of
x and adding it to g of x. So for our functions,

3520
06:13:43,909 --> 06:13:50,957
that means we take x plus one and add x squared,
I can rearrange that as the function x squared

3521
06:13:50,957 --> 06:14:01,899
plus x plus one. So f plus g evaluated on
x means x squared plus x plus one. And if

3522
06:14:01,900 --> 06:14:08,810
I wanted to evaluate f plus g, on the number
two, that would be two squared plus two plus

3523
06:14:08,810 --> 06:14:18,218
one, or seven. Similarly, the notation f minus
g of x means the function we get by taking

3524
06:14:18,218 --> 06:14:25,490
f of x and subtracting g of x. So that would
be x plus one minus x squared. And if I wanted

3525
06:14:25,490 --> 06:14:35,390
to take f minus g evaluated at one, that would
be one plus one minus one squared, or one,

3526
06:14:35,389 --> 06:14:44,599
the notation F dot g of x, which is sometimes
also written just as f g of x. That means

3527
06:14:44,599 --> 06:14:52,969
we take f of x times g of x. In other words,
x plus one times x squared, which could be

3528
06:14:52,970 --> 06:15:02,888
simplified as x cubed plus x squared. The
notation f divided by g of x means I take

3529
06:15:02,887 --> 06:15:11,919
f of x and divided by g of x. So that would
be x plus one divided by x squared. In this

3530
06:15:11,919 --> 06:15:19,717
figure, the blue graph represents h of x.
And the red graph represents the function

3531
06:15:19,718 --> 06:15:26,690
p of x, we're asked to find h minus p of zero.

3532
06:15:26,689 --> 06:15:32,259
We don't have any equations to work with,
but that's okay. We know that for any x, h

3533
06:15:32,259 --> 06:15:40,877
minus p of x is defined as h of x minus p
of x. So for x equals zero, H minus p of zero

3534
06:15:40,878 --> 06:15:48,968
is going to be h of zero minus p of zero.
Using the graph, we can find h of zero by

3535
06:15:48,968 --> 06:15:55,830
finding the value of zero on the x axis, and
finding the corresponding y value for the

3536
06:15:55,830 --> 06:16:04,120
function h of x. So that's about 1.8. Now
P of zero, we can find similarly by looking

3537
06:16:04,119 --> 06:16:10,090
for zero on the x axis, and finding the corresponding
y value for the function p of x, and that's

3538
06:16:10,090 --> 06:16:18,939
a y value of one 1.8 minus one is 0.8. So
that's our approximate value for H minus p

3539
06:16:18,939 --> 06:16:28,699
of zero. If we want to find P times h of negative
three, again, we can rewrite that as P of

3540
06:16:28,700 --> 06:16:34,899
negative three times h of negative three.
And using the graphs, we see that for an x

3541
06:16:34,899 --> 06:16:45,190
value of negative three, the y value for P
is two. And the x value of negative three

3542
06:16:45,189 --> 06:16:50,180
corresponds to a y value of negative two for
H.

3543
06:16:50,180 --> 06:16:55,610
Two times negative two is negative four. So
that's our value for P times h of negative

3544
06:16:55,610 --> 06:16:56,610
three.

3545
06:16:56,610 --> 06:17:03,878
In this video, we saw how to add two functions,
subtract two functions, multiply two functions,

3546
06:17:03,878 --> 06:17:13,290
and divide two functions in the following
way. When you compose two functions, you apply

3547
06:17:13,290 --> 06:17:19,940
the first function, and then you apply the
second function to the output of the first

3548
06:17:19,939 --> 06:17:27,349
function. For example, the first function
might compute population size from time in

3549
06:17:27,349 --> 06:17:35,807
years. So its input would be time in years,
since a certain date, as output would be number

3550
06:17:35,808 --> 06:17:45,620
of people in the population. The second function
g, might compute health care costs as a function

3551
06:17:45,619 --> 06:17:53,919
of population size. So it will take population
size as input, and its output will be healthcare

3552
06:17:53,919 --> 06:18:00,189
costs. If you put these functions together,
that is compose them, then you'll go all the

3553
06:18:00,189 --> 06:18:07,770
way from time in years to healthcare costs.
This is your composition, g composed with

3554
06:18:07,770 --> 06:18:16,409
F. The composition of two functions, written
g with a little circle, f of x is defined

3555
06:18:16,409 --> 06:18:26,919
as follows. g composed with f of x is G evaluated
on f of x, we can think of it schematically

3556
06:18:26,919 --> 06:18:36,009
and diagram f x on a number x and produces
a number f of x, then g takes that output

3557
06:18:36,009 --> 06:18:45,137
f of x and produces a new number, g of f of
x. Our composition of functions g composed

3558
06:18:45,137 --> 06:18:51,950
with F is the function that goes all the way
from X to g of f of x. Let's work out some

3559
06:18:51,950 --> 06:18:58,920
examples where our functions are defined by
tables of values. If we want to find g composed

3560
06:18:58,919 --> 06:19:08,189
with F of four, by definition, this means
g of f of four. To evaluate this expression,

3561
06:19:08,189 --> 06:19:15,029
we always work from the inside out. So we
start with the x value of four, and we find

3562
06:19:15,029 --> 06:19:23,739
f of four, using the table of values for f
of x, when x equals four, f of x is seven,

3563
06:19:23,740 --> 06:19:33,128
so we can replace F of four with the number
seven. Now we need to evaluate g of seven,

3564
06:19:33,128 --> 06:19:39,420
seven becomes our new x value in our table
of values for G, the x value of seven corresponds

3565
06:19:39,419 --> 06:19:47,949
to the G of X value of 10. So g of seven is
equal to 10. We found that g composed with

3566
06:19:47,950 --> 06:19:56,387
F of four is equal to 10. If instead we want
to find f composed with g of four, well, we

3567
06:19:56,387 --> 06:20:04,547
can rewrite that as f of g of four Again work
from the inside out. Now we're trying to find

3568
06:20:04,547 --> 06:20:10,250
g of four. So four is our x value. And we
use our table of values for G to see that

3569
06:20:10,250 --> 06:20:20,430
g of four is one. So we replaced you a four
by one. And now we need to evaluate f of one.

3570
06:20:20,430 --> 06:20:29,817
Using our table for F values, f of one is
eight. Notice that when we've computed g of

3571
06:20:29,817 --> 06:20:36,907
f of four, we got a different answer than
when we computed f of g of four. And in general,

3572
06:20:36,907 --> 06:20:43,509
g composed with F is not the same thing as
f composed with g. Please pause the video

3573
06:20:43,509 --> 06:20:49,808
and take a moment to compute the next two
examples. We can replace f composed with F

3574
06:20:49,809 --> 06:20:56,708
of two by the equivalent expression, f of
f of two. Working from the inside out, we

3575
06:20:56,707 --> 06:21:06,127
know that f of two is three, and f of three
is six. If we want to find f composed with

3576
06:21:06,128 --> 06:21:17,010
g of six, rewrite that as f of g of six, using
the table for g, g of six is eight. But F

3577
06:21:17,009 --> 06:21:25,109
of eight, eight is not on the table as an
x value for the for the f function. And so

3578
06:21:25,110 --> 06:21:35,148
there is no F of eight, this does not exist,
we can say that six is not in the domain,

3579
06:21:35,148 --> 06:21:42,970
for F composed with g. Even though it was
in the domain of g, we couldn't follow all

3580
06:21:42,970 --> 06:21:49,780
the way through and get a value for F composed
with g of six. Next, let's turn our attention

3581
06:21:49,779 --> 06:21:54,399
to the composition of functions that are given
by equations.

3582
06:21:54,400 --> 06:22:03,110
p of x is x squared plus x and q of x as negative
2x. We want to find q composed with P of one.

3583
06:22:03,110 --> 06:22:14,047
As usual, I can rewrite this as Q of P of
one and work from the inside out. P of one

3584
06:22:14,047 --> 06:22:21,567
is one squared plus one, so that's two. So
this is the same thing as Q of two. But Q

3585
06:22:21,567 --> 06:22:28,009
of two is negative two times two or negative
four. So this evaluates to negative four.

3586
06:22:28,009 --> 06:22:35,237
In this next example, we want to find q composed
with P of some arbitrary x, or rewrite it

3587
06:22:35,238 --> 06:22:44,370
as usual as Q of p of x and work from the
inside out. Well, p of x, we know the formula

3588
06:22:44,369 --> 06:22:52,047
for that. That's x squared plus x. So I can
replace my P of x with that expression. Now,

3589
06:22:52,047 --> 06:22:59,919
I'm stuck with evaluating q on x squared plus
x. Well, Q of anything is negative two times

3590
06:22:59,919 --> 06:23:08,239
that thing. So q of x squared plus x is going
to be negative two times the quantity x squared

3591
06:23:08,240 --> 06:23:13,808
plus x, what I've done is I've substituted
in the whole expression x squared plus x,

3592
06:23:13,808 --> 06:23:20,010
where I saw the X in this formula for q of
x, it's important to use the parentheses here.

3593
06:23:20,009 --> 06:23:24,297
So that will be multiplying negative two by
the whole expression and not just by the first

3594
06:23:24,297 --> 06:23:32,147
piece, I can simplify this a bit as negative
2x squared minus 2x. And that's my expression

3595
06:23:32,148 --> 06:23:40,478
for Q composed with p of x. Notice that if
I wanted to compute q composed with P of one,

3596
06:23:40,477 --> 06:23:45,309
which I already did in the first problem,
I could just use this expression now, negative

3597
06:23:45,310 --> 06:23:52,740
two times one squared minus two and I get
negative four, just like I did before. Let's

3598
06:23:52,740 --> 06:24:00,340
try another one. Let's try p composed with
q of x. First I read write this P of q of

3599
06:24:00,340 --> 06:24:07,009
x. Working from the inside out, I can replace
q of x with negative 2x. So I need to compute

3600
06:24:07,009 --> 06:24:15,419
P of negative 2x. Here's my formula for P.
to compute P of this expression, I need to

3601
06:24:15,419 --> 06:24:23,657
plug in this expression everywhere I see an
x in the formula for P. So that means negative

3602
06:24:23,657 --> 06:24:29,819
2x squared plus negative 2x. Again, being
careful to use parentheses to make sure I

3603
06:24:29,819 --> 06:24:40,457
plug in the entire expression in forex. let
me simplify. This is 4x squared minus 2x.

3604
06:24:40,457 --> 06:24:50,407
Notice that I got different expressions for
Q of p of x, and for P of q of x. Once again,

3605
06:24:50,407 --> 06:24:57,720
we see that q composed with P is not necessarily
equal to P composed with Q. Please pause the

3606
06:24:57,720 --> 06:25:06,850
video and try this last example yourself.
rewriting. And working from the inside out,

3607
06:25:06,849 --> 06:25:13,750
we're going to replace p of x with its expression
x squared plus x. And then we need to evaluate

3608
06:25:13,750 --> 06:25:22,090
p on x squared plus x. That means we plug
in x squared plus x, everywhere we see an

3609
06:25:22,090 --> 06:25:29,657
x in this formula, so that's x squared plus
x quantity squared plus x squared plus x.

3610
06:25:29,657 --> 06:25:36,809
Once again, I can simplify by distributing
out, that gives me x to the fourth plus 2x

3611
06:25:36,810 --> 06:25:45,020
cubed plus x squared plus x squared plus x,
or x to the fourth plus 2x cubed plus 2x squared

3612
06:25:45,020 --> 06:25:52,189
plus x. In this last set of examples, we're
asked to go backwards, we're given a formula

3613
06:25:52,189 --> 06:25:57,919
for a function of h of x. But we're supposed
to rewrite h of x as a composition of two

3614
06:25:57,919 --> 06:26:04,359
functions, F and G. Let's think for a minute,
which of these two functions gets applied

3615
06:26:04,360 --> 06:26:13,619
first, f composed with g of x, let's see,
that means f of g of x. And since we evaluate

3616
06:26:13,619 --> 06:26:21,250
these expressions from the inside out, we
must be applying g first, and then F. In order

3617
06:26:21,250 --> 06:26:27,779
to figure out what what f and g could be,
I like to draw a box around some thing inside

3618
06:26:27,779 --> 06:26:32,779
my expression for H, so I'm going to draw
a box around x squared plus seven, then whatever's

3619
06:26:32,779 --> 06:26:38,529
inside the box, that'll be my function, g
of x, the first function that gets applied,

3620
06:26:38,529 --> 06:26:43,849
whatever happens to the box, in this case,
taking the square root sign, that becomes

3621
06:26:43,849 --> 06:26:46,869
my outside function, my second function f.

3622
06:26:46,869 --> 06:26:54,750
So here, we're gonna say g of x is equal to
x squared plus seven, and f of x is equal

3623
06:26:54,750 --> 06:27:00,939
to the square root of x, let's just check
and make sure that this works. So I need to

3624
06:27:00,939 --> 06:27:08,069
check that when I take the composition, f
composed with g, I need to get the same thing

3625
06:27:08,069 --> 06:27:17,029
as my original h. So let's see, if I do f
composed with g of x, well, by definition,

3626
06:27:17,029 --> 06:27:23,397
that's f of g of x, working from the inside
out, I can replace g of x with its formula

3627
06:27:23,398 --> 06:27:30,978
x squared plus seven. So I need to evaluate
f of x squared plus seven. That means I plug

3628
06:27:30,977 --> 06:27:37,500
in x squared plus seven, into the formula
for for F. So that becomes the square root

3629
06:27:37,500 --> 06:27:44,128
of x squared plus seven to the it works because
it matches my original equation. So we found

3630
06:27:44,128 --> 06:27:48,180
a correct answer a correct way of breaking
h down as a composition of two functions.

3631
06:27:48,180 --> 06:27:54,680
But I do want to point out, this is not the
only correct answer. I'll write down my formula

3632
06:27:54,680 --> 06:27:59,738
for H of X again, and this time, I'll put
the box in a different place, I'll just box

3633
06:27:59,738 --> 06:28:08,128
the x squared. If I did that, then my inside
function of my first function, g of x would

3634
06:28:08,128 --> 06:28:17,450
be x squared. And my second function is what
happens to the box. So my f of x is what happens

3635
06:28:17,450 --> 06:28:23,898
to the box, and the box gets added seven to
it, and taking the square root. So in other

3636
06:28:23,898 --> 06:28:32,260
words, f of x is going to be the square root
of x plus seven. Again, I can check that this

3637
06:28:32,259 --> 06:28:39,727
works. If I do f composed with g of x, that's
f of g of x. So now g of x is x squared. So

3638
06:28:39,727 --> 06:28:46,390
I'm taking f of x squared. When I plug in
x squared for x, I do in fact, get the square

3639
06:28:46,390 --> 06:28:53,207
root of x squared plus seven. So this is that
alternative, correct solution. In this video,

3640
06:28:53,207 --> 06:29:00,590
we learn to evaluate the composition of functions.
by rewriting it and working from the inside

3641
06:29:00,590 --> 06:29:10,170
out. We also learn to break apart a complicated
function into a composition of two functions

3642
06:29:10,169 --> 06:29:16,339
by boxing one piece of the function and letting
the first function applied in the composition.

3643
06:29:16,340 --> 06:29:21,270
Let that be the inside of the box, and the
second function applied in the composition

3644
06:29:21,270 --> 06:29:33,540
be whatever happens to the box.

3645
06:29:33,540 --> 06:29:38,580
The inverse of a function undoes what the
function does. So the inverse of tying your

3646
06:29:38,580 --> 06:29:46,750
shoes would be to untie them. And the inverse
of the function that adds two to a number

3647
06:29:46,750 --> 06:29:53,720
would be the function that subtracts two from
a number. This video introduces inverses and

3648
06:29:53,720 --> 06:30:00,378
their properties. Suppose f of x is a function
defined by this chart. In other words, Have

3649
06:30:00,378 --> 06:30:08,790
two is three, f of three is five, f of four
is six, and f of five is one, the inverse

3650
06:30:08,790 --> 06:30:17,817
function for F written f superscript. Negative
1x undoes what f does. Since f takes two to

3651
06:30:17,817 --> 06:30:26,157
three, F inverse takes three, back to two.
So we write this f superscript, negative one

3652
06:30:26,157 --> 06:30:38,430
of three is to. Similarly, since f takes three
to five, F inverse takes five to three. And

3653
06:30:38,430 --> 06:30:46,860
since f takes four to six, f inverse of six
is four. And since f takes five to one, f

3654
06:30:46,860 --> 06:30:54,128
inverse of one is five. I'll use these numbers
to fill in the chart. Notice that the chart

3655
06:30:54,128 --> 06:31:00,450
of values when y equals f of x and the chart
of values when y equals f inverse of x are

3656
06:31:00,450 --> 06:31:07,208
closely related. They share the same numbers,
but the x values for f of x correspond to

3657
06:31:07,207 --> 06:31:14,647
the y values for f inverse of x, and the y
values for f of x correspond to the x values

3658
06:31:14,648 --> 06:31:21,138
for f inverse of x. That leads us to the first
key fact inverse functions reverse the roles

3659
06:31:21,137 --> 06:31:29,439
of y and x. I'm going to plot the points for
y equals f of x in blue. Next, I'll plot the

3660
06:31:29,439 --> 06:31:35,849
points for y equals f inverse of x in red.
Pause the video for a moment and see what

3661
06:31:35,849 --> 06:31:40,349
kind of symmetry you observe in this graph.
How are the blue points related to the red

3662
06:31:40,349 --> 06:31:46,737
points, you might have noticed that the blue
points and the red points are mirror images

3663
06:31:46,738 --> 06:31:55,250
over the mirror line, y equals x. So our second
key fact is that the graph of y equals f inverse

3664
06:31:55,250 --> 06:31:58,047
of x can be obtained from the graph of y equals
f

3665
06:31:58,047 --> 06:31:59,047
of x

3666
06:31:59,047 --> 06:32:06,727
by reflecting over the line y equals x. This
makes sense, because inverses, reverse the

3667
06:32:06,727 --> 06:32:15,770
roles of war annex. In the same example, let's
compute f inverse of f of two, this open circle

3668
06:32:15,770 --> 06:32:23,878
means composition. In other words, we're computing
f inverse of f of two, we compute this from

3669
06:32:23,878 --> 06:32:33,350
the inside out. So that's f inverse of three.
Since F of two is three, and f inverse of

3670
06:32:33,349 --> 06:32:44,717
three, we see as to similarly, we can compute
f of f inverse of three. And that means we

3671
06:32:44,718 --> 06:32:52,690
take f of f inverse of three. Since f inverse
of three is two, that's the same thing as

3672
06:32:52,689 --> 06:33:02,039
computing F of two, which is three. Please
pause the video for a moment and compute these

3673
06:33:02,040 --> 06:33:11,148
other compositions. You should have found
that in every case, if you take f inverse

3674
06:33:11,148 --> 06:33:16,290
of f of a number, you get back to the very
same number you started with. And similarly,

3675
06:33:16,290 --> 06:33:20,738
if you take f of f inverse of any number,
you get back to the same number you started

3676
06:33:20,738 --> 06:33:29,909
with. So in general, f inverse of f of x is
equal to x, and f of f inverse of x is also

3677
06:33:29,909 --> 06:33:37,430
equal to x. This is the mathematical way of
saying that F and n f inverse undo each other.

3678
06:33:37,430 --> 06:33:43,229
Let's look at a different example. Suppose
that f of x is x cubed. Pause the video for

3679
06:33:43,229 --> 06:33:48,409
a moment, and guess what the inverse of f
should be. Remember, F inverse undoes the

3680
06:33:48,409 --> 06:33:57,180
work that F does. You might have guessed that
f inverse of x is going to be the cube root

3681
06:33:57,180 --> 06:34:05,680
function, we can check that this is true by
looking at f of f inverse of x, that's F of

3682
06:34:05,680 --> 06:34:09,200
the cube root of function, which means the
cube root function

3683
06:34:09,200 --> 06:34:16,680
cubed, which gets us back to x. Similarly,
if we compute f inverse of f of x, that's

3684
06:34:16,680 --> 06:34:19,610
the cube root of x cubed.

3685
06:34:19,610 --> 06:34:25,600
And we get back to excellence again. So the
cube root function really is the inverse of

3686
06:34:25,599 --> 06:34:30,699
the cubing function. When we compose the two
functions, we get back to the number that

3687
06:34:30,700 --> 06:34:37,440
we started with. It'd be nice to have a more
systematic way of finding inverses of functions

3688
06:34:37,439 --> 06:34:44,547
besides guessing and checking. One method
uses the fact that inverses reverse the roles

3689
06:34:44,547 --> 06:34:50,520
of y and x. So if we want to find the inverse
of the function, f of x equals five minus

3690
06:34:50,520 --> 06:34:59,840
x over 3x. We can write it as y equals five
minus x over 3x. Reverse the roles of y and

3691
06:34:59,840 --> 06:35:09,957
x To get x equals five minus y over three
y, and then solve for y. To solve for y, let's

3692
06:35:09,957 --> 06:35:19,250
multiply both sides by three y. Bring all
terms with y's in them to the left side, and

3693
06:35:19,250 --> 06:35:28,369
alternate without wizened them to the right
side, factor out the why. And divide to isolate

3694
06:35:28,369 --> 06:35:40,047
why this gives us f inverse of x as five over
3x plus one. Notice that our original function

3695
06:35:40,047 --> 06:35:46,307
f and our inverse function, f inverse are
both rational functions, but they're not the

3696
06:35:46,308 --> 06:35:53,080
reciprocals of each other. And in general,
f inverse of x is not usually equal to one

3697
06:35:53,080 --> 06:36:01,350
over f of x. This can be confusing, because
when we write two to the minus one, that does

3698
06:36:01,349 --> 06:36:08,750
mean one of our two, but f to the minus one
of x means the inverse function and not the

3699
06:36:08,750 --> 06:36:16,047
reciprocal. It's natural to ask us all functions
have inverse functions. That is for any function

3700
06:36:16,047 --> 06:36:23,119
you might encounter. Is there always a function
that its is its inverse? In fact, the answer

3701
06:36:23,119 --> 06:36:30,279
is no. See, if you can come up with an example
of a function that does not have an inverse

3702
06:36:30,279 --> 06:36:37,977
function. The word function here is key. Remember
that a function is a relationship between

3703
06:36:37,977 --> 06:36:47,559
x values and y values, such that for each
x value in the domain, there's only one corresponding

3704
06:36:47,560 --> 06:36:57,388
y value. One example of a function that does
not have an inverse function is the function

3705
06:36:57,387 --> 06:37:01,067
f of x equals x squared.

3706
06:37:01,067 --> 06:37:02,237
To see that,

3707
06:37:02,238 --> 06:37:09,840
the inverse of this function is not a function.
Note that for the x squared function, the

3708
06:37:09,840 --> 06:37:17,957
number two and the number negative two, both
go to number four. So if I had an inverse,

3709
06:37:17,957 --> 06:37:27,029
you would have to send four to both two and
negative two, the inverse would not be a function,

3710
06:37:27,029 --> 06:37:34,217
it might be easier to understand the problem,
when you look at a graph of y equals x squared.

3711
06:37:34,218 --> 06:37:40,878
Recall that inverse functions reverse the
roles of y and x and flip the graph over the

3712
06:37:40,878 --> 06:37:47,840
line y equals x. But when I flipped the green
graph over the line y equals x, I get this

3713
06:37:47,840 --> 06:37:53,599
red graph. This red graph is not the graph
of a function because it violates the vertical

3714
06:37:53,599 --> 06:37:59,859
line test. The reason that violates the vertical
line test is because the original green function

3715
06:37:59,860 --> 06:38:10,000
violates the horizontal line test, and has
2x values with the same y value. In general,

3716
06:38:10,000 --> 06:38:14,950
a function f has an inverse function if and
only if the graph of f satisfies the horizontal

3717
06:38:14,950 --> 06:38:21,690
line test, ie every horizontal line intersects
the graph. In it most one point, pause the

3718
06:38:21,689 --> 06:38:26,869
video for a moment and see which of these
four graphs satisfy the horizontal line test.

3719
06:38:26,869 --> 06:38:34,579
In other words, which of the four corresponding
functions would have an inverse function?

3720
06:38:34,580 --> 06:38:42,160
You may have found that graphs A and B violate
the horizontal line test. So their functions

3721
06:38:42,159 --> 06:38:48,860
would not have inverse functions. But graph
C and D satisfy the horizontal line test.

3722
06:38:48,860 --> 06:38:54,740
So these graphs represent functions that do
have inverses. functions that satisfy the

3723
06:38:54,740 --> 06:39:02,399
horizontal line test are sometimes called
One to One functions. Equivalently a function

3724
06:39:02,399 --> 06:39:10,739
is one to one, if for any two different x
values, x one and x two, the y value is f

3725
06:39:10,739 --> 06:39:18,530
of x one and f of x two are different numbers.
Sometimes, as I said, f is one to one, if,

3726
06:39:18,529 --> 06:39:27,509
whenever f of x one is equal to f of x two,
then x one has to equal x two. As our last

3727
06:39:27,509 --> 06:39:32,939
example, let's try to find P inverse of x,
where p of x is the square root of x minus

3728
06:39:32,939 --> 06:39:40,707
two drawn here. If we graph P inverse on the
same axis as p of x, we get the following

3729
06:39:40,707 --> 06:39:49,457
graph simply by flipping over the line y equals
x. If we try to solve the problem algebraically

3730
06:39:49,457 --> 06:39:56,728
we can write y equal to a squared of x minus
two, reverse the roles of y and x and solve

3731
06:39:56,728 --> 06:40:07,579
for y by squaring both sides adding two. Now
if we were to graph y equals x squared plus

3732
06:40:07,580 --> 06:40:13,290
two, that would look like a parabola, it would
look like the red graph we've already drawn

3733
06:40:13,290 --> 06:40:22,670
together with another arm on the left side.
But we know that our actual inverse function

3734
06:40:22,669 --> 06:40:31,779
consists only of this right arm, we can specify
this algebraically by making the restriction

3735
06:40:31,779 --> 06:40:40,897
that x has to be bigger than or equal to zero.
This corresponds to the fact that on the original

3736
06:40:40,898 --> 06:40:47,530
graph for the square root of x, y was only
greater than or equal to zero. Looking more

3737
06:40:47,529 --> 06:40:55,820
closely at the domain and range of P and P
inverse, we know that the domain of P is all

3738
06:40:55,820 --> 06:41:01,650
values of x such that x minus two is greater
than or equal to zero. Since we can't take

3739
06:41:01,650 --> 06:41:07,250
the square root of a negative number. This
corresponds to x values being greater than

3740
06:41:07,250 --> 06:41:14,069
or equal to two, or an interval notation,
the interval from two to infinity. The range

3741
06:41:14,069 --> 06:41:20,419
of P, we can see from the graph is all y value
is greater than or equal to zero, or the interval

3742
06:41:20,419 --> 06:41:29,047
from zero to infinity. Similarly, based on
the graph, we see the domain of P inverse

3743
06:41:29,047 --> 06:41:34,137
is x values greater than or equal to zero,
the interval from zero to infinity. And the

3744
06:41:34,137 --> 06:41:40,500
range of P inverse is Y values greater than
or equal to two, or the interval from two

3745
06:41:40,500 --> 06:41:45,907
to infinity. If you look closely at these
domains and ranges, you'll notice that the

3746
06:41:45,907 --> 06:41:53,919
domain of P corresponds exactly to the range
of P inverse, and the range of P corresponds

3747
06:41:53,919 --> 06:41:57,569
to the domain of P inverse.

3748
06:41:57,569 --> 06:42:04,387
This makes sense, because inverse functions
reverse the roles of y and x. The domain of

3749
06:42:04,387 --> 06:42:10,977
f inverse of x is the x values for F inverse,
which corresponds to the y values or the range

3750
06:42:10,977 --> 06:42:18,349
of F. The range of f inverse is the y values
for F inverse, which correspond to the x values

3751
06:42:18,349 --> 06:42:28,559
or the domain of f. In this video, we discussed
five key properties of inverse functions.

3752
06:42:28,560 --> 06:42:36,920
inverse functions, reverse the roles of y
and x. The graph of y equals f inverse of

3753
06:42:36,919 --> 06:42:47,349
x is the graph of y equals f of x reflected
over the line y equals x. When we compose

3754
06:42:47,349 --> 06:42:56,099
F with F inverse, we get the identity function
y equals x. And similarly, when we compose

3755
06:42:56,099 --> 06:43:04,217
f inverse with F, that brings x to x. In other
words, F and F inverse undo each other. The

3756
06:43:04,218 --> 06:43:18,218
function f of x has an inverse function if
and only if the graph of y equals f of x satisfies

3757
06:43:18,218 --> 06:43:31,110
the horizontal line test. And finally, the
domain of f is the range of f inverse and

3758
06:43:31,110 --> 06:43:39,000
the range of f is the domain of f inverse.
These properties of inverse functions will

3759
06:43:39,000 --> 06:43:45,457
be important when we study exponential functions
and their inverses logarithmic functions.