[00:17] This video is about the exponent rules, rules
that govern expressions like two to the fifth,
[00:23] or x to the n.
[00:26] two to the fifth is just shorthand for two
times two times two times two times two, written
[00:32] five times. And similarly x to the n is just
x multiplied by itself. And times
[00:40] when we write these expressions, the number
on the bottom that's being multiplied by itself
[00:47] is called the base. And the number at the
top telling us how many times we're multiplying
[00:54] the base by itself is called the exponent.
[00:58] Sometimes the exponent is also called the
power.
[01:02] The product rule says that a 5x to the power
of n times x to the power of m, that's the
[01:10] same thing as x to the n plus m power. In
other words, when I multiply two expressions
[01:19] with the same base,
[01:23] then I can add their exponents.
[01:27] For example, if I have two cubed times two
to the fourth, that's equal to two to the
[01:33] seventh. And that makes sense, because two
cubed times two to the fourth, means I multiply
[01:40] two by itself three times. And then I multiply
that by two multiplied by itself four times.
[01:46] And in the end, I have to multiplied by itself
seven times,
[01:53] which is two to the seventh,
[01:56] I'm just adding up the number of times as
multiplied in each piece, to get the number
[02:01] of times as multiplied total.
[02:04] The quotient rule says that if I have x to
the n power divided by x to the m power, that's
[02:10] equal to x to the n minus m power. In other
words, if I divide two expressions with the
[02:17] same base,
[02:19] then I can subtract their exponents.
[02:23] For example, three to the six divided by three
squared is going to be three to the six minus
[02:31] two, or three to the fourth. And this makes
sense, because three to the six means I multiply
[02:37] three by itself six times, and then I divide
that by three multiplied by itself twice.
[02:45] So when I cancel out threes, I have four threes
left,
[02:50] notice that I have to subtract the number
of threes on the bottom, from the number of
[02:54] threes at the top to get my number of threes
remaining. That's why subtract my exponents.
[03:01] The power rule tells us if I have x to the
n power raised to the m power, that's the
[03:08] same thing as x to the n times M power. In
other words, when I raise a power to a power,
[03:17] I get to multiply the exponents.
[03:21] For example, five to the fourth cubed is equal
to five to the four times three, or five to
[03:29] the 12th. And this makes sense, because five
to the fourth cubed can be thought of as five
[03:34] to the fourth times five to the fourth, times
five to the fourth, expanding this out some
[03:40] more, that's five times five times five times
five times the same thing times the same thing
[03:49] again.
[03:51] So I have three groups of four, or five, which
is a total of three times four, or 12. fives.
[04:03] The next rule involves what happens when I
raise a number, or a variable to the zeroeth
[04:09] power, it turns out that anything to the zeroeth
power is equal to one.
[04:14] Usually, this is just taken as a definition.
But here's why it makes sense to me.
[04:20] If you have something like two cubed divided
by two cubed, Well, certainly that has to
[04:27] equal one, anything divided by itself is just
one. But using the quotient rule,
[04:35] we know that this is the same thing as two
to the three minus three, because when we
[04:40] divide two things with the same base, we get
to subtract their exponents. Therefore, this
[04:47] is the same thing as two to the zero. So two
to the zero has to equal one in order to make
[04:53] it work with the quotient rule.
[04:55] And the same argument shows that anything
to the zero power
[05:00] has to be equal to one.
[05:02] What happens when we take something to a negative
power,
[05:07] x to the n is equal to one over x to the n.
Again, most people just take this as a definition
[05:15] of a negative exponent. But here's why it
makes sense.
[05:20] If I take something like five to seven times
five to the negative seven, then buy the product
[05:26] roll the SAS to equal five to the seven plus
negative seven, which is five to the zero,
[05:33] and we just said that that is equal to one.
[05:35] Now I have the equation of five to the seventh
times five to the negative seventh equals
[05:41] one, if I divide both sides by five, the seventh,
I get that five to the negative seventh has
[05:47] to equal one over five to the seventh. So
that's where this rule about negative exponents
[05:52] comes from. That has to be true in order to
be consistent with the product rule.
[05:58] Finally, let's look at a fractional exponents.
What does an expression like x to the one
[06:03] over N really mean? Well, it means the nth
root of x, for example, 64 to the 1/3 power
[06:14] means the cube root of 64, which happens to
be four, and nine to the one half means the
[06:20] square root of nine, which is usually written
without that little superscript up there.
[06:26] Now, the square root of nine is just three.
[06:29] fractional exponents also makes sense. For
example, if I have five to the 1/3, and I
[06:39] cube that, then by the power role, that's
equal to five to the 1/3, times three,
[06:46] which is just five to the one or five. So
in other words, five to the 1/3, is the number
[06:55] that when you cube it, you get five. And that's
exactly what's meant by the cube root of five,
[07:01] the cube root of five is also a number that
when you cube it, you get five.
[07:07] The next rule tells us we can distribute an
exponent over a product.
[07:12] In other words, if we have a product, x times
y, all raised to the nth power, that's equal
[07:18] to x to the n times y to the N.
[07:22] For example, five times seven,
[07:27] all raised to the third power is equal to
five cubed times seven cubed. And this makes
[07:33] sense, because five times seven, all raised
to the cube power can be expanded as five
[07:39] times seven, times five times seven, times
five times seven. But if I rearrange the order
[07:46] of multiplication, this is the same thing
as five times five times five, times seven
[07:51] times seven times seven, or five cubed times
seven cubed.
[07:56] Similarly, we could distribute an exponent
over a quotient, if we have the quotient X
[08:02] over Y, all raised to the n power, that's
the same as x to the n over y to the N. For
[08:09] example, two sevenths raised to the fifth
power is the same thing as two to the fifth
[08:15] over seven to the fifth. This makes sense,
because two sevens to the fifth can be expanded
[08:23] as two sevens multiply by itself five times,
which can be written rewritten as two multiplied
[08:31] by itself five times divided by seven multiplied
by itself five times, and that's two to the
[08:37] fifth, over seven to the fifth, as wanted.
[08:43] We've seen that we can distribute an exponent
[08:47] over multiplication, and division.
[08:51] But be careful, because we cannot distribute
next bowknot.
[08:56] over addition, or subtraction, for example,
a plus b to the n is not generally equal to
[09:06] 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
[09:14] b to the n. And if you're not sure, just try
an example with numbers.
[09:19] For example, two plus three squared is not
the same thing as two squared plus three squared,
[09:28] and two minus three squared is definitely
not equal to two squared minus three squared.
[09:36] In this video, I gave eight exponent rules,
which I'll list again here. There's the product
[09:42] rule,
[09:43] the quotient rule, the power rule,
[09:46] the zero exponent, the negative exponent,
the fractional exponent,
[09:55] and the two rules involving distributing exponents.
[10:00] Across multiplication,
[10:02] and division.
[10:06] In another video, I'll use these exponent
rules to rewrite and simplify expressions
[10:11] involving exponents.
[10:13] In this video, I'll work out some examples
of simplifying expressions using exponent
[10:20] rules.
[10:21] I'll start by reviewing the exponent rules.
[10:22] The product rule says that when you multiply
two expressions with the same base, you add
[10:30] the exponents. The quotient rule says that
when you divide two expressions with the same
[10:36] base, you subtract the exponents. The power
rule says that when you take a power to a
[10:42] power, you multiply the exponents. The power
of zero rule says that anything to the zero
[10:50] power is one, as long as the base is not zero.
[10:54] Since zero to the zero is undefined, it doesn't
make sense.
[10:59] negative exponents to evaluate x to the minus
n, we take the reciprocal one over x to the
[11:07] n. To evaluate a fractional exponent, like
x to the one over n, we take the nth root
[11:16] of x,
[11:17] we can distribute an exponent over a product,
a times b to the n is equal to A to the N
[11:24] times b to the n. And we can distribute an
exponent over a quotient a over b to the n
[11:31] is a to the n over b to the n.
[11:36] In the rest of this video, we'll use these
exponent rules to simplify expressions.
[11:40] For our first example, we want to simplify
three times x to the minus two divided by
[11:47] x to the fourth, there's several possible
ways to proceed.
[11:51] For example, we could use the negative exponent
rule dr x to the minus two as one over x squared,
[12:00] all that gets divided by x to the fourth,
still,
[12:03] notice that we only take the reciprocal of
the x squared, the three stays where it is.
[12:09] And that's because the exponent of negative
two only applies to the x not to the three.
[12:16] Now if we think of three as three over one,
we have a product of two fractions and our
[12:23] numerator. And so we evaluate that by taking
the product of the numerators times the product
[12:30] of the denominators, which is three over x
squared, all divided by x to the fourth,
[12:36] I can think of x to the fourth as x to the
fourth over one. So now I have a fraction
[12:45] of a fraction, which I can evaluate by multiplying
by the reciprocal, that simplifies to three
[12:52] times one divided by x squared times x to
the fourth, which is three over x to the six,
[13:00] using the product rule. Since x squared times
x to the fourth is equal to x to the two plus
[13:07] four, or x to the six.
[13:12] An alternate way of solving this problem
[13:20] is to start by using the quotient rule,
[13:21] I can rewrite this as three times x to the
minus two over x to the fourth, and by the
[13:25] quotient rule, that's three times x to the
minus two minus four, or three times x to
[13:31] the minus six.
[13:32] Now using the negative exponent roll, x to
the minus six is one over x to the six. And
[13:40] this product of fractions simplifies to three
over x to the six, the same answer I got before.
[13:49] The second problem can be solved in similar
ways. Please pause the video and try it before
[14:00] going on.
[14:01] One way to simplify would be to use the negative
exponent rule first, and rewrite y to the
[14:04] minus five as one over wide the fifth.
[14:08] thinking of this as a fraction, divided by
a fraction, I can multiply by the reciprocal
[14:18] and get four y cubed y to the fifth over one
by the product rule. The numerator here is
[14:28] four y to the eight. And so my final answer
is just for one of the eight.
[14:35] Alternatively, I could decide to use the quotient
rule first.
[14:41] As in the previous problem, I can write this
as for y cubed minus negative five by the
[14:48] quotient rule. And so that's for y to the
eighth as before.
[14:53] I'd like to show you one more method to solve
these two problems, kind of a shortcut method
[14:59] before we
[15:00] To go on, that shortcut relies on the principle
that a negative exponent in the numerator
[15:08] corresponds to a positive exponent in the
denominator. For example, the x to the negative
[15:14] two in the numerator here, after some manipulations
became an X to the positive two in the denominator.
[15:22] Furthermore, a negative exponent in the denominator
[15:27] is equivalent to a positive exponent in the
numerator.
[15:32] That's what happened when we had the y to
the negative five in the denominator, and
[15:39] translated into a y to the positive five in
the numerator.
[15:42] Sometimes people like to talk about this principle,
by saying that you can pass a factor
[15:52] across the fraction bar
[15:54] by switching the sine of the exponent that
is making a positive exponent negative, or
[16:04] a negative exponent positive.
[16:07] Let's see how this principle gives us a shortcut
for solving these two problems.
[16:11] In the first problem, 3x to the minus two
over x to the four, we can move the negative
[16:20] exponent in the numerator and make it a positive
exponent the denominator, so we get three
[16:25] over x to the four plus two or x to the six.
[16:32] In the second example, for y cubed over y
to the minus five, we can change the Y to
[16:40] the minus five in the denominator into a y
to the five in the numerator and get our final
[16:45] answer of four y to the three plus five or
eight.
[16:51] We'll use this principle again in the next
problems.
[16:54] In this example, notice that I have I have
y's in the numerator and the denominator,
[17:01] and also Z's in the numerator and the denominator.
In order to simplify, I'm going to try to
[17:09] get all my y's either in the numerator or
the denominator. And similarly for the Z's.
[17:13] Since I have more y's in the denominator,
let me move this y to the three downstairs
[17:19] and make it a y to the negative three. I'm
using the principle here that a positive exponent
[17:26] in the numerator corresponds to a negative
exponent in the denominator.
[17:31] Now since I have a positive exponent, z in
the numerator and a negative exponent denominator,
[17:37] and I want to get rid of negative exponents,
I'm going to pass the Z's to the numerator
[17:44] as e to the minus two in the denominator,
it comes as e to the plus two in the numerator.
[17:51] Notice that my number seven doesn't move.
And when I do any of these manipulations,
[17:57] because it doesn't have an exponent, and the
exponent of negative two, for example, only
[17:59] applies to the Z not to the seven.
[18:01] Now that I've got all my Z's in the numerator
and all my y's in the denominator, it's easy
[18:10] to clean this up using the product rule.
[18:13] And I have my simplified expression.
[18:16] In this last example, we have a complicated
expression raised to a fractional power.
[18:22] I'm going to start by simplifying the expression
inside the parentheses.
[18:26] I can bring all my y's downstairs and all
my x's upstairs and get rid of negative exponents
[18:34] at the same time.
[18:35] In other words, I can rewrite this as 25x
to the fourth,
[18:40] I'll bring the Y to the minus five downstairs
and make it y to the fifth on the denominator,
[18:48] bring the x to the minus six upstairs and
make it x to the sixth the numerator, and
[18:54] then I still have the Y cubed on the denominator,
all that's raised to the three halves power.
[19:00] Using the product rule, I can rewrite the
expression on the inside of the parentheses
[19:07] as 25x to the 10th over y to the eighth.
[19:12] Recall that we're allowed to distribute an
exponent across a product or across a quotient.
[19:21] When I distribute my three halves power,
[19:27] I get 25 to the three halves times x to the
10th to the three halves divided by y to the
[19:36] eighth to the three halves.
[19:38] Now the power rule tells me when I have a
power to a power, I get to multiply the exponents.
[19:46] So I can rewrite this as 25 to the three halves
times x to the 10 times three halves of our
[19:53] y to the eight times three halves. In other
words, 25 to the three halves times x to the
[20:01] 15th over y to the 12th. Finally, I need to
rewrite 25 to the three halves. Since three
[20:10] halves can be thought of as three times one
half, or as one half times three, I can write
[20:17] 25 to the three halves as 25 to the three
times one half, or as 25 to the one half times
[20:27] three.
[20:28] Well, using the power rule in reverse, I can
think of this as 25 cubed to the one half,
[20:33] or as 25 to the one half cubed. Since when
I take a power to a power, I multiply the
[20:39] exponents
[20:40] 25 cubed to the one half might be hard to
evaluate, since 25 cubed is a huge number,
[20:47] but 25 to the one half is just the square
root of 25. So I have the square root of 25
[20:58] cubed, or five cubed, which is 125.
[21:02] Therefore, my original expression is going
to be 120 5x to the 15 over y to the 12th.
[21:09] In this video, we use the exponent rules to
simplify complicated expressions.
[21:18] This video goes through a few tricks for simplifying
expressions with radicals in them.
[21:23] Recall that this notation means the nth root
of x, so this notation here means the cube
[21:30] root of eight, the number that when you cube
it, you get eight, that number would be two,
[21:38] when we write the root sign without a little
number, that just means the square root so
[21:44] the two is implied.
[21:47] In this case, the square root of 25 is five
since five squared is 25.
[21:54] Let's start by reviewing some rules for radical
expressions. First, if we have the radical
[22:00] of a product, we can rewrite that as the product
of two radicals.
[22:06] For example, the square root of nine times
16 is the same thing as the square root of
[22:14] nine times the square root of 16, you can
check that both of these evaluate to 12.
[22:21] Similarly, it's possible to distribute a radical
sine across division, the radical of a divided
[22:29] by b is the same thing as the radical of A
divided by the radical of B. For example,
[22:37] the cube root of 64 over eight is the same
thing as the cube root of 64 over the cube
[22:45] root of eight, and you can check that both
of these evaluated to
[22:48] you have to be a little bit careful though,
because it's not okay to distribute a radical
[22:57] sign across addition. In general, the nth
root of a plus b is not equal to the nth root
[23:03] of A plus the nth root of b. And similarly,
it's not okay to distribute a radical across
[23:11] subtraction.
[23:14] If you're ever in doubt, you can always check
with simple examples. For example, the square
[23:19] root of one plus one is not the same thing
as the square root of one plus the square
[23:24] root of one,
[23:26] the right side evaluates to one plus one or
two, and the left side is square root of two
[23:32] and the irrational number.
[23:34] The second expression to show that that fails,
I don't think it'll work to use the square
[23:40] root of one minus one it'll actually hold
in that case, but I can show it's false by
[23:44] using say the square root of two minus one,
which does not equal the square root of two
[23:49] minus the square root of one.
[23:53] You might notice that these Rules for Radicals,
the ones that hold and the ones that don't
[23:57] hold, remind you of rules for exponents. And
that's no coincidence. Because radicals can
[24:04] be written in terms of exponents. For example,
if we look at the first rule, we can rewrite
[24:09] this, the nth root of A times B is the same
thing as the one of our nth power. And by
[24:19] exponent rules, I can distribute an exponent
across multiplication. And so this radical
[24:26] rule can be restated completely in terms of
an exponent rule. Similarly, the second rule
[24:34] can be restated in terms of exponents as a
or b to the one over n is equal to A to the
[24:41] one over n divided by b to the one over n.
We can use the relationship between radicals
[24:46] and the exponents to rewrite a to the m over
n. a to the m over N is the same thing as
[24:54] a to the m with the N through taken. That's
also the same thing
[25:00] As the nth root of A, all taken to the nth
power to see whether that's true, think about
[25:06] exponent rules. So a to the m over N is the
same thing as a to the m, taken to the one
[25:13] over nth power. That's because when we take
a power to a power, we multiply exponents,
[25:19] and M times one over n is equal to M over
n.
[25:25] But a one over nth power is the same thing
as an nth root. And therefore, this expression
[25:31] is the same thing as this expression. And
that proves the first equivalence. The second
[25:39] equivalence can we prove similarly, by writing
a to the m over n as a to the one over n times
[25:48] M. Again, this is works because when I take
the power to the power, I'm multiply exponents,
[25:55] one over n times M is the same thing as M
over n. But now, these two expressions are
[26:02] the same, because the one over nth power is
the same as the nth root.
[26:08] One mnemonic for remembering these relationships
is flower over root. So flour is like power,
[26:15] and root is like root, so that tells us we
can write a fractional exponent, the M becomes
[26:20] the power, and the n becomes the root in either
of these two orders.
[26:26] Now let's use these rules in some examples.
If we want to compute 25 to the negative three
[26:33] halves power, well, first I'll use my exponent
rule to rewrite that negative exponent as
[26:39] one over 25 to the three halves power.
[26:43] Next, I'll use the power of a root mnemonic
to rewrite this as 25 to the third power square
[26:52] rooted, or, as 25 square rooted to the third
power.
[26:59] I wrote the two's there for the square root
for emphasis, but most of the time, people
[27:04] will omit this and just write the square root
without a little number there.
[27:09] Now, I could use either of these two equivalent
expressions to continue, but I'd rather use
[27:15] this one because it's easier to compute without
a calculator. The square root of 25 is just
[27:20] five, five cubed is 125. So my answer is one
over 125.
[27:28] If I tried to compute the cube of 25, first,
I'd get a huge number. In general, it's usually
[27:34] easier to compute the route before the power
when you're working without a calculator.
[27:41] Now let's do an example simplifying a more
complicated expression with with exponent
[27:46] cynet. I want to take the square root of all
this stuff. And since I don't really like
[27:52] negative exponents, I'm first going to rewrite
this as the square root of 60x squared y to
[27:58] the sixth over z to the 11th. So I'll change
that negative exponent to a positive exponent
[28:04] by moving this, this factor to the denominator.
[28:07] Now, when you're asked to simplify radical
expression, that generally means to pull as
[28:13] much as possible, out of the radical side.
[28:17] To pull things out of the square root side,
I'm going to factor my numbers and try to
[28:23] rewrite everything in terms of squares as
much as possible. Since the square root of
[28:28] a square, those two operations undo each other.
So I'll show you what I mean first, our factor
[28:34] 60. So 60 is going to be two squared times
three times five. And I'll just copy everything
[28:43] over for now.
[28:47] Now I'll break things up into squares as much
as possible. So I've got a two squared, and
[28:52] three times five, I've already got an x squared,
I write the wider the six as y squared times
[28:57] y squared times y squared. And I'll write
the Z to the 11th as z squared times z squared,
[29:03] I guess five times 12345 times when extra
z, that should add up to z to the 11th. I
[29:12] add all those exponents together.
[29:15] Now I know that I can distribute my radical
sign across multiplication and division. So
[29:20] I'll write this with a zillion different radicals
here.
[29:25] And every time I see the square root of something
squared, I can just cancel those square roots
[29:32] and the squares out and get what's what's
left here. So So after doing that cancellation,
[29:37] I get two times the square root of three times
five times x times y times y times y over
[29:46] z times itself, I guess five times times the
square root of z. And now I can clean that
[29:53] up with exponents. I'll write that as the
square root of 15 I guess two times a squared
[29:59] of 15
[30:00] times x times y cubed over z to the fifth
the square root of z.
[30:07] I'm gonna leave this example as is. But sometimes
people prefer to rewrite radical expressions
[30:15] without radical signs in the denominator.
That's called rationalizing the denominator.
[30:20] I won't do it here, but I'll show you how
to do it in the next example.
[30:24] This example asks us to rationalize the denominator,
that means to rewrite as an equivalent expression
[30:31] without radical signs in the denominator.
[30:36] To get rid of the radical sine and the denominator,
I want to multiply my denominator by square
[30:44] root of x. But I can't just multiply the denominator
willy nilly by something unless I multiply
[30:49] the numerator by the same thing. So then just
multiply my expression by one in a fancy forum
[30:55] and I don't change the value of my expression.
[30:57] Now, if I just multiply together numerators
3x squared of x and multiply denominators
[31:06] squared of x 10 squared of x is the square
root of x squared squared of x squared is
[31:12] just x.
[31:14] Now I can cancel my access from the numerator
denominator, and my final answer is three
[31:19] times the square root of x. I rationalize
my denominator, and the process got a nicer
[31:24] looking expression.
[31:26] In this video, we went over the Rules for
Radicals. And we simplified some radical expressions
[31:34] by working with fractional exponents,
[31:37] pulling things out of the radical sign
[31:41] and rationalizing the denominator.
[31:46] This video goes over some common methods of
factoring. Recall that factoring an expression
[31:51] means to write it as a product. So we could
factor the number 30, by writing it as six
[31:56] times five, we could factor it more completely
by writing it as two times three times five.
[32:04] As another example, we could factor the expression
x squared plus 5x plus six by writing it as
[32:12] x plus two times x plus three.
[32:16] In this video, I'll go over how I get from
here to here, how I know how to do the factoring.
[32:23] But for right now, I just want to review how
I can go backwards how I can check that the
[32:29] factoring is correct. And that's just by multiplying
out or distributing. If I distribute x plus
[32:35] two times x plus three, then I multiply x
by x, that gives me x squared, x times three
[32:40] gives me 3x. Two times x gives me 2x. And
two times three gives me six. So that simplifies
[32:48] to x squared plus 5x plus six, which checks
out with what I started with. So you can think
[32:54] of factoring as the opposite of distributing
out. And you can always check your factoring
[33:00] by distributing or multiplying out
[33:01] a bit of terminology, when I think of an expression
as a sum of a bunch of things, then the things
[33:09] I sum up are called the terms. But if I think
of the same expression as a product of things,
[33:17] then the things that I multiply together are
called factors.
[33:22] Now let's get started on techniques of factoring.
When I have to factor something, I always
[33:28] like to start by pulling out the greatest
common factor, the greatest common factor
[33:33] means the largest thing that divides each
of the terms.
[33:38] In this first example, the largest thing that
divides both 15 and 25x is five.
[33:46] So the GCF is five. So I pull the five out,
and then I divide each of the terms by that
[33:53] number. And so I get three plus 5x.
[33:58] Pause the video for a moment and see if you
can find the greatest common factor of x squared
[34:04] y and y squared x cubed.
[34:08] The biggest thing that divides both x squared
y and y squared x cubed is going to be x squared
[34:15] times y.
[34:18] One way to find this is to look for the power
of x that smallest in each of these terms.
[34:24] So that's x squared. And the power of y that
smallest in each of these terms is just y
[34:29] to the one or y.
[34:31] Now if I factor out the x squared y from each
of the terms, that's like dividing each term
[34:38] by x squared y, if I divide the first term
by x squared y, I just get one. If I divide
[34:44] the second term by x squared y, I'm going
to be left with an X and a Y. I'll write this
[34:50] out on the side just to make it more clear,
y squared x cubed over x squared y. That's
[34:57] like two y's on the end.
[34:59] 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
[35:07] a Y. So I'll write the x, y here, and I factored
my expression. As always, I can check my answer
[35:16] by multiplying out. So if I multiply out my
factored expression, I get x squared y is
[35:23] the first term and the second term, I get
x there. Now let's see three X's multiplied
[35:28] together and two y's multiplied together.
And that checks out with what I started with.
[35:35] The next technique of factoring, I'd like
to go over his factoring by grouping. In this
[35:40] example, notice that we have four terms, factoring
by grouping is a handy method to look at.
[35:46] If you have four terms in your expression,
you need to factor
[35:50] in order to factor by grouping, I'm first
going to factor out the greatest common factor
[35:56] of the first two terms, and then separately,
factor out the greatest common factor of the
[36:00] last two terms. The greatest common factor
of x cubed, and 3x squared is x squared. So
[36:08] I factor out the x squared, and I get x plus
three. And now the greatest common factor
[36:14] of forex and 12 is just four. So I factor
out the four from those two terms.
[36:23] Notice that the factor of x plus three now
appears in both pieces. So I can factor out
[36:31] the greatest common factor of x plus three,
and I'll factor it out on the left side instead
[36:36] of the right. And now I have an x squared
from this first piece, and I have a four from
[36:45] this second piece. And that completes my factoring
by grouping. You might wonder if we could
[36:52] factor further by factoring the expression
x squared plus four. But in fact, as we'll
[36:57] see later, this expression, which is a sum
of two squares, x squared plus two squared
[37:03] does not factor any further over the integers.
[37:06] Next, we'll do some factoring of quadratics.
a quadratic is an expression with a squared
[37:13] term,
[37:15] just a term with x in it, and a constant term
with no x's in it.
[37:20] I'd like to factor this expression as a product
of x plus or minus some number times x plus
[37:29] or minus some other number.
[37:32] The key idea is that if I can find those two
numbers, then if I were to distribute out
[37:37] this expression, those two numbers would have
to multiply to give me my constant term of
[37:41] eight. And these two numbers would end up
having to add to give me my negative six,
[37:48] because when I multiply out, this number will
be a coefficient of x, and this number will
[37:53] be also another coefficient of x, they'll
add together to the negative six.
[37:59] So if I look at all the pairs of numbers that
multiply together to give me eight, so that
[38:04] could be one and eight, two, and four, four
and two, but that's really the same thing
[38:11] as I had before. And that's sort of the same
thing I had before. I shouldn't forget the
[38:16] negatives, I could have negative one, negative
eight, or I could have negative two, negative
[38:20] four, those alternate multiply together to
give me eight. Now I just have to find, see
[38:25] if there's a pair of these numbers that add
to negative six, and it's not hard to see
[38:31] that these ones will work. So now I can write
out my factoring as that would be x minus
[38:39] two times x minus four. And it's always a
good idea to check by multiplying out, I'm
[38:45] going to get x squared minus 4x minus 2x,
plus eight. And that works out to just what
[38:53] I want. Now this second examples a bit more
complicated, because now my leading coefficient,
[39:00] my coefficient of x squared is not just one,
it's the number 10.
[39:04] Now, there are lots of different methods for
approaching a problem like this. And I'm just
[39:08] going to show you one method, my favorite
method that uses factoring by grouping, but
[39:13] but to start out, I'm going to multiply my
coefficient of x squared by my constant term,
[39:18] so I'm multiplying 10 by negative six, that
gives me negative 60. And I'll also take my
[39:25] coefficient of x the number 11. And write
that down here. Now I'm going to look for
[39:31] two numbers that multiply to give me negative
60. And add to give me 11. You might notice
[39:38] that this is exactly what we were doing in
the previous problem. It's just here, we didn't
[39:42] have to multiply the coefficient of x squared
by eight, because the coefficient of x squared
[39:47] was just one. So to find the two numbers that
multiply to negative 60 and add to 11, you
[39:53] might just be able to come up with them in
your head thinking about it, but if not, you
[39:58] can figure it out. Pretty simple.
[40:00] thematically by writing out all the factors,
pairs of factors that multiply to negative
[40:05] 60. So I can start with negative one and 60,
negative two and 30, negative three and 20.
[40:12] And keep going like this until I have found
factors that actually add together to give
[40:19] me the number 11. And, and now that I look
at it, I've already found them. 15 minus four,
[40:28] gives me 11. So I don't have to continue with
my chart of factors. Now, once I found those
[40:34] factors, I write out my expression 10x squared,
but instead of writing 11x, I write negative
[40:40] 4x plus 15x. Now I copy down the negative
six, notice that negative 4x plus 15x equals
[40:49] 11x. That's how I chose those numbers. And
so this expression is evaluates is the same
[40:55] as, as this expression, I haven't changed
my expression. But I have turned it into something
[41:01] that I can apply factoring by grouping on
Look, I've got four terms here. And so if
[41:05] I factor out my greatest common factor of
my first two terms, that's let's see, I think
[41:11] it's 2x. So I factor out the 2x, I get 5x
minus two, and then I factor out the greatest
[41:18] common factor of 15x and negative six, that
would be three, and I get a 5x minus two,
[41:25] again, this is working beautifully. So I have
a 5x minus two in each part. And so I put
[41:31] the 5x minus two on the right, and I put what's
left from these terms in here. So that's 2x
[41:38] plus three. And I have factored my expression.
[41:43] There are a couple special kinds of expressions
that appear frequently, that it's handy to
[41:50] just memorize the formula for. So the first
one is the difference of squares. If you see
[41:55] something of the form a squared minus b squared,
then you can factor that as a plus b times
[42:02] a minus b. And let's just check that that
works. If I do a plus b times a minus b and
[42:08] multiply that out, I get a squared minus a
b plus b A minus B squared, and those middle
[42:17] two terms cancel out. So it gives me back
the difference of squares just like I want
[42:21] it. So for this first example, I if I think
of x squared minus 16 as x squared minus four
[42:30] squared, then I can see that's a difference
of squares. And I can immediately write it
[42:34] as x plus four times x minus four. And the
second example, nine p squared minus one,
[42:41] that's the same thing as three p squared minus
one squared. So that's three p plus one times
[42:50] three p minus one.
[42:54] Notice that if I have a sum of squares,
[42:57] for example,
[43:01] x squared plus four, which is x squared plus
two squared, then that does not factor.
[43:09] The difference of squares formula doesn't
apply. And there is no formula that applies
[43:13] for a sum of squares. There is, however, a
formula for both a difference of cubes and
[43:19] a sum of cubes. The difference of cubes formula,
a cubed minus b cubed is a minus b times a
[43:27] squared plus a b plus b squared. The formula
for the sum of cubes is pretty much the same,
[43:34] you just switch the negative and positive
sign here in here. So that gives us a plus
[43:41] b times a squared minus a b plus b squared.
As usual, you can check these formulas by
[43:49] multiplying out. Let's look at one example
of using these formulas. Y cubed plus 27 is
[43:55] actually a sum of two cubes because it's y
cubed plus three cubed. So I can factor it
[44:03] using the sum of cubes formula by plugging
in y for a and three for B. That gives me
[44:10] y plus three times y squared minus y times
three plus three squared. And I can clean
[44:18] that up a little bit to read y plus three
times y squared minus three y plus nine.
[44:23] So in this video, we went over several methods
of factoring. We did factoring out the greatest
[44:31] common factor. We did factoring by grouping.
We did factoring quadratics. And we did a
[44:40] difference of squares. And we did a difference
and a psalm of cubes
[44:47] and more complicated problems, you may need
to apply several these techniques in order
[44:54] to get through a single problem. For example,
you might need to start by pulling out a greatest
[44:58] common factor and then
[45:00] Add to a factoring of quadratics, or something
similar.
[45:03] Now go forth and factor.
[45:07] This video gives some additional examples
of factoring. Please pause the video and decide
[45:13] which of these first five expressions factor
and which one does not.
[45:20] The first expression can be factored by pulling
out a common factor of x from each term.
[45:27] So that becomes x times x plus one.
[45:34] The second example can be factors as a difference
of two squares, since x squared minus 25 is
[45:41] something squared, minus something else squared.
And we know that anytime we have something
[45:44] like a squared minus b squared, that's a plus
b times a minus b. So we can factor this as
[45:55] X plus five times x minus five.
[45:58] The third one is a sum of two squares, there's
no way to factor a sum of two squares over
[46:05] real numbers. So this is the one that does
not factor.
[46:09] just for completeness, let's look at the next
to this next one does factor by grouping.
[46:16] When we factor by grouping, we pull up the
biggest common factor out of the first two
[46:23] terms, that would be an x squared, that becomes
x squared times x plus two. And then we factor
[46:29] as much as we can add the next two terms,
that would be a three times x plus two. Notice
[46:36] that the x plus two factor now occurs in both
of the resulting terms. So we can pull that
[46:42] x plus two out and get x plus two times x
squared plus three,
[46:51] we can't factor any further because x squared
plus three doesn't factor.
[46:57] Finally, we have a quadratic, this also factors.
And I like to factor these also using a factoring
[47:04] by grouping trick. So first, what I do is
I multiply the coefficient of x squared and
[47:12] the constant term, five times eight is 40.
I'll write that on the top of my x. Now I
[47:19] take the coefficient of the x term, that's
negative 14, and I write that on the bottom
[47:24] part of the x. Now I'm looking for two numbers
that multiply to 40 and add to negative 14.
[47:32] Sometimes I can just guess numbers like this,
but if not, I start writing out factors of
[47:37] 40. So factors of 40, I could do one times
40. Well, now I just noticed, I'm trying to
[47:46] add to a negative number. So if I use two
positive factors, there's no way there's going
[47:51] to add to a negative number. It's better for
me to use negative numbers, that factor 40,
[47:56] a negative times a negative still multiplies
to 40. But they have a chance of adding to
[48:02] a negative number. So but negative one and
negative 40, of course, don't work, they don't
[48:06] add to negative 14, they add to negative 41.
So let me try some other factors. The next
[48:11] biggest number that divides 40, besides one
is two, so I'll try negative two and negative
[48:15] 20. Those add to negative 22. That doesn't
work. Next one that divides 40 would be four,
[48:22] so I'll try negative four and negative 10.
Aha, we have a winner. So negative four plus
[48:29] negative 10 is negative 14, negative four
times negative 10 is positive 40. We've got
[48:35] it. Alright, so the next step is to use factoring
by grouping, we're going to first split up
[48:41] this negative 14x as negative 4x minus 10x.
And carry down the eight and the 5x squared.
[48:50] Notice that this works, because I picked negative
four and negative 10 to add up to negative
[48:57] 14, so so negative 4x minus 10x, will add
up to negative 14x. So I've got the same expression,
[49:04] just just expand it out a little bit. Now
I have four terms, I can do factoring by grouping,
[49:09] so I can group the first two terms and factor
out the biggest thing I can that will be n
[49:13] x times 5x minus four. And now I'll factor
the biggest thing I can out of these two numbers,
[49:19] including the the negative. So that becomes,
let's see, I can factor out a negative two
[49:26] and that becomes 5x minus four since negative
two times minus four is eight. All right,
[49:33] I've got the same 5x plus four in both my
terms, so factoring by grouping is going swimmingly,
[49:40] I can factor out the 5x minus four from both
those terms and I get the x minus two, and
[49:45] I factored this quadratic. If I want to, of
course, I can always check my work by distributing
[49:51] out by multiplying out. So a check here would
be multiplying 5x times x is 5x.
[49:59] squared 5x minus 10 to minus two is minus
10x minus four times x is minus 4x. And minus
[50:08] four times minus two is plus eight. So let's
see, this does check out to exactly what it
[50:13] should be. So that was the method of factoring
a quadratic.
[50:22] And all of these factors except for the sum
of squares.
[50:27] So we saw that factoring by grouping is handy
for factoring this expression here. It was
[50:35] also handy for factoring the quadratic indirectly,
after splitting up
[50:39] the middle term
[50:44] into two terms. So how can you tell when a
an expression is is appropriate to factor
[50:52] by grouping, there's, there's an easy way
to tell that it might be a candidate, and
[50:57] that's that it has four terms.
[51:00] So if you see four terms, or in the case of
quadratic, you can split it up into four terms,
[51:05] then that's a good candidate for factoring
by grouping because you can group the first
[51:09] two terms group the second two terms, factoring
by grouping all always work on, on expressions
[51:16] with four terms. But, but that's like the
first thing to look for. So let's just review
[51:21] what are the same main techniques of factoring.
We saw these on the previous page, we saw
[51:25] there was pull out common factors. There's
difference of squares.
[51:32] There's factoring by grouping.
[51:36] There's factoring quadratics.
[51:41] And one more that I didn't mention is factoring
sums and differences of cubes.
[51:49] that uses the formulas, aq minus b cubed is
a minus b times a squared plus a b plus b
[51:57] squared. And a cubed plus b cubed is a plus
b times a squared minus a b plus b squared.
[52:07] One important tip when factoring,
[52:11] I always recommend doing this first, pull
out the common factors first.
[52:17] That'll simplify things and making the rest
of factoring easier. One more tip is that
[52:21] you might need to do several these factoring
techniques in one problem, for example,
[52:28] you might have to first pull out a common
factor, then factor a difference of squares.
[52:31] And then you might notice that one of your
factors is itself a difference of squares,
[52:35] and you have to apply a difference of squares
again, so don't stop when you factor a little
[52:41] bit, keep factoring as far as you can go.
[52:43] Here are some extra examples of factoring
quadratics. For you to practice, please pause
[52:49] the video and give these a try.
[52:52] For the first one, let's multiply two times
negative 14. That gives us negative 28. And
[53:01] then we'll bring the three down in the bottom
of the x. Now we're looking for two numbers
[53:06] that multiply to negative 28 and add to three.
Well, to multiply two numbers to get a negative
[53:14] 28, we'll need one of them to be negative
and one of them to be positive. So to be one,
[53:23] negative 128, or one, negative 28, those don't
work. Let's see negative 214 or two, negative
[53:26] 14, those don't work. Hey, I just noticed
the positive number had better be bigger than
[53:30] the negative number, so they add to a positive
number. Let's see what comes next. How about
[53:37] negative four times seven, four times negative
seven, I think four, negative four times seven
[53:43] will work. So I'll write those here at negative
four, seven,
[53:49] copy down the two z squared. And I'll split
up the three z into negative four z plus seven
[53:58] z and then minus 14. Now factoring by grouping,
pull out a to z, that becomes z minus two,
[54:05] pull out a seven and that becomes z minus
two again, looking good. I've got two z plus
[54:12] seven times z minus two as my factored expression.
[54:18] My second expression, I could work at the
same way, drawing my axe and factoring by
[54:23] grouping that kind of thing. But it's actually
going to be easier if I notice first that
[54:28] I can pull out a common factor from all of
my terms, though, that'll make things a lot
[54:32] simpler to deal with. So notice that a five
divides each of these terms, and in fact,
[54:36] I'm going to go ahead and pull out the negative
five because I don't like having negatives
[54:39] in front of my squared term. So I'm going
to pull out a common factor of negative five.
[54:44] Again, it would work if I forgot to do this,
but it would be a lot more complicated. So
[54:49] negative five v squared, this becomes minus,
this becomes plus nine V, since nine times
[54:56] negative five is negative 45. And this becomes
minus
[54:59] 10 cents native 10 times negative five is
positive 50. Now I can start my x and my factoring
[55:06] by grouping, or I can use kind of a shortcut
method, which you may have seen before. So
[55:10] I can just put these here, and then I know
that whatever numbers go here, they're gonna
[55:15] have to multiply to the negative 10. And they're
gonna have to add to the nine. So that would
[55:21] be plus 10, and a minus one will do the trick.
[55:26] Those are all my factoring examples for today.
I hope you enjoy your snow morning and have
[55:32] a chance to spend some time working in ALEKS.
Bye.
[55:37] This video is about working with rational
expressions. A rational expression is a fraction
[55:43] usually with variables in it, something like
x plus two over x squared minus three is a
[55:49] rational expression. In this video, we'll
practice adding, subtracting, multiplying
[55:55] and dividing rational expressions and simplifying
them to lowest terms.
[56:01] We'll start with simplifying to lowest terms.
Recall that if you have a fraction with just
[56:07] numbers in it, something like 21 over 45,
we can reduce it to lowest terms by factoring
[56:14] the numerator
[56:17] and factoring the denominator
[56:22] and then canceling common factors.
[56:26] So in this example, the three is cancel, and
our fraction reduces to seven over 15.
[56:35] If we want to reduce a rational expression
with the variables and add to lowest terms,
[56:40] we proceed the same way. First, we'll factor
the numerator, that's three times x plus two,
[56:47] and then factor the denominator. In this case
of factors 2x plus two times x plus two, we
[56:54] could also write that as x plus two squared.
Now we cancel the common factors. And we're
[57:00] left with three over x plus two. Definitely
a simpler way of writing that rational expression.
[57:08] Next, let's practice multiplying and dividing.
Recall that if we multiply two fractions with
[57:15] just numbers in them, we simply multiply the
numerators and multiply the denominators.
[57:20] So in this case, we would get four times two
over three times five or 8/15.
[57:28] If we want to divide two fractions, like in
the second example, then we can rewrite it
[57:35] as multiplying by the reciprocal of the fraction
on the denominator. So here, we get four fifths
[57:43] times three halves, and that gives us 12 tenths.
But actually, we could reduce that fraction
[57:50] to six fifths,
[57:53] we use the same rules when we compute the
product or quotient of two rational expressions
[58:00] with the variables. And then here, we're trying
to divide two rational expressions. So instead,
[58:06] we can multiply by the reciprocal. I call
this flipping and multiplying.
[58:12] And now we just multiply the numerators.
[58:18] And multiply the denominators.
[58:21] It might be tempting at this point to multiply
out to distribute out the numerator and the
[58:28] denominator. But actually, it's better to
leave it in this factored form and factored
[58:32] even more completely. That way, we'll be able
to reduce the rational expression to cancel
[58:38] the common factors. So let's factor even more
the x squared plus x factors as x times x
[58:46] plus one, and x squared minus 16. And that's
a difference of two squares, that's x plus
[58:52] four times x minus four, the denominator is
already fully factored, so we'll just copy
[58:59] it over. And now we can cancel common factors
here and here, and we're left with x times
[59:06] x minus four. This is our final answer.
[59:12] Adding and subtracting fractions is a little
more complicated because we first have to
[59:16] find a common denominator. A common denominator
is an expression that both denominators divided
[59:23] into, it's usually best of the long run to
use the least common denominator, which is
[59:29] the smallest expression that both denominators
divided into.
[59:34] In this example, if we just want a common
denominator, we could use six times 15, which
[59:39] is 90 because both six and 15 divided evenly
into 90. But if we want the least common denominator,
[59:46] the best way to do that is to factor the two
denominators. So six is two times 315 is three
[59:55] times five, and then put together only the
factors we need for
[59:59] Both six and 50 into divider numbers. So if
we just use two times three times five, which
[01:00:07] is 30, we know that two times three will divide
it, and three times five will also divide
[01:00:14] it. And we won't be able to get a denominator
any smaller, because we need the factors two,
[01:00:19] three, and five, in order to ensure both these
numbers divided. Once we have our least common
[01:00:24] denominator, we can rewrite each of our fractions
in terms of that denominator. So seven, six,
[01:00:31] I need to get a 30 in the denominator, so
I'm going to multiply that by five over five,
[01:00:38] and multiply by the factors that are missing
from the current denominator in order to get
[01:00:45] my least common denominator of 30. For the
second fraction, for 15th 15 times two is
[01:00:54] 30. So I'm going to multiply by two over two,
[01:00:57] I can rewrite this as 3530 s minus 8/30. And
now that I have a common denominator, I can
[01:01:08] just subtract my two numerators. And I get
27/30.
[01:01:15] If I factor, I can reduce this
[01:01:20] to three squared over two times five, which
is nine tenths. The process for finding the
[01:01:28] sum of two rational expressions with variables
in them follows the exact same process. First,
[01:01:33] we have to find the least common denominator,
I'll do that by factoring the two denominators.
[01:01:39] So 2x plus two factors as two times x plus
1x squared minus one, that's a difference
[01:01:45] of two squares. So that's x plus one times
x minus one. Now for the least common denominator,
[01:01:51] I'm going to take all the factors, I need
to get an expression that each of these divides
[01:01:57] into, so I need the factor two, I need the
factor x plus one, and I need the factor x
[01:02:02] minus one, I don't have to repeat the factor
x plus one I just need to have at one time.
[01:02:08] And so I will get my least common denominator
two times x plus one times x minus one, I'm
[01:02:14] not going to bother multiplying this out,
it's actually better to leave it in factored
[01:02:17] form to help me simplify later. Now I can
rewrite each of my two rational expressions
[01:02:24] by multiplying by whatever's missing from
the denominator in terms of the least common
[01:02:29] denominator. So what I mean is, I can rewrite
three over 2x plus two, I'll write the 2x
[01:02:36] plus two is two times x plus one, I'll write
it in factored form. And then I noticed that
[01:02:41] compared to the least common denominator,
I'm missing the factor of x minus one. So
[01:02:45] I multiply the numerator and the denominator
by x minus one, I just need it in the denominator.
[01:02:52] But I can't get away with just multiplying
by the denominator without changing my expression,
[01:02:56] I have to multiply by it on the numerator
and the denominator. So I'm just multiplying
[01:03:00] by one and a fancy form and not changing the
value here. So now I do the same thing for
[01:03:06] the second rational expression, I'll I'll
write the denominator in factored form to
[01:03:10] make it easier to see what's missing from
the denominator. What's missing in this denominator,
[01:03:16] compared to my least common denominator is
just the factor two, so multiply the numerator
[01:03:21] and the denominator by two. Now I can rewrite
everything. So the first rational expression
[01:03:27] becomes three times x minus one over two times
x plus 1x minus one, and the second one becomes
[01:03:36] five times two over two times x plus 1x minus
one, notice that I now have a common denominator.
[01:03:44] So I can just add together my numerators.
So I get three times x minus one plus 10 over
[01:03:51] two times x plus 1x minus one. I'd like to
simplify this. And the best way to do that
[01:03:58] is to leave the denominator in factored form.
But I do have to multiply out the numerator
[01:04:04] so that I can add things together. So I get
3x minus three plus 10 over two times x plus
[01:04:12] 1x minus one, or 3x plus seven over two times
x plus 1x minus one. Now 3x plus seven doesn't
[01:04:22] factor. And there's therefore no factors that
I can cancel out. So this is already reduced.
[01:04:29] As much as it can be. This is my final answer.
[01:04:33] In this video, we saw how to simplify rational
expressions to lowest terms by factoring and
[01:04:40] canceling common factors.
[01:04:42] We also saw how to multiply rational expressions
by multiplying the numerator and multiplying
[01:04:47] the denominator, how to divide rational expressions
by flipping and multiplying and how to add
[01:04:53] and subtract rational expressions by writing
them in terms of the least common denominator
[01:05:00] This video is about solving quadratic equations.
a quadratic equation is an equation that contains
[01:05:07] the square of the variable, say x squared,
but no higher powers of x.
[01:05:11] The standard form for a quadratic equation
is the form a x squared plus b x plus c equals
[01:05:20] zero, where A, B and C
[01:05:25] represent real numbers. And a is not zero
so that we actually have an x squared term.
[01:05:31] Let me give you an example. 3x squared plus
7x minus two equals zero is a quadratic equation
[01:05:40] in standard form, here a is three, B is seven,
and C is minus two. The equation 3x squared
[01:05:49] equals minus 7x plus two is also a quadratic
equation, it's just not in standard form.
[01:05:56] The key steps to solving quadratic equations
are usually to write the equation in standard
[01:06:02] form, and then either factor it
[01:06:07] or use the quadratic formula, which I'll show
you later in this video.
[01:06:14] Let's start with the example y squared equals
18 minus seven y. Here our variable is y,
[01:06:21] and we need to rewrite this quadratic equation
in standard form, we can do this by subtracting
[01:06:26] 18 from both sides and adding seven y to both
sides. That gives us the equation y squared
[01:06:33] minus 18 plus seven y equals zero. And I can
rearrange a little bit to get y squared plus
[01:06:40] seven y minus 18 equals zero. Now I've got
my equation in standard form. Next, I'm going
[01:06:48] to try to factor it. So I need to look for
two numbers that multiply to negative 18 and
[01:06:54] add to seven.
[01:06:57] two numbers that work are nine and negative
two, so I can factor my expression on the
[01:07:02] left as y plus nine times y minus two equals
zero. Now, anytime you have two quantities
[01:07:12] that multiply together to give you zero, either
the first quantity has to be zero, or the
[01:07:17] second quantity has to be zero, or I suppose
they both could be zero. In some situations,
[01:07:22] this is really handy, because that means that
I know that either y plus nine equals zero,
[01:07:27] or y minus two equals zero. So I can as my
next step, set my factors equal to zero. So
[01:07:36] y plus nine equals zero, or y minus two equals
zero, which means that y equals negative nine,
[01:07:43] or y equals two.
[01:07:46] It's not a bad idea to check that those answers
actually work by plugging them into the original
[01:07:51] equation, negative nine squared, does that
equal 18 minus seven times negative nine,
[01:07:56] and you can work out that it does. And similarly,
two squared equals 18 minus seven times two.
[01:08:04] In the next example, let's find solutions
of the equation w squared equals 121. This
[01:08:11] is a quadratic equation, because it's got
a square of my variable W, I can rewrite it
[01:08:17] in standard form by subtracting 121 from both
sides.
[01:08:24] Notice that A is equal to one b is equal to
zero because there's no w term, and c is equal
[01:08:31] to negative 121 in the standard form, a W
squared plus BW plus c equals zero. Next step,
[01:08:39] I'm going to try to factor this expression.
So since 121, is 11 squared, this is a difference
[01:08:47] of two squares, and it factors as w plus 11,
times w minus 11 is equal to zero. If I set
[01:08:56] the factors equal to zero,
[01:08:59] I get w plus 11 equals zero or w minus 11
equals zero. So w equals minus 11, or w equals
[01:09:07] 11. In this example, I could have solved the
equation more simply, I could have instead
[01:09:14] said that if W squared is 121, and then w
has to equal plus or minus the square root
[01:09:20] of 121. In other words, W is plus or minus
11.
[01:09:26] If you saw the equation this way, it's important
to remember the plus or minus since minus
[01:09:30] 11 squared equals 121, just like 11 squared
does.
[01:09:36] Now let's find the solutions for the equation.
x times x plus two equals seven. Some people
[01:09:41] might be tempted to say that, oh, if two numbers
multiply to equal seven, then one of them
[01:09:46] better equal one and the other equals seven
or maybe negative one and negative seven.
[01:09:51] But that's faulty reasoning in this case,
because x and x plus two don't have to be
[01:09:57] whole numbers. They could be crazy.
[01:10:00] fractions or even irrational numbers. So instead,
let's rewrite this equation in standard form.
[01:10:08] To do that, I'm first going to multiply out.
So x times x is x squared x times two is 2x.
[01:10:16] That equals seven, and I'll subtract the seven
from both sides to get x squared plus 2x minus
[01:10:23] seven is zero. Now I'm looking to factor it.
So I need two numbers that multiply to negative
[01:10:28] seven and add to two, since the only way to
factor negative seven is as negative one times
[01:10:34] seven or seven times negative one, it's easy
to see that there are no whole numbers that
[01:10:39] will do will work. So there's no way to factor
this expression over the integers. Instead,
[01:10:46] let's use the quadratic equation. So we have
our leading coefficient of x squared is one.
[01:10:53] So A is one, B is two, and C is minus seven.
And we're going to plug that into the equation
[01:11:00] quadratic equation, which goes x equals negative
b plus or minus the square root of b squared
[01:11:06] minus four, I see all over two
[01:11:10] different people have different ways of remembering
this formula, I'd like to remember it by seeing
[01:11:15] it x equals negative b plus or minus the square
root of b squared minus four, I see oh over
[01:11:22] to a, but you can use any pneumonic you like.
Anyway, plugging in here, we have x equals
[01:11:28] negative two plus or minus the square root
of two squared minus four times one times
[01:11:34] negative seven support and remember the negative
seven there to all over two times one.
[01:11:40] Now two squared is four, and four times one
times negative seven is negative 28. So this
[01:11:48] whole quantity under the square root sign
becomes four minus negative 28, or 32. So
[01:11:55] I can rewrite this as x equals negative two
plus or minus the square root of 32, all over
[01:12:00] two.
[01:12:01] Since 32, is 16 times two and 16 is a perfect
square, I can rewrite this as negative two
[01:12:11] plus or minus the square root of 16 times
the square root of two over two, which is
[01:12:15] negative two plus or minus four times the
square root of two over two.
[01:12:19] Next, I'm going to split out my fraction
[01:12:26] as negative two over two plus or minus four
square root of two over two, and then simplify
[01:12:32] those fractions. This becomes negative one
plus or minus two square root of two. So my
[01:12:38] answers are negative one plus two square root
of two and negative one minus two square root
[01:12:45] of two. And if I need a decimal answer for
any reason, I could work this out on my calculator.
[01:12:52] As our final example, let's find all real
solutions for the equation, one half y squared
[01:12:57] equals 1/3 y minus two. I'll start as usual
by putting it in standard form. So that gives
[01:13:04] me one half y squared minus 1/3 y plus two
equals zero, I could go ahead and start trying
[01:13:12] to factor or use the quadratic formula right
now. But I find fractional coefficients kind
[01:13:17] of annoying. So I'd like to get rid of them.
By doing what I call clearing the denominator,
[01:13:22] that means I'm going to multiply the whole
entire equation by the least common denominator.
[01:13:27] In this case, the least common denominator
is two times three or six. So I'll multiply
[01:13:33] the whole equation by six have to make sure
I multiplied both sides of the equation, but
[01:13:37] in this case, six times zero is just zero.
And when I distribute the six, I get three
[01:13:42] y squared minus two y plus 12 equals zero.
[01:13:47] Now, I could try to factor this, but I think
it's easier probably just to plunge in and
[01:13:51] use the quadratic formula.
[01:13:54] So I get x equals negative B, that's negative
negative two or two, plus or minus the square
[01:14:01] root of b squared, minus four times a times
c, all over to a.
[01:14:12] Working out the stuff in a square root sign,
negative two squared is four. And here we
[01:14:17] have, let's see 144.
[01:14:21] So this simplifies to x equals to plus or
minus the square root of four minus 144. That's
[01:14:28] negative 140. All of our sex. Well, if you're
concerned about that negative number under
[01:14:35] the square root sign, you should be we can't
take the square root of a negative number
[01:14:40] and get an N get a real number is our answer.
There's no real number whose square is a negative
[01:14:45] number. And therefore, our conclusion is we
have no real solutions to this quadratic equation.
[01:14:54] In this video, we solve some quadratic equations
by first writing them in standard form and
[01:14:59] then
[01:15:00] Either factoring or using the quadratic formula.
[01:15:03] In some examples, factoring doesn't work,
it's not possible to factor the equation.
[01:15:08] But in fact, using the quadratic formula will
always work even if it's also possible to
[01:15:13] solve it by factoring. So you can't really
lose by using the quadratic formula. It's
[01:15:19] just sometimes it'll be faster to factor instead.
[01:15:23] This video is about solving rational equations.
A rational equation, like this one is an equation
[01:15:29] that has rational expressions in that, in
other words, an equation that has some variables
[01:15:34] in the denominator.
[01:15:35] There are several different approaches for
solving a rational equation, but they all
[01:15:41] start by finding the least common denominator.
In this example, the denominators are x plus
[01:15:46] three and x, we can think of one as just having
a denominator of one.
[01:15:52] Since the denominators don't have any factors
in common, I can find the least common denominator
[01:15:57] just by multiplying them together.
[01:16:00] My next step is going to be clearing the denominator.
[01:16:04] By this, I mean that I multiply both sides
of my equation by this least common denominator,
[01:16:12] x plus three times x, I multiply on the left
side of the equation, and I multiply by the
[01:16:19] same thing on the right side of the equation.
[01:16:22] Since I'm doing the same thing to both sides
of the equation, I don't change the the value
[01:16:28] of the equation. Multiplying the least common
denominator on both sides of the equation
[01:16:33] is equivalent to multiplying it by all three
terms in the equation, I can see this when
[01:16:38] I multiply out,
[01:16:41] I'll rewrite the left side the same as before,
pretty much. And then I'll distribute the
[01:16:47] right side to get x plus three times x times
one plus x plus three times x times one over
[01:16:56] x. So I've actually multiplied the least common
denominator by all three terms of my equation.
[01:17:01] Now I can have a blast canceling things. The
x plus three cancels with the x plus three
[01:17:07] on the denominator.
[01:17:10] The here are nothing cancels out because there's
no denominator, and here are the x in the
[01:17:15] numerator cancels with the x in the denominator.
[01:17:18] So I can rewrite my expression as x squared
equals x plus three times x times one plus
[01:17:26] x plus three. Now I'm going to simplify.
[01:17:29] So I'll leave the x squared alone on this
side, I'll distribute out x squared plus 3x
[01:17:36] plus x plus three, hey, look, the x squared
is cancel on both sides. And so I get zero
[01:17:44] equals 4x plus three, so 4x is negative three,
and x is negative three fourths. Finally,
[01:17:51] I'm going to plug in my answer to check. This
is a good idea for any kind of equation. But
[01:17:57] it's especially important for a rational equation
because occasionally for rational equations,
[01:18:01] you'll get what's called extraneous solution
solutions that don't actually work in your
[01:18:05] original equation because they make the denominator
zero. Now, in this example, I don't think
[01:18:09] we're going to get any extraneous equations
because negative three fourths is not going
[01:18:16] to make any of these denominators zero, so
it should work out fine when I plug in. If
[01:18:20] I plug in,
[01:18:22] I get this, I can simplify
[01:18:29] the denominator here, negative three fourths
plus three, three is 12 fourths, as becomes
[01:18:34] nine fourths. And this is one I'll flip and
multiply to get minus four thirds. So here,
[01:18:42] I can simplify my complex fraction, it ends
up being negative three nights, and one minus
[01:18:49] four thirds is negative 1/3. So that all seems
to check out.
[01:18:55] And so my final answer is x equals negative
three fourths.
[01:19:00] This next example looks a little trickier.
And it is, but the same approach will work.
[01:19:05] First off, find the least common denominator.
So here, my denominators are c minus five,
[01:19:13] c plus one, and C squared minus four c minus
five, I'm going to factor that as C minus
[01:19:22] five times c plus one. Now, my least common
denominator needs to have just enough factors
[01:19:29] to that each of these denominators divided
into it. So I need the factor c minus five,
[01:19:35] I need the factor c plus one. And now I've
already got all the factors I need for this
[01:19:39] denominator. So here is my least common denominator.
Next step is to clear the denominators.
[01:19:47] So I do this by multiplying both sides of
the equation by my least common denominator.
[01:19:53] In fact, I can just multiply each of the three
terms by this least common denominator
[01:20:01] I went ahead and wrote my third denominator
in factored form to make it easier to see
[01:20:06] what cancels. Now canceling time dies, this
dies. And both of those factors die. cancelling
[01:20:16] out the denominator is the whole point of
multiplying by the least common denominator,
[01:20:21] you're multiplying by something that's big
enough to kill every single denominator, so
[01:20:25] you don't have to deal with denominators anymore.
[01:20:28] Now I'm going to simplify by multiplying out.
[01:20:32] So I get, let's see, c plus one times four
c, that's four c squared plus four c, now
[01:20:41] I get minus just c minus five, and then over
here, I get three c squared plus three,
[01:20:51] I can rewrite the minus quantity c minus five
is minus c plus five.
[01:21:00] And now I can subtract the three c squared
from both sides to get just a C squared over
[01:21:06] here, and the four c minus c that becomes
a three C.
[01:21:15] And finally, I can subtract the three from
both sides to get c squared plus three c plus
[01:21:21] two equals zero. got myself a quadratic equation
that looks like a nice one that factors. So
[01:21:28] this factors to C plus one times c plus two
equals zero. So either c plus one is zero,
[01:21:36] or C plus two is zero. So C equals negative
one, or C equals negative two.
[01:21:44] Now let's see, we need to still check our
answers.
[01:21:49] Without even going to the trouble of calculating
anything, I can see that C equals negative
[01:21:54] one is not going to work, because if I plug
it in to this denominator here, I get a denominator
[01:22:01] zero, which doesn't make sense. So C equals
minus one is an extraneous solution, it doesn't
[01:22:09] actually satisfy my original equation. And
so I can just cross it right out, C equals
[01:22:16] negative two.
[01:22:19] I can go if I go ahead, and that doesn't make
any of my denominators zero. So if I haven't
[01:22:23] made any mistakes, it should satisfy my original
equation, but, but I'll just plug it in to
[01:22:30] be sure.
[01:22:33] And after some simplifying,
[01:22:36] I get a true statement.
[01:22:39] So my final answer is C equals negative two.
In this video, we saw the couple of rational
[01:22:46] equations using the method of finding the
least common denominator and then clearing
[01:22:53] the denominator,
[01:22:55] we cleared the denominator by multiplying
both sides of the equation by the least common
[01:22:59] denominator or equivalently. multiplying each
of the terms by that denominator.
[01:23:04] There's another equivalent method that some
people prefer, it still starts out the same,
[01:23:10] we find the least common denominator, but
then we write all the fractions
[01:23:16] over that least common denominator. So in
this example, we'd still use the least common
[01:23:22] denominator of x plus three times x. But our
next step would be to write each of these
[01:23:27] rational expressions over that common denominator
by multiplying the top and the bottom by the
[01:23:32] appropriate things. So one, in order to get
the common denominator of x plus 3x, I need
[01:23:38] to multiply the top and the bottom by x plus
three times x, one over x, I need to multiply
[01:23:44] the top and the bottom just by x plus three
since that's what's missing from the denominator
[01:23:49] x. Now, if I simplify a little bit,
[01:23:53] let's say this is x squared over that common
denominator, and
[01:24:00] here I have just x plus three times x over
that denominator, and here I have x plus three
[01:24:09] over that common denominator. Now add together
my fractions on the right side, so they have
[01:24:17] a common denominator.
[01:24:18] So this is x plus three times x plus x plus
three. And now I have two fractions that have
[01:24:27] that are equal that have the same denominator,
therefore their numerators have to be equal
[01:24:32] also. So the next step is to set the numerators
equal.
[01:24:37] So I get x squared is x plus three times x
plus x plus three. And if you look back at
[01:24:46] the previous way, we solve this equation,
you'll recognize this equation. And so from
[01:24:52] here on we just continue as before.
[01:24:57] When choosing between these two methods, I
personally tend
[01:25:00] prefer the clear the denominators method,
because it's a little bit less writing, you
[01:25:03] don't have to get rid of those denominators
earlier, you don't have to write them as many
[01:25:07] times. But some people find this one a little
bit easier to remember, a little easier to
[01:25:12] understand either of these methods is fine.
[01:25:15] One last caution, don't forget at the end,
to check your solutions and eliminate any
[01:25:21] extraneous solutions.
[01:25:23] These will be solutions that make the denominators
of your original equations go to zero.
[01:25:30] This video is about solving radical equations,
that is equations like this one that have
[01:25:36] square root signs in them, or cube roots or
any other kind of radical.
[01:25:40] When I see an equation with a square root
in it, I really want to get rid of the square
[01:25:45] root. But it'll be easiest to get rid of the
square root. If I first isolate the square
[01:25:51] root. In other words, I want to get the term
with the square root and that on one side
[01:25:56] of the equation by itself, and everything
else on the other side of the equation. If
[01:26:01] I start with my original equation, x plus
the square root of x equals 12. And I subtract
[01:26:07] x from both sides, then that does isolate
the square root term on the left side with
[01:26:12] everything else on the right. Once I've isolated
the term with the square root, I want to get
[01:26:19] rid of the square root. And I'll do that by
squaring both sides
[01:26:27] of my equation. So I'll take the square root
of x equals 12 minus x and square both sides.
[01:26:36] Now the square root of x squared is just x,
taking the square root and then squaring those
[01:26:42] operations undo each other
[01:26:45] to work out 12 minus x squared, write it out
and distribute 12 times 12 is 144 12 times
[01:26:54] minus x is minus 12x. I get another minus
12x from here.
[01:27:02] And finally minus x times minus x is positive
x squared. So I can combine my minus 12 x's,
[01:27:09] that's minus 24x. And now I can subtract x
from both sides to get zero equals 144 minus
[01:27:18] 25x plus x squared. That's a quadratic equation,
I'll rewrite it in a little bit more standard
[01:27:26] form here.
[01:27:27] So now I've got a familiar quadratic equation
with no radical signs left in my equation,
[01:27:34] I'll just proceed to solve it like I usually
do for a quadratic, I'll try to factor it.
[01:27:38] So I'm going to look for two numbers that
multiply to 144 and add to minus 25. I know
[01:27:46] I'm going to need negative numbers to get
to negative 25. And in fact, I'll need two
[01:27:53] negative numbers. So they still multiply to
a positive number. So I'll start listing some
[01:27:57] factors of 144, I could have negative one
and a negative 144, negative two, and negative
[01:28:04] 72, negative four, negative 36, and so on.
Once I've listed out the possible factors,
[01:28:12] it's not hard to find the two that add to
negative 25. So that's negative nine and negative
[01:28:17] 16. So now I can factor in my quadratic equation
as x minus nine times x minus 16 equals zero,
[01:28:28] that means that x minus nine is zero or x
minus 16 is zero. So x equals nine or x equals
[01:28:34] 16. I'm almost done. But there's one last
very important step. And that's to check the
[01:28:41] Solutions so that we can eliminate any extraneous
solutions and extraneous solution as a solution
[01:28:49] that we get that does not actually satisfy
our original equation and extraneous solutions
[01:28:54] can happen when you're solving equations with
radicals in them. So let's first check x equals
[01:29:00] nine. If we plug in to our original equation,
we get nine plus a squared of nine and we
[01:29:06] want that to equal 12. Well, the square root
of nine is three, and nine plus three does
[01:29:11] indeed equal 12. So that solution checks out.
Now, let's try x equals 16. Plugging in, we
[01:29:18] get 16 plus a squared of 16. And that's supposed
to equal 12. Well, that says 16 plus four
[01:29:25] is supposed to equal 12. But that most definitely
is not true. And so x equals 16. Turns out
[01:29:31] to be extraneous solution, and our only solution
is x equals nine
[01:29:47] This next equation might not look like an
equation involving radicals. But in fact,
[01:30:26] we can think of a fractional exponent as being
a radical in disguise. Let's start by doing
[01:30:32] the same thing we did on the previous problem
by isolating This time, we'll isolate the
[01:30:38] part of the equation
[01:30:41] that involves the fractional exponent. So
I'll start with the original equation, two
[01:30:46] times P to the four fifths equals 1/8. And
I'll divide both sides by two or equivalently,
[01:30:53] I can multiply both sides by one half, that
gives me p to the four fifths equals 1/16.
[01:31:00] And I've effectively isolated the part of
the equation with the fractional exponent
[01:31:05] as much as possible. Now, in the previous
example, the next step was to get rid of the
[01:31:12] radical. In this example, we're going to get
rid of the fractional exponent. And I'm going
[01:31:18] to actually do this in two stages. First,
I'm going to raise both sides to the fifth
[01:31:24] power.
[01:31:26] That's because when I take an exponent to
an exponent, I'm multiply my exponents. And
[01:31:34] so that becomes just p to the fourth equals
1/16 to the fifth power. Now, I'm going to
[01:31:41] get rid of the fourth power by raising both
sides to the 1/4 power, or by taking the fourth
[01:31:48] root, there's something that you need to be
careful about though, when taking
[01:31:53] an even root, or the one over an even number
power, you always have to consider plus or
[01:32:01] minus your answer.
[01:32:04] It's kind of like when you write x squared
equals four, and you take the square root
[01:32:10] of both sides, x could equal plus or minus
the square root of four, right, x could equal
[01:32:16] plus or minus two, since minus two squared
is four, just as well as two squared. So that's
[01:32:23] why when you take an even root, or a one over
an even power of both sides, you always need
[01:32:32] to include the plus or minus sign, when it's
an odd root or one over an odd power, you
[01:32:39] don't need to do that. If you had something
like x cubed equals negative eight, then x
[01:32:46] equals the cube root of negative eight, which
is negative two would be your only solution,
[01:32:51] you don't need to do the plus or minus because
positive two wouldn't work.
[01:32:55] So that aside, explains why we need this plus
or minus power. And now p to the four to the
[01:33:02] 1/4. When I raise a power to a power, I multiply
by exponents, so that's just p to the one,
[01:33:09] which is equal to plus or minus 1/16 to the
fifth power to the 1/4 power.
[01:33:17] Now I just need to simplify this expression,
I don't really want to raise 1/16 to the fifth
[01:33:23] power, because 16 to the fifth power is like
a really huge number. So I think I'm actually
[01:33:28] going to rewrite this first
[01:33:32] as P equals plus or minus 1/16. I'll write
it back as the 5/4 power again.
[01:33:42] And as I continue to solve using my exponent
rules,
[01:33:49] I'm going to prefer to write this as 1/16
to the fourth root to the fifth power, because
[01:33:58] it's going to be easier to take the fourth
root, let's see the fourth root of 1/16 is
[01:34:03] the same thing as the fourth root of one over
the fourth root of 16. Raise all that to the
[01:34:08] fifth power. fourth root of one is just one
and the fourth root of 16 is to raise that
[01:34:14] to the fifth power, that's just going to be
one to the fifth over two to the fifth, which
[01:34:19] is plus or minus 130 seconds.
[01:34:24] The last step is to check answers.
[01:34:28] So I have the two answers p equals 130 seconds,
and P equals minus 130 seconds. And if I plug
[01:34:35] those both in
[01:34:36] 1/32 to the fourth fifth power.
[01:34:43] That gives me two times one to the fourth
power over 32 to the fourth power, which is
[01:34:51] two times one over 32/5 routed to the fourth
power. fifth root of 32 is two
[01:34:59] Raisa to the fourth power, I get 16. So this
is two times 1/16, which is 1/8, just as we
[01:35:07] wanted in the original equation up here. Similarly,
we can check that the P equals negative 1/32
[01:35:16] actually does satisfy the equation, I'll leave
that step to the viewer.
[01:35:24] So our two solutions are p equals one over
32, and P equals minus one over 32. I do want
[01:35:30] to point out an alternate approach to getting
rid of the fractional exponent, we could have
[01:35:35] gotten rid of it all in one fell swoop by
raising both sides of our equation to the
[01:35:41] five fourths power.
[01:35:45] five fourths is the reciprocal of four fifths.
So when I use my exponent rules, and say that
[01:35:51] when I raise the power to the power, I multiply
my exponents, that gives me p to the four
[01:35:56] fifths times five fourths is plus or minus
1/16 to the five fourths, in other words,
[01:36:03] P to the One Power, which is just P is plus
or minus 1/16, to the five fourths, so that's
[01:36:10] an alternate and possibly faster way to get
the solution. Once again, the plus or minus
[01:36:16] comes from the fact that when we take the
5/4 power, we're really taking an even root
[01:36:22] a fourth root, and so we need to consider
both positive and negative answers.
[01:36:27] This video is about solving radical equations,
that is equations like this one that have
[01:36:33] square root signs in them, or cube roots or
any other kind of radical.
[01:36:39] In this video, we solved radical equations
by first isolating the radical sign, or the
[01:36:45] fractional exponent,
[01:36:48] and then removing the radical sine or the
fractional exponent
[01:36:54] by either squaring both sides or taking the
reciprocal power of both sides.
[01:37:01] This video is about solving equations with
absolute values in them.
[01:37:06] Recall that the absolute value of a positive
number is just the number, but the absolute
[01:37:12] value of a negative number is its opposite.
In general, I think of the absolute value
[01:37:17] of a number as representing its distance from
zero on the number line,
[01:37:23] the number four,
[01:37:25] and the number of negative four are both at
distance for from zero, and so the absolute
[01:37:32] value of both of them is four.
[01:37:35] Similarly, if I write the equation, the absolute
value of x is three, that means that x has
[01:37:44] to be three units away from zero on the number
line.
[01:37:50] And so X would have to be either negative
three, or three. Let's start with the equation
[01:37:56] three times the absolute value of x plus two
equals four.
[01:38:00] I'd like to isolate the absolute value part
of the equation, I can do this by starting
[01:38:08] with my original equation,
[01:38:14] subtracting two from both sides
[01:38:17] and dividing both sides by three.
[01:38:23] Now I'll think in terms of distance on a number
line, the absolute value of x is two thirds
[01:38:31] means that x is two thirds away from zero.
So x could be here for here at negative two
[01:38:40] thirds or two thirds.
[01:38:43] And the answer to my equation is x is negative
two thirds, or two thirds,
[01:38:50] I can check my answers by plugging in
[01:38:52] three times the added value of negative two
thirds plus two, I need to check it that equals
[01:39:00] four. Well, the absolute value of negative
two thirds is just two thirds. So this is
[01:39:04] three times two thirds plus two, which works
out to four.
[01:39:10] Similarly, if I plug in positive two thirds,
it also works out to give me the correct answer.
[01:39:20] The second example is a little different,
because the opposite value sign is around
[01:39:24] a more complicated expression, not just around
the X.
[01:39:29] I would start by isolating the absolute value
part.
[01:39:35] But it's already isolated. So I'll just go
ahead and jump to thinking about distance
[01:39:41] on the number line. So on my number line,
the whole expression 3x plus two is supposed
[01:39:49] to be at a distance of four from zero.
[01:39:53] So that means that 3x plus two is here at
four or 3x plus two
[01:39:59] though is it negative four, all right, those
as equations
[01:40:03] 3x plus two equals four, or 3x plus two is
minus four. And then I can solve.
[01:40:12] So this becomes 3x equals two, or x equals
two thirds. And over here, I get 3x equals
[01:40:20] minus six, or x equals minus two. Finally,
I'll check my answers.
[01:40:29] I'll leave it to you to verify that they both
work.
[01:40:32] A common mistake on absolute value equations
is to get rid of the absolute value signs
[01:40:38] like we did here, and then just solve for
one answer, instead of solving for both answers.
[01:40:44] Another mistake sometimes people make is,
once they get the first answer, they just
[01:40:48] assume that the negative of that works also.
[01:40:52] But that doesn't always work. In the first
example, our two answers were both the negatives
[01:40:57] of each other. But in our second examples,
or two answers, were not just the opposites
[01:41:02] of each other one was two thirds and the other
was negative two.
[01:41:06] In this third example, let's again, isolate
the absolute value part of the equation.
[01:41:12] So starting with our original equation,
[01:41:16] we can subtract 16 from both sides,
[01:41:22] and divide both sides by five or equivalently,
multiply by 1/5.
[01:41:31] Now let's think about distance on the number
line,
[01:41:35] we have an absolute value needs to equal negative
three. So that means whatever is inside the
[01:41:41] absolute value sign needs to be at distance
negative three from zero, well, you can't
[01:41:48] be at distance negative three from zero. Another
way of thinking about this is you can't have
[01:41:52] the absolute value of something and end up
with a negative number if the value is always
[01:41:57] positive, or zero. So this equation doesn't
actually make sense, and there are no solutions
[01:42:05] to this equation.
[01:42:07] In this video, we solved absolute value equations.
In many cases, an absolute value equation
[01:42:15] will have two solutions. But in some cases,
it'll have no solutions. And occasionally,
[01:42:20] it'll have just one solution.
[01:42:25] This video is about interval notation, and
easy and well known way to record inequalities.
[01:42:30] Before dealing with interval notation, it
is important to know how to deal with inequalities.
[01:42:39] Our first example of an inequality is written
here, all numbers between one and three, not
[01:42:45] including one and three.
[01:42:46] First, we used to write down our variable
x, that it is important to locate the key
[01:42:53] values for this problem that is one and three.
Now here we see that x is between these two
[01:43:01] numbers, meaning we will have one inequality
statement on each side of the variable. Here
[01:43:07] it says not including one and three, which
means that we do not have an or equal to sign
[01:43:12] beneath each inequality. Here we will put
one the lowest key value, and here we will
[01:43:18] put three the highest key value. Next, we're
going to graph this inequality on a number
[01:43:24] line.
[01:43:27] Here we write our key values one, and three.
Because it is not including one, and three,
[01:43:34] we have an empty circle around each number.
Because it isn't the numbers between we have
[01:43:40] a line connecting them.
[01:43:41] The last step of this problem is writing this
inequality in interval notation.
[01:43:50] writing things in interval notation is kind
of like writing an ordered pair. Here we will
[01:43:54] put one and then here we will put three.
[01:43:57] Next we need to put brackets around these
numbers.
[01:44:03] For this problem, it is not including one
in three, which means we will use soft brackets.
[01:44:09] However, if it were including one and three,
we will use hard brackets.
[01:44:15] It is important to note for interval notation,
that the smallest value always goes on the
[01:44:22] left and the biggest value always goes on
the right.
[01:44:29] You also include a comma between your two
key values.
[01:44:32] Now let's work on problem B. Once again we
write our variable x. Again it is between
[01:44:41] negative four and two but this time it is
including a negative four into this requires
[01:44:47] us to have the or equal to assign below the
inequality. Then we put our lowest key value
[01:44:53] here and their highest here. The number line
graph for this problem is slightly different.
[01:44:59] We still
[01:45:00] have our key values negative four and two.
But instead of an open circle, like we have
[01:45:07] right here, we instead use a closed circle,
representing that it is including negative
[01:45:15] four and two, you complete this with a line
in between. Last we write this in interval
[01:45:21] notation. In the last problem, I said that
for numbers, including the outside values,
[01:45:27] we would use these hard brackets. Now we are
actually using this. So we have or hard bracket
[01:45:35] on each side because isn't including foreign
zoo, then we put our smallest value on the
[01:45:40] left, and our highest value on the right,
with the comma in between.
[01:45:46] You now know how to correctly write these
two types of intervals.
[01:45:51] Now we are going to practice transforming
these from inequality notation to interval
[01:45:58] notation, and vice versa. This is slightly
more difficult than our previous examples
[01:46:05] of having two soft brackets or two hard brackets,
we'll have two different types. Now on the
[01:46:11] left side will have a hard bracket because
it is including the three. However, on the
[01:46:18] right side, it is a soft bracket because it
is it it is not including the one that we
[01:46:24] put a comma in the middle, a lower key value
and our higher key value. For the next problem,
[01:46:31] we're taking an equation already written an
interval notation and putting it back into
[01:46:37] inequality notation.
[01:46:38] For the second problem, we can see are key
values as being five and negative infinity.
[01:46:47] Now, this sounds a little bit weird, but let's
just set up or an equality, we have our variable
[01:46:54] x, which is less than because of the soft
bracket five, and greater than without the
[01:47:02] or equal to because it has a soft bracket
there to infinity. But because x is always
[01:47:09] greater than negative infinity, we can take
out this part, leaving us with x is less than
[01:47:16] five.
[01:47:17] It is important to note that a soft bracket
always accompanies infinity. This is because
[01:47:25] infinity is not a real number, so we cannot
include it just go as far up to it as we can.
[01:47:32] The next problem is a little tricky, because
we don't see another key value over here.
[01:47:38] But let's just start off the equation as a
soft bracket on this side due to the absence
[01:47:43] of an order equal to sign. And negative 15
is the lower key value.
[01:47:51] On the other side, we put the highest possible
number infinity, and then close it off with
[01:47:57] a soft bracket. We can see the relation of
this to the previous problem. And how it goes
[01:48:05] into this problem when x is greater than a
key value instead of less than a key value.
[01:48:12] Once again, we have our sauce bracket for
infinity, which is always true.
[01:48:19] Part D brings up an important point about
which number goes on the left. For all of
[01:48:26] our other problems, we have had the inequalities
pointing left
[01:48:30] instead of rights. When seeing Part D, you
might think oh, four is on the left, so it
[01:48:41] would go here, and zero is on the right who
would go here. But that is not true.
[01:48:47] When writing inequalities, you must always
have the lower value on the left, which for
[01:48:53] this problem is zero. To fix this, we can
simply flip this inequality. So it reads like
[01:49:01] this, it is still identical just written in
an easier form to transform it into interval
[01:49:06] notation, then we can see that zero has a
soft bracket, because it is not including
[01:49:13] zero, but r comma four or other key value
and a hard backup because it is or equal to
[01:49:19] four.
[01:49:21] For interval notation, you must always have
the smaller number on the left side.
[01:49:29] This was our video on interval notation, an
alternative way to write inequalities.
[01:49:36] This video is about solving inequalities that
have absolute value signs in them.
[01:49:41] Let's look at the inequality absolute value
of x is less than five on the number line.
[01:49:45] Thinking of absolute value as distance. This
means that the distance between x and zero
[01:49:52] is less than five units. So x has to live
somewhere in between negative five and five
[01:50:01] We can express this as an inequality without
absolute value signs by saying negative five
[01:50:06] is less than x, which is less than five. Or
we can use interval notation, soft bracket
[01:50:12] negative five, five, soft bracket.
[01:50:16] Both of these formulations are equivalent
to the original one, but don't involve the
[01:50:20] absolute value signs.
[01:50:23] In the second example, we're looking for the
values of x for which the absolute value of
[01:50:28] x is greater than or equal to five.
[01:50:32] on the number line, this means that the distance
of x from the zero, it's got to be bigger
[01:50:39] than or equal to five units.
[01:50:42] A distance bigger than five units means that
x has to live somewhere over here, or somewhere
[01:50:48] over here, where it's farther than five units
away from zero. Course x could also have a
[01:50:54] distance equal to five units. So I'll fill
in that dot and shade in the other parts of
[01:51:00] the number line that satisfy my inequality.
[01:51:04] Now I can rewrite the inequality without the
absolute value symbols by saying that x is
[01:51:10] less than or equal to negative five, or x
is greater than or equal to five. I could
[01:51:17] also write this in interval notation,
[01:51:20] soft bracket, negative infinity, negative
five, hard bracket. And the second part is
[01:51:26] hard bracket five infinity soft bracket, I
combine these with a u for union. Because
[01:51:33] I'm trying to describe all these points on
the number line together with all these other
[01:51:38] points.
[01:51:39] Let's take this analysis a step further with
a slightly more complicated problem. Now I
[01:51:44] want the absolute value of three minus two
t to be less than four.
[01:51:49] And an absolute value less than four means
a distance less than four on the number line.
[01:51:56] But it's not the variable t that lives in
here at a distance of less than four from
[01:52:00] zero, it's the whole expression, three minus
two t.
[01:52:04] So three minus two t, live somewhere in here.
And I can rewrite this as an inequality without
[01:52:13] absolute value signs by saying negative four
is less than three minus two t is less than
[01:52:21] four. Now I have a compound inequality that
I can solve the usual way. First, I subtract
[01:52:27] three from all three sides to get negative
seven is less than negative two t is less
[01:52:33] than one. And now I'll divide all three sides
by negative two. Since negative two is a negative
[01:52:40] number, this reverses the directions of the
inequalities.
[01:52:46] Simplifying, I get seven halves is greater
than t is greater than negative one half.
[01:52:52] So my final answer on the number line looks
like all the stuff between negative a half
[01:52:58] and seven halves.
[01:53:01] But not including the endpoints, and an interval
notation, I can write this soft bracket negative
[01:53:07] a half, seven, half soft bracket,
[01:53:09] please pause the video and try the next problem
on your own.
[01:53:14] thinking in terms of distance, this inequality
says that the distance between the expression
[01:53:19] three minus two t and zero is always bigger
than four. Let me draw this on the number
[01:53:25] line.
[01:53:27] If three minus two t has a distance bigger
than four from zero, then it can't be in this
[01:53:33] region that's near zero, it has to be on the
outside in one of these two regions.
[01:53:39] That is three minus two t is either less than
negative four, or three minus two t is bigger
[01:53:48] than four. I solve these two inequalities
separately, the first one, subtracting three
[01:53:54] from both sides, I get negative two t is less
than negative seven divided by negative two,
[01:54:01] I get T is bigger than seven halves. And then
on the other side, I get negative two t is
[01:54:08] greater than one. So t is less than negative
one half, I had to flip inequalities in the
[01:54:15] last step due to dividing by a negative number.
So let's check this out on the number line
[01:54:21] again, the first piece says that t is greater
than seven halves. I'll draw that over here.
[01:54:28] And the second piece says that t is less than
negative one half. I'll draw that down here.
[01:54:41] Because these two statements are joined with
an or I'm looking for the t values that are
[01:54:46] in this one, or in this one. That is I want
both of these regions put together. So an
[01:54:52] interval notation, this reads negative infinity
to negative one half soft bracket union soft
[01:55:00] bracket
[01:55:01] Seven halves to infinity.
[01:55:03] This last example looks more complicated.
But if I simplify first and isolate the absolute
[01:55:09] value part, it looks pretty much like the
previous ones. So I'll start by subtracting
[01:55:14] seven from both sides. And then I'll divide
both sides by two. Now I'll draw my number
[01:55:20] line.
[01:55:21] And I'm looking for this expression for x
plus five, to always have distance greater
[01:55:28] than or equal to negative three from zero,
wait a second, distance greater than equal
[01:55:33] to negative three, well, distance is always
greater than or equal to negative three is
[01:55:36] always greater than equal to zero. So this,
in fact, is always true.
[01:55:43] And so the answer to my inequality is all
numbers between negative infinity and infinity.
[01:55:50] In other words, all real numbers.
[01:55:55] Once solving absolute value inequalities,
it's good to think about distance. And absolute
[01:56:01] value of something that's less than a number
means that whatever's inside the absolute
[01:56:08] value signs is close to zero.
[01:56:12] On the other hand, an absolute value is something
and being greater than a number means that
[01:56:18] whatever's inside the absolute value sign
is far away from zero, because its distance
[01:56:25] from zero is bigger than that certain number.
[01:56:29] drawing these pictures on the number line
is a helpful way to rewrite the absolute value
[01:56:34] and equality as an inequality that doesn't
contain an absolute value sign. In this case,
[01:56:40] it would be negative three is less than x
plus two is less than three. And in the other
[01:56:47] case, it would be either x plus two is less
than negative three, or x plus two is greater
[01:56:53] than three.
[01:56:55] This video is about solving linear inequalities.
Those are inequalities like this one that
[01:57:01] involve a variable here x, but don't involve
any x squared or other higher power terms.
[01:57:09] The good news is, we can solve linear inequalities,
just like we solve linear equations by distributing,
[01:57:15] adding and subtracting terms to both sides
and multiplying and dividing by numbers on
[01:57:19] both sides. The only thing that's different
is that if you multiply or divide by a negative
[01:57:26] number, then you need to reverse the direction
of the inequality. For example, if we had
[01:57:32] the inequality, negative x is less than negative
five, and we wanted to multiply both sides
[01:57:39] by negative one to get rid of the negative
signs, we'd have to also switch or reverse
[01:57:45] the inequality. With this caution in mind,
let's look at our first example.
[01:57:50] Since our variable x is trapped in parentheses,
I'll distribute the negative five to free
[01:57:56] it from the parentheses.
[01:57:58] That gives me negative 5x minus 10 plus three
is greater than eight.
[01:58:04] Negative 10 plus three is negative seven,
so I'll rewrite this as negative 5x minus
[01:58:10] seven is greater than eight. Now I'll add
seven to both sides to get negative 5x is
[01:58:16] greater than 15. Now I'd like to divide both
sides by negative five. Since negative five
[01:58:23] is a negative number that reverses the inequality.
So I get x is less than 15 divided by negative
[01:58:31] five. In other words, x is less than negative
three.
[01:58:36] If I wanted to graph this on a number line,
I just need to put down a negative three,
[01:58:40] an open circle around it, and shade in to
the left,
[01:58:44] I use an open circle, because x is strictly
less than negative three and can't equal negative
[01:58:51] three. If I wanted to write this in interval
notation, I've added a soft bracket negative
[01:58:56] infinity, negative three soft bracket. Again,
the soft bracket is because the negative three
[01:59:01] is not included. This next example is an example
of a compound inequality. It has two parts.
[01:59:10] Either this statement is true, or this statement
is true. I want to find the values of x that
[01:59:15] satisfy either one. I'll solve this by working
out each part separately, and then combining
[01:59:21] them at the end. For the inequality on the
left, I'll copy it over here. I'm going to
[01:59:28] add four to both sides.
[01:59:32] then subtract x from both sides
[01:59:36] and then divide both sides by two.
[01:59:40] I didn't have to reverse the inequality sign
because I divided by a positive number
[01:59:45] on the right side, I'll copy the equation
over, subtract one from both sides, and divide
[01:59:51] both sides by 696 is the same as three halves.
Now I'm looking for the x values that make
[01:59:59] this statement.
[02:00:00] row four, make this statement true. Let me
graph this on a number line.
[02:00:07] x is less than or equal to negative two, means
I put a filled in circle there and graph everything
[02:00:14] to the left. X is greater than three halves
means I put a empty circle here and shaded
[02:00:22] and everything to the right.
[02:00:25] My final answer includes both of these pieces,
which I'll reshade in green, because x is
[02:00:33] allowed to be an either one or the other.
Finally, I can write this in interval notation.
[02:00:38] The first piece on the number line can be
described as soft bracket negative infinity,
[02:00:44] negative two hard bracket. And the second
piece can be described as soft bracket three
[02:00:50] halves, infinity soft bracket to indicate
that x can be in either one of these pieces,
[02:00:57] I use the union side, which is a U.
[02:01:01] That means that my answer includes all x values
in here together with all x values in here.
[02:01:09] This next example is also a compound inequality.
This time I'm joined by an ad, the and means
[02:01:15] I'm looking for all y values that satisfy
both this and this at the same time. Again,
[02:01:22] I can solve each piece separately
[02:01:25] on the left, to isolate the Y, I need to multiply
by negative three halves on both sides. So
[02:01:34] that gives me Why is less than negative 12
times negative three halves, the greater than
[02:01:40] flip to a less than because I was multiplying
by negative three halves, which is a negative
[02:01:46] number.
[02:01:47] By clean up the right side, I get why is less
than 18.
[02:01:54] On the right side, I'll start by subtracting
two from both sides.
[02:01:59] And now I'll divide by negative four, again,
a negative number, so that flips the inequality.
[02:02:07] So that's why is less than three over negative
four. In other words, y is less than negative
[02:02:13] three fourths. Again, I'm looking for the
y values that make both of these statements
[02:02:19] true at the same time.
[02:02:22] Let me graph this on the number line,
[02:02:25] the Y is less than 18. I can graph that by
drawing the number 18.
[02:02:34] I don't want to include it. So I use an empty
circle and I shaded everything to the left
[02:02:41] the statement y is less than negative three
fourths, I need to draw a negative three fourths.
[02:02:47] And again, I don't include it, but I do include
everything on the left. Since I want the y
[02:02:54] values for which both of these statements
are true, I need the y values that are in
[02:02:59] both colored blue and colored red. And so
that would be this part right here. I'll just
[02:03:05] draw it above so you can see it easily. So
that would be all the numbers from negative
[02:03:10] three fourths and lower, not including negative
three, four, since those are the parts of
[02:03:15] the number line that have both red and blue
colors on them. In interval notation, my final
[02:03:21] answer will be soft bracket negative infinity
to negative three fourths soft bracket.
[02:03:26] As my final example, I have an inequality
that has two inequality signs in it negative
[02:03:34] three is less than or equal to 6x minus two
is less than 10. I can think of this as being
[02:03:40] a compound inequality with two parts, negative
3x is less than or equal to 6x minus two.
[02:03:47] And at the same time 6x minus two is less
than 10. I could solve this into pieces as
[02:03:54] before. But instead, it's a little more efficient
to just solve it all at once, by doing the
[02:04:00] same thing to all three sides. So as a first
step, I'll add two to all three sides. That
[02:04:08] gives me negative one is less than or equal
to 6x is less than 12. And now I'm going to
[02:04:14] divide all three sides by six to isolate the
x. So that gives me negative one, six is less
[02:04:20] than or equal to x is less than two.
[02:04:24] If we solved it, instead, in two pieces above,
we'd end up with the same thing because we
[02:04:29] get negative one six is less than or equal
to x from this piece, and we'd get x is less
[02:04:35] than two on this piece. And because of the
and statement, that's the same thing as saying
[02:04:40] negative one, six is less than or equal to
x, which is less than two.
[02:04:45] Either way we do it. Let's see if what it
looks like on the number line. So on the number
[02:04:50] line, we're looking for things that are between
two and negative one six, including the negative
[02:04:58] one, six
[02:04:59] But not including the to interval notation,
we can write this as hard bracket, negative
[02:05:05] 162 soft bracket.
[02:05:08] In this video, we solve linear inequalities,
including some compound inequalities, joined
[02:05:13] by the conjunctions and, and or,
[02:05:18] remember when we're working with and we're
looking for places on the number line where
[02:05:22] both statements are true.
[02:05:24] That is, we're looking for the overlap on
the number line.
[02:05:30] In this case, the points on the number line
that are colored both red and blue at the
[02:05:35] same time.
[02:05:36] When we're working with oral statements, we're
looking for places where either one or the
[02:05:43] other statement is true or both
[02:05:46] on the number line,
[02:05:49] this corresponds to points that are colored
either red or blue, or both. And in this picture,
[02:05:54] that will actually correspond to the entire
number line.
[02:05:57] In this video, we'll solve inequalities involving
polynomials like this one, and inequalities
[02:06:04] involving rational expressions like this one.
[02:06:09] Let's start with a simple example. Maybe a
deceptively simple example. If you see the
[02:06:14] inequality, x squared is less than four, you
might be very tempted to take the square root
[02:06:19] of both sides and get something like x is
less than two as your answer. But in fact,
[02:06:26] that doesn't work.
[02:06:27] To see why it's not correct, consider the
x value of negative 10.
[02:06:35] Negative 10 satisfies the inequality, x is
less than two since negative 10 is less than
[02:06:43] two. But it doesn't satisfy the inequality
x squared is less than four, since negative
[02:06:51] 10 squared is 100, which is not less than
four. So these two inequalities are not the
[02:06:58] same. And it doesn't work to solve a quadratic
inequality just to take the square root of
[02:07:04] both sides, you might be thinking part of
why this reasoning is wrong, as we've ignored
[02:07:09] the negative two option, right? If we had
the equation, x squared equals four, then
[02:07:15] x equals two would just be one option, x equals
negative two would be another solution. So
[02:07:21] somehow, our solution to this inequality should
take this into account. In fact, a good way
[02:07:27] to solve an inequality involving x squares
or higher power terms, is to solve the associated
[02:07:33] equation first. But before we even do that,
I like to pull everything over to one side,
[02:07:40] so that my inequality has zero on the other
side.
[02:07:44] So for our equation, I'll subtract four from
both sides to get x squared minus four is
[02:07:49] less than zero.
[02:07:50] Now, I'm going to actually solve the associated
equation, x squared minus four is equal to
[02:07:58] zero, I can do this by factoring to x minus
two times x plus two is equal to zero. And
[02:08:06] I'll set my factors equal to zero, and I get
x equals two and x equals minus two.
[02:08:12] Now, I'm going to plot the solutions to my
equation on the number line. So I write down
[02:08:17] negative two and two, those are the places
where my expression x squared minus four is
[02:08:23] equal to zero.
[02:08:26] Since I want to find where x squared minus
four is less than zero, I want to know where
[02:08:30] this expression x squared minus four is positive
or negative, a good way to find that out is
[02:08:36] to plug in test values. So first, a plug in
a test value in this area of the number line,
[02:08:43] something less than negative to say x equals
negative three.
[02:08:47] If I plug in negative three into x squared
minus four, I get negative three squared minus
[02:08:53] four, which is nine minus four, which is five,
that's a positive number. So at negative three,
[02:09:02] the expression x squared minus four is positive.
And in fact, everywhere on this region of
[02:09:08] the number line, my expression is going to
be positive, because it can jump from positive
[02:09:12] to negative without going through a place
where it's zero, I can figure out whether
[02:09:17] x squared minus four is positive or negative
on this region, and on this region of the
[02:09:21] number line by plugging in test value similar
way,
[02:09:25] evaluate the plug in between negative two
and two, a nice value is x equals 00 squared
[02:09:31] minus four, that's negative four and negative
number. So I know that my expression x squared
[02:09:36] minus four is negative on this whole interval.
Finally, I can plug in something like x equals
[02:09:43] 10, something bigger than two, and I get 10
squared minus four. Without even computing
[02:09:49] that I can tell that that's going to be a
positive number. And that's all that's important.
[02:09:53] Again, since I want x squared minus four to
be less than zero, I'm looking for the places
[02:09:57] on this number line where I'm getting
[02:09:59] negatives. So I will share that in on my number
line. It's in here, not including the endpoints.
[02:10:06] Because the endpoints are where my expression
x squared minus four is equal to zero, and
[02:10:10] I want it strictly less than zero,
[02:10:12] I can write my answer as an inequality, negative
two is less than x is less than two, or an
[02:10:19] interval notation as soft bracket negative
two to soft bracket.
[02:10:25] Our next example, we can solve similarly,
first, we'll move everything to one side,
[02:10:32] so that our inequality is x cubed minus 5x
squared minus 6x is greater than or equal
[02:10:39] to zero. Next, we'll solve the associated
equation by factoring. So first, I'll write
[02:10:47] down the equation. Now I'll factor out an
x. And now I'll factor the quadratic. So the
[02:10:55] solutions to my equation are x equals 0x equals
six and x equals negative one,
[02:11:01] I'll write the solutions to the equation on
the number line.
[02:11:05] So that's negative one, zero, and six. That's
where my expression x times x minus six times
[02:11:17] x plus one is equal to zero.
[02:11:21] But I want to find where it's greater than
or equal to zero. So again, I can use test
[02:11:28] values, I can plug in, for example, x equals
negative two, either to this version of expression,
[02:11:35] or to this factored version. Since I only
care whether my answer is positive or negative,
[02:11:40] it's sometimes easier to use the factored
version. For example, when x is negative two,
[02:11:45] this factor is negative.
[02:11:48] But this factor, x minus six is also negative
when I plug in negative two for x.
[02:11:55] Finally, x plus one, when I plug in negative
two for x, that's negative one, that's also
[02:12:03] negative. And a negative times a negative
times a negative gives me a negative number.
[02:12:08] If I plug in something, between negative one
and zero, say x equals negative one half,
[02:12:16] then I'm going to get a negative for this
factor, a negative for this factor, but a
[02:12:21] positive for this third factor.
[02:12:25] Negative times negative times positive gives
me a positive
[02:12:28] for a test value between zero and six, let's
try x equals one.
[02:12:34] Now I'll get a positive for this factor a
negative for this factor, and a positive for
[02:12:41] this factor.
[02:12:43] positive times a negative times a positive
gives me a negative. Finally, for a test value
[02:12:50] bigger than six, we could use say x equals
100. That's going to give me positive, positive
[02:12:56] positive.
[02:12:59] So my product will be positive.
[02:13:03] Since I want values where my expression is
greater than or equal to zero, I want the
[02:13:08] places where it equals zero.
[02:13:12] And the places where it's positive.
[02:13:16] So my final answer will be close bracket negative
one to zero, close bracket union, close bracket
[02:13:24] six to infinity.
[02:13:25] As our final example, let's consider the rational
inequality, x squared plus 6x plus nine divided
[02:13:34] by x minus one is less than or equal to zero.
[02:13:38] Although it might be tempting to clear the
denominator and multiply both sides by x minus
[02:13:43] one, it's dangerous to do that, because x
minus one could be a positive number. But
[02:13:49] it could also be a negative number. And when
you multiply both sides by a negative number,
[02:13:53] you have to reverse the inequality. Although
it's possible to solve the inequality this
[02:13:59] way, by thinking of cases where x minus one
is less than zero or bigger than zero, I think
[02:14:04] it's much easier just to solve the same way
as we did before. So we'll start by rewriting
[02:14:10] so that we move all terms to the left and
have zero on the right. Well, that's already
[02:14:15] true. So the next step would be to solve the
associated equation.
[02:14:20] That is x squared plus 6x plus nine over x
minus one is equal to zero.
[02:14:29] That would be where the numerator is 0x squared
plus 6x plus nine is equal to zero. So we're
[02:14:35] x plus three squared is zero, or x equals
negative three. There's one extra step we
[02:14:42] have to do for rational expressions. And that's
we need to find where the expression does
[02:14:48] not exist. That is, let's find where the denominator
is zero. And that said, x equals one.
[02:14:55] I'll put all those numbers on the number line.
The places where am I
[02:14:59] rational expression is equal to zero, and
the place where my rational expression doesn't
[02:15:04] exist, then I can start in with test values.
For example, x equals minus four, zero and
[02:15:13] to work. If I plug those values into this
expression here, I get a negative answer a
[02:15:20] negative answer and a positive answer. The
reason I need to conclude the values on my
[02:15:26] number line where my denominators zero is
because I can my expression can switch from
[02:15:32] negative to positive by passing through a
place where my rational expression doesn't
[02:15:36] exist, as well as passing by passing flew
through a place where my rational expression
[02:15:41] is equal to zero.
[02:15:43] Now I'm looking for where my original expression
was less than or equal to zero. So that means
[02:15:49] I want the places on the number line where
my expression is equal to zero, and also the
[02:15:56] places where it's negative.
[02:15:59] So my final answer is x is less than one,
or an interval notation, negative infinity
[02:16:06] to one.
[02:16:08] In this video, we solved polynomial and rational
inequalities by making a number line and you
[02:16:14] think test values to make a sine chart.
[02:16:19] The distance formula can be used to find the
distance between two points, if we know their
[02:16:26] coordinates,
[02:16:27] if this first point has coordinates, x one,
y one, and the second point has coordinates
[02:16:34] x two, y two, then the distance between them
is given by the formula, the square root of
[02:16:42] x two minus x one squared plus y two minus
y one squared.
[02:16:49] This formula actually comes from the Pythagorean
Theorem, let me draw a right triangle with
[02:16:55] these two points as two of its vertices,
[02:17:01] then the length of this side is the difference
between the 2x coordinates. Similarly, the
[02:17:09] length of this vertical side is the difference
between the two y coordinates.
[02:17:14] Now that Pythagoras theorem says that for
any right triangle, with sides labeled A and
[02:17:20] B, and hypotony is labeled C, we have that
a squared plus b squared equals c squared.
[02:17:27] Well, if we apply that to this triangle here,
[02:17:31] this hypotony News is the distance between
the two points that we're looking for.
[02:17:38] So but Tyrion theorem says, the square of
this side length, that is x two minus x one
[02:17:45] squared, plus the square of this side length,
which is y two minus y one squared has to
[02:17:55] equal the square of the hypothesis, that is
d squared.
[02:18:01] taking the square root of both sides of this
equation, we get the square root of x two
[02:18:07] minus x one squared plus y two minus y one
squared is equal to d,
[02:18:13] we don't have to worry about using plus or
minus signs when we take the square root because
[02:18:18] distance is always positive.
[02:18:21] So we've derived our distance formula. Now
let's use it as an example.
[02:18:26] Let's find the distance between the two points,
negative one five, for negative one, five,
[02:18:34] maybe over here, and for two,
[02:18:39] this notation just means that P is the point
with coordinates negative one five, and q
[02:18:48] is the point with coordinates for two.
[02:18:52] We have the distance formula,
[02:18:54] let's think of P being the point with coordinates
x one y one, and Q being the point with coordinates
[02:19:02] x two y two.
[02:19:04] But as we'll see, it really doesn't matter
which one is which, plugging into our formula,
[02:19:10] we get d is the square root of n x two minus
x one. So that's four minus negative one squared,
[02:19:23] and then we add
[02:19:26] y two minus y one, so that's two minus five
squared.
[02:19:37] Working out the arithmetic a little bit for
minus negative one, that's four plus one or
[02:19:41] five squared, plus negative three squared.
So that's the square root of 25 plus nine
[02:19:49] are the square root of 34. Let's see what
would have happened if we'd call this first
[02:19:55] point, x two y two instead, and a second.
[02:20:00] Why'd x one y one, then we would have gotten
the same distance formula, but we would have
[02:20:06] taken negative one minus four
[02:20:12] and added the difference of y's squared. So
that's why two five minus two squared.
[02:20:22] That gives us the square root of negative
five squared plus three squared, which works
[02:20:27] out to the same thing.
[02:20:31] In this video, we use the distance formula
to find the distance between two points. When
[02:20:36] writing down the distance formula, sometimes
students forget whether this is a plus or
[02:20:40] a minus and whether these are pluses or minuses.
One way to remember is to think that the distance
[02:20:45] formula comes from the Pythagorean Theorem.
That's why there's a plus in here.
[02:20:51] And then a minus in here, because we're finding
the difference between the coordinates to
[02:20:56] find the lengths of the sides.
[02:20:59] The midpoint formula helps us find the coordinates
of the midpoint of a line segment, as long
[02:21:05] as we know the coordinates of the endpoints.
[02:21:09] So let's call the coordinates of this endpoint
x one, y one, and the coordinates of this
[02:21:15] other endpoint x two, y two,
[02:21:17] the x coordinate of the midpoint is going
to be exactly halfway in between the x coordinates
[02:21:25] of the endpoints. To get a number halfway
in between two other numbers, we just take
[02:21:31] the average.
[02:21:34] Similarly, the y coordinate of this midpoint
is going to be exactly halfway in between
[02:21:42] the y coordinates of the endpoints. So the
y coordinate of the midpoint is going to be
[02:21:47] the average of those y coordinates.
[02:21:53] So we see that the coordinates of the midpoint
are x one plus x two over two, y one plus
[02:22:00] y two over two.
[02:22:04] Let's use this midpoint formula in an example,
[02:22:07] we want to find the midpoint of the segment
between the points, negative one, five,
[02:22:17] and four
[02:22:20] to
[02:22:24] me draw the line segment between them. So
visually, the midpoint is going to be somewhere
[02:22:29] around here. But to find its exact coordinates,
we're going to use x one plus x two over two,
[02:22:36] and y one plus y two over two, where this
first point is coordinates, x one, y one,
[02:22:44] and the second point has coordinates x two
y two, it doesn't actually matter which point
[02:22:49] you decide is x one y one, which one is x
two, y two, the formula will still give you
[02:22:54] the same answer for the midpoint.
[02:22:56] So let's see, I take my average of my x coordinates.
So that's negative one, plus four over two,
[02:23:06] and the average of my Y coordinates, so that's
five plus two over two, and I get three halves,
[02:23:15] seven halves, as the coordinates of my midpoint.
[02:23:19] In this video, we use the midpoint formula
to find the midpoint of a line segment, just
[02:23:25] by taking the average of the x coordinates
and the average of the y coordinates. This
[02:23:30] video is about graphs and equations of circles.
Suppose we want to find the equation of a
[02:23:36] circle of radius five, centered at the point
three, two
[02:23:42] will look something like this.
[02:23:47] For any point, x, y on the circle, we know
that
[02:23:53] the distance of that point xy from the center
is equal to the radius that is five from the
[02:24:00] distance formula, that distance of five is
equal to the square root of the difference
[02:24:07] of the x coordinates. That's x minus three
[02:24:11] squared plus the difference of the y coordinates,
that's y minus two squared.
[02:24:20] And if I square both sides of that equation,
I get five squared is this square root squared.
[02:24:31] In other words, 25 is equal to x minus three
squared plus y minus two squared. Since the
[02:24:40] square root and the squared undo each other.
[02:24:43] A lot of times people will write the X minus
three squared plus y minus two squared on
[02:24:49] the left side of the equation and the 25 on
the right side of the equation. This is the
[02:24:53] standard form for the equation of the circle.
[02:24:58] The same reasoning can be generalized
[02:24:59] Find the general equation for a circle with
radius R centered at a point HK.
[02:25:07] For any point with coordinates, x, y on the
circle, the distance between the point with
[02:25:14] coordinates x, y and the point with coordinates
HK is equal to the radius r. So we have that
[02:25:22] the distance is equal to r, but by the distance
formula, that's the square root of the difference
[02:25:28] between the x coordinates x minus h squared
plus the difference in the y coordinates y
[02:25:36] minus k squared. squaring both sides as before,
we get r squared is the square root squared,
[02:25:48] canceling the square root and the squared,
and rearranging the equation, that gives us
[02:25:53] x minus h squared plus y minus k squared equals
r squared. That's the general formula for
[02:26:01] a circle with radius r, and center HK.
[02:26:06] Notice that the coordinates h and k are subtracted
here, the two squares are added because they
[02:26:13] are in the distance formula, and the radius
is squared on the other side, if you remember
[02:26:18] this general formula, that makes it easy to
write down the equation for a circle, for
[02:26:23] example, if we want the equation for a circle,
[02:26:29] of radius six, and center at zero, negative
three,
[02:26:35] then we have our equal six, and h k is our
zero negative three. So plugging into the
[02:26:45] formula,
[02:26:47] we get x minus zero squared, plus y minus
negative three squared equals six squared,
[02:26:57] or simplified this is x squared plus y plus
three squared equals 36.
[02:27:04] Suppose we're given an equation like this
one, and we want to decide if it's the equation
[02:27:09] of a circle, and if so what's the center was
the radius.
[02:27:13] Well, this equation matches the form for a
circle, x minus h squared plus y minus k squared
[02:27:21] equals r squared. If we let h, b five, Why
be negative sex since that way, subtracting
[02:27:30] a negative six is like adding a six,
[02:27:34] five must be our r squared. So that means
that R is the square root of five. So this
[02:27:41] is our radius. And our center is the point
five, negative six. And this is indeed the
[02:27:49] equation for a circle,
[02:27:52] which we could then graph by putting down
the center and estimating the radius, which
[02:27:57] is a little bit more than two.
[02:28:03] This equation might not look like the equation
of a circle, but it actually can be transformed.
[02:28:08] To look like the equation of a circle, we're
trying to make it look like something of the
[02:28:13] form x minus h squared plus y minus k squared
equals r squared.
[02:28:19] First, I'd like to get rid of the coefficients
in front of the x squared and the y squared,
[02:28:24] so I'm going to divide everything by nine.
This gives me x squared plus y squared plus
[02:28:30] 8x minus two y plus four equals zero. Next,
I'm going to group the x terms together the
[02:28:38] x squared and the 8x. So write x squared plus
8x. I'll group the y terms together the Y
[02:28:45] squared minus two y.
[02:28:49] And I'll subtract the four over to the other
side.
[02:28:54] This still doesn't look much like the equation
of a circle. But as the next step, I'm going
[02:28:59] to do something called completing the square,
I'm going to take this coefficient of x eight
[02:29:05] divided by two and then square it. In other
words, eight over two squared, that's four
[02:29:12] squared, which is 16. I'm going to add 16
to both sides of my equation.
[02:29:18] So the 16 here,
[02:29:22] and on the other side,
[02:29:24] now I'm going to do the same thing to the
coefficient of y, negative two divided by
[02:29:31] two is negative one, square that and I get
one. So now I'm going to add a one to both
[02:29:38] sides, but I'll put it near the y terms. So
add a one there. And then to keep things balanced,
[02:29:44] I have to add it to the other side. On the
right side, I have the number 13. On the left
[02:29:50] side, I can wrap up this expression x squared
plus 8x plus 16 into x plus four squared.
[02:29:59] To convince you, that's correct. If we multiply
out x plus four times x plus four, we get
[02:30:07] x squared plus 4x plus 4x plus 16, or x squared
plus 8x plus 16, the same thing we have right
[02:30:16] here. Similarly, we can wrap up y squared
minus two y plus one as
[02:30:25] y minus one squared. Again, I'll work out
the distribution down here, that's y squared
[02:30:32] minus y minus y plus one, or y squared minus
two y plus one, just like we have up here.
[02:30:40] If you're wondering how I knew to use four
and minus one, the four came from taking half
[02:30:46] of eight, and the minus one came from taking
half the negative two. Now we have an equation
[02:30:53] for a circle and standard form. And we can
read off the center, which is negative four,
[02:31:01] negative one, and the radius, which is the
square root of 13.
[02:31:10] It might seem like magic that this trick of
taking half of this coefficient and squaring
[02:31:14] it and adding it to both sides lets us wrap
up this expression into this expression a
[02:31:20] perfect square. But to see why that works,
let's take a look at what happens if we expand
[02:31:26] out x minus h squared, the thing that we're
trying to get to. So if we expand that out,
[02:31:33] x minus h squared, which is x minus h times
x minus h is, which multiplies out to x squared
[02:31:43] minus HX minus h x plus h squared, or x squared
minus two h x plus h squared. Now if we start
[02:31:53] out with this part, with some x squared term
and some coefficient of x, we're trying to
[02:32:01] decide what to add on in order to wrap it
up into x minus h squared. The thing to add
[02:32:06] on is h squared here, which comes from half
of this coefficient squared, right, because
[02:32:13] half of negative two H is a negative H, square
that you get the H squared. And then when
[02:32:18] we do wrap it up, it's half of this coefficient
of x, that appears right here.
[02:32:26] This trick of completing the square is really
handy for turning an equation for a circle
[02:32:30] in disguise, into the standard equation for
the circle. In this video, we found the standard
[02:32:37] equation for a circle x minus h squared plus
y minus k squared equals r squared, where
[02:32:44] r is the radius, and h k is the center.
[02:32:52] We also showed a method of completing the
square.
[02:32:56] When you have an equation for a circle in
disguise, completing the square will help
[02:33:01] you rewrite it into the standard form.
[02:33:06] This video is about graphs and equations of
lines.
[02:33:10] Here we're given the graph of a line, we want
to find the equation, one standard format
[02:33:16] for the equation of a line is y equals mx
plus b. here am represents the slope, and
[02:33:25] B represents the y intercept, the y value
where the line crosses the y axis, the slope
[02:33:34] is equal to the rise over the run. Or sometimes
this is written as the change in y values
[02:33:41] over the change in x values. Or in other words,
y two minus y one over x two minus x one,
[02:33:49] where x one y one and x two y two are points
on the line.
[02:33:56] While we could use any two points on the line,
to find the slope, it's convenient to use
[02:34:01] points where the x and y coordinates are integers,
that is points where the line passes through
[02:34:08] grid points. So here would be one convenient
point to use. And here's another convenient
[02:34:13] point to use.
[02:34:15] The coordinates of the first point are one,
two, and the next point, this is let's say,
[02:34:22] five, negative one. Now I can find the slope
by looking at the rise over the run. So as
[02:34:29] I go through a run of this distance, I go
through a rise of that distance especially
[02:34:35] gonna be a negative rise or a fall because
my line is pointing down. So let's see counting
[02:34:41] off squares. This is a run of 1234 squares
and a rise of 123. So negative three, so my
[02:34:48] slope is going to be
[02:34:52] negative three over four.
[02:34:56] I got that answer by counting squares, but
I could have also gotten it by
[02:35:00] Looking at the difference in my y values over
the difference in my x values, that is, I
[02:35:05] could have done
[02:35:08] negative one minus two, that's from my difference
in Y values, and divide that by my difference
[02:35:15] in x values,
[02:35:18] which is five minus one, that gives me negative
three over four, as before.
[02:35:26] So my M is negative three fourths.
[02:35:30] Now I need to figure out the value of b, my
y intercept, well, I could just read it off
[02:35:36] the graph, it looks like approximately 2.75.
But if I want to be more accurate, I can again
[02:35:41] use a point that has integer coordinates that
I know it's exact coordinates. So either this
[02:35:47] point or that point, let's try this point.
And I can start off with my equation y equals
[02:35:53] mx plus b, that is y equals negative three
fourths x plus b. And I can plug in the point
[02:36:02] one, two, for my x and y. So that gives me
two equals negative three fourths times one
[02:36:12] plus b, solving for B. Let's see that's two
equals negative three fourths plus b. So add
[02:36:20] three fourths to both sides, that's two plus
three fourths equals b. So b is eight fourths
[02:36:28] plus three fourths, which is 11. Four switches
actually, just what I eyeballed it today.
[02:36:33] So now I can write out my final equation for
my line y equals negative three fourths x
[02:36:40] plus 11 fourths by plugging in for m and b.
Next, let's find the equation for this horizontal
[02:36:48] line.
[02:36:49] a horizontal line has slope zero. So if we
think of it as y equals mx plus b, m is going
[02:36:57] to be zero. In other words, it's just y equals
b, y is some constant. So if we can figure
[02:37:03] out what that that constant y value is, it
looks like it's
[02:37:08] two, let's see this three, three and a half,
we can just write down the equation directly,
[02:37:14] y equals 3.5.
[02:37:18] For a vertical line, like this one, it doesn't
really have a slope. I mean, if you tried
[02:37:24] to do the rise over the run,
[02:37:26] there's no run. So you'd I guess you'd be
dividing by zero and get an infinite slope.
[02:37:32] But But instead, we just think of it as an
equation of the form x equals something. And
[02:37:39] in this case, x equals negative two, notice
that all of the points on our line have the
[02:37:45] same x coordinate of negative two and the
y coordinate can be anything.
[02:37:50] So this is how we write the equation for a
vertical line.
[02:37:54] In this example, we're not shown a graph of
the line, we're just get told that it goes
[02:37:59] through two points. But knowing that I go
through two points is enough to find the equation
[02:38:04] for the line. First, we can find the slope
by taking the difference in Y values over
[02:38:12] the difference in x values. So that's negative
three minus two over four minus one, which
[02:38:20] is negative five thirds.
[02:38:24] So we can use the
[02:38:26] standard equation for the line, this is called
the slope intercept form.
[02:38:34] And we can plug in negative five thirds. And
we can use one point, either one will do will
[02:38:42] still get the same final answer. So let's
use one two and plug that in to get two equals
[02:38:48] negative five thirds times one plus b. And
so B is two plus five thirds, which is six
[02:38:57] thirds plus five thirds, which is 11 thirds.
So our equation is y equals negative five
[02:39:03] thirds x plus 11 thirds.
[02:39:07] This is method one.
[02:39:11] method two uses a slightly different form
of the equation, it's called the point slope
[02:39:17] form and it goes y minus y naught is equal
to m times x minus x naught, where x naught
[02:39:25] y naught is a point on the line and again
is the slope. So we calculate the slope the
[02:39:33] same way by taking a difference in Y values
over a difference in x values. But then, we
[02:39:40] can simply plug in any point.
[02:39:43] For example, the point one two will work we
can plug one in for x naught and two in for
[02:39:53] Y not in this point slope form. That gives
us y
[02:39:59] minus
[02:40:00] Two is equal to minus five thirds x minus
one.
[02:40:06] Notice that these two equations, while they
may look different, are actually equivalent,
[02:40:11] because if I distribute the negative five
thirds, and then add the two to both sides,
[02:40:25] I get the same equation as above.
[02:40:28] So we've seen two ways of finding the equation
for the line using the slope intercept form.
[02:40:37] And using the point slope form.
[02:40:40] In this video, we saw that you can find the
equation for a line, if you know the slope.
[02:40:48] And you know one point,
[02:40:52] you can also find the equation for the line
if you know two points, because you can use
[02:40:57] the two points to get the slope and then plug
in one of those points. To figure out the
[02:41:03] rest of the equation.
[02:41:06] We saw two standard forms for the equation
of a line
[02:41:08] the slope intercept form y equals mx plus
b, where m is the slope, and B is the y intercept.
[02:41:19] And the point slope form y minus y naught
equals m times x minus x naught, where m again
[02:41:27] is the slope and x naught y naught is a point
on the line.
[02:41:35] This video is about finding parallel and perpendicular
lines. Suppose we have a line of slope three
[02:41:43] fourths, in other words, the rise
[02:41:47] over the run
[02:41:50] is three over four,
[02:41:52] than any other line that's parallel to this
line will have the same slope, the same fries
[02:42:01] over run.
[02:42:03] So that's our first fact to keep in mind,
parallel lines have the same slopes. Suppose
[02:42:09] on the other hand, we want to find a line
perpendicular to our original line with its
[02:42:15] original slope of three fourths.
[02:42:18] A perpendicular line, in other words, align
at a 90 degree angle to our original line
[02:42:25] will have a slope, that's the negative reciprocal
of the opposite reciprocal of our original
[02:42:32] slope. So we take the reciprocal of three
for us, that's four thirds, and we make it
[02:42:37] its opposite by changing it from positive
to negative.
[02:42:41] So I'll write this as a principle that perpendicular
lines have opposite reciprocal slopes, to
[02:42:49] get the hang of what it means to be an opposite
reciprocal, let's look at a few examples.
[02:42:54] So here's the original slope, and this will
be the opposite reciprocal, which would represent
[02:43:01] the slope of our perpendicular line. So for
example, if one was to the opposite reciprocal,
[02:43:07] reciprocal of two is one half opposite means
I change the sign, if the first slope turned
[02:43:14] out to be, say, negative 1/3, the reciprocal
of 1/3 is three over one or just three. And
[02:43:23] the opposite means I change it from a negative
to a positive, so my perpendicular slope would
[02:43:29] be would be three.
[02:43:32] One more example, if I started off with a
slope of, say, seven halves, then the reciprocal
[02:43:38] of that would be two sevenths. And I change
it to an opposite reciprocal, by changing
[02:43:43] the positive to a negative.
[02:43:47] Let's use these two principles in some examples.
[02:43:52] In our first example, we need to find the
equation of a line that's parallel to this
[02:43:56] slump, and go through the point negative three
two. Well, in order to get started, I need
[02:44:02] to figure out the slope of this line. So let
me put this line into a more standard form
[02:44:07] the the slope intercept form. So I'll start
with three y minus 4x plus six equals zero,
[02:44:14] I'm going to solve for y to put it in this
this more standard form. So I'll say three
[02:44:19] y equals 4x minus six and then divide by three
to get y equals four thirds x minus six thirds
[02:44:29] is divided all my turns by three, I can simplify
that a little bit y equals four thirds x minus
[02:44:34] two. Now I can read off the slope of my original
line, and one is four thirds. That means my
[02:44:43] slope of my new line, my parallel line will
also be four thirds. So my new line will have
[02:44:51] the equation y equals four thirds ax plus
b for some B, of course, B will probably not
[02:44:58] be negative two like it was for the
[02:45:00] first line, it'll be something else determined
by the fact that it goes through this point
[02:45:04] negative three to, to figure out what b is,
I plug in the point negative three to four,
[02:45:10] negative three for x, and two for y. So that
gives me two equals four thirds times negative
[02:45:16] three plus b, and I'll solve for b. So let's
see. First, let me just simplify. So two equals,
[02:45:25] this is negative 12 thirds plus b, or in other
words, two is negative four plus b. So that
[02:45:32] means that B is going to be six, and by my
parallel line will have the equation y equals
[02:45:38] four thirds x plus six.
[02:45:42] Next, let's find the equation of a line that's
perpendicular to a given line through and
[02:45:49] go through a given point. Again, in order
to get started, I need to find what the slope
[02:45:54] of this given line is. So I'll rewrite it,
well, first, I'll just copy it over 6x plus
[02:45:59] three, y equals four. And then I can put it
in the standard slope intercept form by solving
[02:46:06] for y. So three, y is negative 6x plus four,
divide everything by three, I get y equals
[02:46:14] negative six thirds x plus four thirds. So
y is negative 2x, plus four thirds. So my
[02:46:22] original slope of my original line is negative
two, which means my perpendicular slope is
[02:46:28] the opposite reciprocal, so I take the reciprocal
of of negative two, that's negative one half
[02:46:34] and I change the sign so that gives me one
half as the slope of my perpendicular line.
[02:46:39] Now, my new line, I know is going to be y
equals one half x plus b for some B, and I
[02:46:47] can plug in my point on my new line, so one
for y and four for x, and solve for b, so
[02:46:57] I get one equals two plus b, So b is negative
one. And so my new line has equation one half
[02:47:04] x minus one.
[02:47:07] These next two examples are a little bit different,
because now we're looking at a line parallel
[02:47:14] to a completely horizontal line, let me draw
this line y equals three, so the y coordinate
[02:47:23] is always three, which means that my line
will be a horizontal line through that height
[02:47:29] at height y equals three, if I want something
parallel to this line, it will also be a horizontal
[02:47:36] line. Since it goes through the point negative
two, one, C, negative two, one has to go through
[02:47:42] that point there,
[02:47:45] it's gonna have always have a y coordinate
of one the same as the y coordinate of the
[02:47:51] point it goes through, so my answer will just
be y equals one.
[02:47:55] In the next example, we want a line that's
perpendicular to the line y equals four another
[02:48:06] horizontal line y equals four, but perpendicular
means I'm going to need a vertical line. So
[02:48:12] I need a vertical line that goes through the
point three, four.
[02:48:16] Okay, and so I'm going to draw a vertical
line there. Now vertical lines have the form
[02:48:27] x equals something for the equation. And to
find what x equals, I just need to look at
[02:48:32] the x coordinate of the point I'm going through
this point three four, so that x coordinate
[02:48:38] is three and all the points on this, this
perpendicular and vertical line have X coordinate
[02:48:45] three so my answer is x equals three.
[02:48:53] In this video will use the fact that parallel
lines have the same slope and perpendicular
[02:48:58] lines have opposite reciprocal slopes, to
find the equations of parallel and perpendicular
[02:49:03] lines.
[02:49:05] This video introduces functions and their
domains and ranges.
[02:49:09] A function is a correspondence between input
numbers usually the x values and output numbers,
[02:49:17] usually the y values, that sends each input
number to exactly one output number. Sometimes
[02:49:25] a function is thought of as a rule or machine
[02:49:28] in which you can feed in x values as input
and get out y values as output. So, non mathematical
[02:49:38] example of a function might be the biological
mother function, which takes as input any
[02:49:44] person and it gives us output their biological
mother.
[02:49:49] This function satisfies the condition that
each input number object person in this case,
[02:49:57] get sent to exactly one output per
[02:50:01] Because if you take any person, they just
have one biological mother. So that rule does
[02:50:07] give you a function. But if I change things
around and just use the the mother function,
[02:50:14] which sends to each person, their mother,
that's no longer going to be a function because
[02:50:19] there are some people who have more than one
mother, right, they could have a biological
[02:50:23] mother and adopted mother, or a mother and
a stepmother, or any number of situations.
[02:50:28] So since there's, there's at least some people
who you would put it as input, and then you'd
[02:50:33] get like more than one possible output that
violates this, this rule of functions, that
[02:50:39] would not be a function. Now, most of the
time, we'll work with functions that are described
[02:50:43] with equations, not in terms of mothers. So
for example, we can have the function y equals
[02:50:48] x squared plus one.
[02:50:51] This can also be written as f of x equals
x squared plus one. Here, f of x is function
[02:50:57] notation that stands for the output value
of y.
[02:51:02] Notice that this notation is not representing
multiplication, we're not multiplying f by
[02:51:07] x, instead, we're going to be putting in a
value for x as input and getting out a value
[02:51:14] of f of x or y. For example, if we want to
evaluate f of two, we're plugging in two as
[02:51:21] input for x either in this equation, or in
that equation. Since F of two means two squared
[02:51:30] plus one, f of two is going to equal five.
Similarly, f of five means I plug in five
[02:51:38] for x, so that's gonna be five squared plus
one, or 26. Sometimes it's useful to evaluate
[02:51:46] a function on a more complicated expression
involving other variables. In this case, remember,
[02:51:52] the functions value on any expression, it's
just what you what you get when you plug in
[02:51:57] that whole expression for x.
[02:52:00] So f of a plus three is going to be the quantity
a plus three squared plus one,
[02:52:09] we could rewrite that as a squared plus six
a plus nine plus one, or a squared plus six
[02:52:17] a plus 10.
[02:52:19] When evaluating a function on a complex expression,
it's important to keep the parentheses when
[02:52:25] you plug in for x. That way, you evaluate
the function on the whole expression. For
[02:52:32] example, it would be wrong
[02:52:35] to write f of a plus three equals a plus three
squared plus one without the parentheses,
[02:52:42] because that would imply that we were just
squaring the three and not the whole expression,
[02:52:47] a plus three.
[02:52:52] Sometimes a function is described with a graph
instead of an equation. In this example, this
[02:52:58] graph is supposed to represent the function
g of x. Not all graphs actually represent
[02:53:04] functions. For example, the graph of a circle
doesn't represent a function. That's because
[02:53:11] the graph of a circle violates the vertical
line test, you can draw a vertical line and
[02:53:15] intersect the graph in more than one point.
[02:53:18] But our graph, it left satisfies the vertical
line test, any vertical line intersects the
[02:53:24] graph, and at most one point, that means is
a function, because every x value will have
[02:53:29] at most one y value that corresponds to it.
Let's evaluate g have to note that two is
[02:53:37] an x value. And we'll use the graph to find
the corresponding y value. So I look for two
[02:53:44] on the x axis, and find the point on the graph
that has that x value that's right here. Now
[02:53:51] I can look at the y value of that point looks
like three, and therefore gf two is equal
[02:53:58] to three. If I try to do the same thing to
evaluate g of five, I run into trouble. Five
[02:54:05] is an x value, I look for it on the x axis,
but there's no point on the graph that has
[02:54:11] that x value.
[02:54:13] Therefore, g of five is undefined, or we can
say it does not exist. The question of what
[02:54:21] x values and y values make sense for a function
leads us to a discussion of domain and range.
[02:54:27] The domain of a function is all possible x
values that makes sense for that function.
[02:54:32] The range is the y values that make sense
for the function.
[02:54:36] In this example, we saw that the x value of
five didn't have a corresponding y value for
[02:54:42] this function. So the x value of five is not
in the domain of our function J. To find the
[02:54:49] x values in the domain, we have to look at
the x values that correspond to points on
[02:54:54] the graph. One way to do that is to take the
shadow or projection of the graph
[02:55:00] onto the x axis and see what x values are
hit. It looks like we're hitting all x values
[02:55:07] starting at negative eight, and continuing
up to four. So our domain
[02:55:15] is the x's between negative eight, and four,
including those endpoints. Or we can write
[02:55:21] this in interval notation as negative eight
comma four with square brackets.
[02:55:27] To find the range of the function, we look
for the y values corresponding to points on
[02:55:31] this graph, we can do that.
[02:55:35] By taking the shadow or projection of the
graph onto the y axis,
[02:55:42] we seem to be hitting our Y values from negative
five, up through three.
[02:55:51] So our range is wise between negative five
and three are an interval notation negative
[02:55:59] five, three with square brackets. If we meet
a function that's described as an equation
[02:56:05] instead of a graph, one way to find the domain
and range are to graph the function. But it's
[02:56:11] often possible to find the domain at least
more quickly, by using algebraic considerations.
[02:56:17] We think about what x values that makes sense
to plug into this expression, and what x values
[02:56:22] need to be excluded, because they make the
algebraic expression impossible to evaluate.
[02:56:27] Specifically, to find the domain of a function,
we need to exclude x values that make the
[02:56:35] denominator zero. Since we can't divide by
zero,
[02:56:38] we also need to exclude x values that make
an expression inside a square root sign negative.
[02:56:47] Since we can't take the square root of a negative
number. In fact, we need to exclude values
[02:56:52] that make the expression inside any even root
negative because we can't take an even root
[02:56:58] of a negative number, even though we can take
an odd root like a cube root of a negative
[02:57:02] number. Later, when we look at logarithmic
functions, we'll have some additional exclusions
[02:57:07] that we have to make. But for now, these two
principles should handle all functions We'll
[02:57:12] see. So let's apply them to a couple examples.
[02:57:16] For the function in part A, we don't have
any square root signs, but we do have a denominator.
[02:57:23] So we need to exclude x values that make the
denominator zero. In other words, we need
[02:57:28] x squared minus 4x plus three to not be equal
to zero.
[02:57:34] If we solve x squared minus 4x plus three
equal to zero, we can do that by factoring.
[02:57:42] And that gives us x equals three or x equals
one, so we need to exclude these values.
[02:57:48] All other x values should be fine. So if I
draw the number line, I can put on one and
[02:57:54] three and just dig out a hole at both of those.
And my domain includes everything else on
[02:58:00] the number line. In interval notation, this
means my domain is everything from negative
[02:58:06] infinity to one, together with everything
from one to three, together with everything
[02:58:12] from three to infinity.
[02:58:15] In the second example, we don't have any denominator
to worry about, but we do have a square root
[02:58:21] sign. So we need to exclude any x values that
make three minus 2x less than zero. In other
[02:58:29] words, we can include all x values for which
three minus 2x is greater than or equal to
[02:58:35] zero. Solving that inequality gives us three
is greater than equal to 2x. In other words,
[02:58:43] x
[02:58:44] is less than or equal to three halves,
[02:58:48] I can draw this on the number line,
[02:58:51] or write it in interval notation.
[02:58:54] Notice that three halves is included, and
that's because three minus 2x is allowed to
[02:59:00] be zero, I can take the square root of zero,
that's just zero. And that's not a problem.
[02:59:05] Finally, let's look at a more complicated
function that involves both the square root
[02:59:10] and denominator. Now there are two things
I need to worry about.
[02:59:14] I need the denominator
[02:59:18] to not be equal to zero, and I need the stuff
inside the square root side to be greater
[02:59:24] than or equal to zero. from our earlier work,
we know the first condition means that x is
[02:59:30] not equal to three, and x is not equal to
one. And the second condition means that x
[02:59:36] is less than or equal to three halves. Let's
try both of those conditions on the number
[02:59:41] line.
[02:59:44] x is not equal to three and x is not equal
to one means we've got everything except those
[02:59:51] two dug out points. And the other condition
x is less than or equal to three halves means
[02:59:56] we can have three halves and everything
[03:00:00] To the left of it. Now to be in our domain
and to be legit for our function, we need
[03:00:05] both of these conditions to be true. So I'm
going to connect these conditions with N and
[03:00:09] N, that means we're looking for a numbers
on the number line that are colored both red
[03:00:13] and blue. So I'll draw that above in purple.
So that's everything from three halves to
[03:00:19] one, I have to dig out one because one was
a problem for the denominator. And then I
[03:00:24] can continue for all the things that are colored
both both colors red and blue. So my final
[03:00:30] domain is going to be, let's see negative
infinity, up to but not including one together
[03:00:37] with one, but not including it to three halves,
and I include three half since that was colored,
[03:00:44] both red and blue. Also,
[03:00:46] in this video, we talked about functions,
how to evaluate them, and how to find domains
[03:00:51] and ranges. This video gives the graphs of
some commonly used functions that I call the
[03:00:57] toolkit functions. The first function is the
function y equals x, let's plot a few points
[03:01:03] on the graph of this function. If x is zero,
y zero, if x is one, y is one, and so on y
[03:01:13] is always equal to x doesn't have to just
be an integer, it can be any real number,
[03:01:19] and we'll connecting the dots, we get a straight
line through the origin.
[03:01:24] Let's look at the graph of y equals x squared.
If x is zero, y is zero. So we go through
[03:01:30] the origin again, if x is one, y is one, and
x is negative one, y is also one,
[03:01:40] the x value of two gives a y value of four
and the x value of negative two gives a y
[03:01:45] value of four also connecting the dots, we
get a parabola.
[03:01:51] That is this, this function is an even function.
[03:01:56] That means it has mirror symmetry across the
y axis, the left side it looks like exactly
[03:02:02] like the mirror image of the right side. That
happens because when you square a positive
[03:02:08] number, like two, you get the exact same y
value as when you square it's a mirror image
[03:02:14] x value of negative two.
[03:02:17] The next function y equals x cubed. I'll call
that a cubic. Let's plot a few points when
[03:02:24] x is zero, y is zero. When x is one, y is
one, when x is negative one, y is negative
[03:02:31] one, two goes with the point eight way up
here and an x value of negative two is going
[03:02:37] to give us negative eight. If I connect the
dots, I get something that looks kind of like
[03:02:42] this.
[03:02:45] This function is what's called an odd function,
because it has 180 degree rotational symmetry
[03:02:53] occur around the origin. If I rotate this
whole graph by 180 degrees, or in other words,
[03:02:58] turn the paper upside down, I'll get exactly
the same shape. The reason this function has
[03:03:04] this odd symmetry is because when I cube a
positive number, and get its y value, I get
[03:03:12] n cube the corresponding negative number to
get its y value, the negative number cubed
[03:03:17] gives us exactly the negative of the the y
value we get with cubing the positive number.
[03:03:25] Let's look at the next example. Y equals the
square root of x.
[03:03:29] Notice that the domain of this function is
just x values bigger than or equal to zero
[03:03:37] because we can't take the square root of a
negative number. Let's plug in a few x values.
[03:03:42] X is zero gives y is 0x is one square root
of one is one, square root of four is two,
[03:03:49] and connecting the dots, I get a function
that looks like this.
[03:03:55] The absolute value function is next. Again,
if we plug in a few points, x is zero goes
[03:04:01] with y equals 0x is one gives us one, the
absolute value of negative one is 122 is on
[03:04:11] the graph, and the absolute value of negative
two is to I'm ending up getting a V shaped
[03:04:17] graph. It also has even or a mirror symmetry.
[03:04:24] Y equals two to the x is what's known as an
exponential function. That's because the variable
[03:04:30] x is in the exponent. If I plot a few points,
[03:04:35] two to the zero is one,
[03:04:41] two to the one is two, two squared is four,
two to the minus one is one half. I'll plot
[03:04:49] these on my graph, you know, let me fill in
a few more points. So let's see. two cubed
[03:04:55] is eight. That's way up here and negative
two gives
[03:04:59] Maybe 1/4 1/8 connecting the dots, I get something
that's shaped like this.
[03:05:07] You might have heard the expression exponential
growth, when talking about, for example, population
[03:05:11] growth, this is function is represents exponential
growth, it grows very, very steeply. Every
[03:05:19] time we increase the x coordinate by one,
we double the y coordinate.
[03:05:26] We could also look at a function like y equals
three dx, or sometimes y equals e to the x,
[03:05:32] where E is just a number about 2.7. These
functions look very similar. It's just the
[03:05:40] bigger bass makes us rise a little more steeply.
[03:05:45] Now let's look at the function y equals one
over x. It's not defined when x is zero, but
[03:05:52] I can plug in x equals one half, one over
one half is to
[03:05:58] whenever one is one, and one or two is one
half. By connect the dots, I get this shape
[03:06:06] in the first quadrant, but I haven't looked
at negative values of x, when x is negative,
[03:06:11] whenever negative one is negative one, whenever
negative half is negative two, and I get a
[03:06:18] similar looking piece in the third quadrant.
[03:06:22] This is an example of a hyperbola.
[03:06:27] And it's also an odd function, because it
has that 180 degree rotational symmetry, if
[03:06:34] I turn the page upside down, it'll look exactly
the same. Finally, let's look at y equals
[03:06:40] one over x squared.
[03:06:42] Again, it's not defined when x is zero, but
I can plug in points like one half or X, let's
[03:06:48] see one over one half squared is one over
1/4, which is four.
[03:06:55] And one over one squared, one over two squared
is a fourth, it looks pretty much like the
[03:07:03] previous function is just a little bit more
extreme rises a little more steeply, falls
[03:07:08] a little more dramatically. But for negative
values of x, something a little bit different
[03:07:12] goes on. For example, one over negative two
squared is just one over four, which is a
[03:07:19] fourth. So I can plot that point there, and
one over a negative one squared, that's just
[03:07:27] one. So my curve for negative values of x
is gonna lie in the second quadrant, not the
[03:07:32] third quadrant. This is an example of an even
function
[03:07:38] because it has perfect mirror symmetry across
the y axis.
[03:07:42] These are the toolkit functions, and I recommend
you memorize the shape of each of them.
[03:07:49] That way, you can draw at least a rough sketch
very quickly without having to plug in points.
[03:07:56] That's all for the graphs of the toolkit functions.
[03:07:58] If we change the equation of a function, then
the graph of the function changes or transforms
[03:08:07] in predictable ways.
[03:08:09] This video gives some rules and examples for
transformations of functions. To get the most
[03:08:14] out of this video, it's helpful if you're
already familiar with the graph of some typical
[03:08:19] functions, I call them toolkit functions like
y equals square root of x, y equals x squared,
[03:08:25] y equals the absolute value of x and so on.
If you're not familiar with those graphs,
[03:08:30] I encourage you to watch my video called toolkit
functions first, before watching this one.
[03:08:36] I want to start by reviewing function notation.
[03:08:39] If g of x represents the function, the square
root of x, then we can rewrite these expressions
[03:08:46] in terms of square roots. For example, g of
x minus two is the same thing as the square
[03:08:53] root of x minus two.
[03:08:55] g of quantity x minus two means we plug in
x minus two, everywhere we see an x, so that
[03:09:05] would be the same thing as the square root
of quantity x minus two.
[03:09:11] In this second example, I say that we're subtracting
two on the inside of the function, because
[03:09:16] we're subtracting two before we apply the
square root function. Whereas on the first
[03:09:21] example, I say that the minus two is on the
outside of the function, we're doing the square
[03:09:26] root function first and then subtracting two.
In this third example, g of 3x, we're multiplying
[03:09:33] by three on the inside of the function. To
evaluate this in terms of square root, we
[03:09:38] plug in the entire 3x for x and the square
root function. That gives us the square root
[03:09:44] of 3x.
[03:09:45] In the next example, we're multiplying by
three on the outside of the function. This
[03:09:51] is just three times the square root of x.
Finally, g of minus x means the square root
[03:09:58] of minus x.
[03:10:00] Now, this might look a little odd because
we're not used to taking the square root of
[03:10:03] a negative number. But remember that if x
itself is negative, like negative to the negative
[03:10:09] x will be negative negative two or positive
two. So we're really be taking the square
[03:10:13] root of a positive number in that case, let
me record which of these are inside and which
[03:10:18] of these are outside of my function.
[03:10:21] In this next set of examples, we're using
the same function g of x squared of x. But
[03:10:26] this time, we're starting with an expression
with square roots in it, and trying to write
[03:10:29] it in terms of g of x. So the first example,
the plus 17 is on the outside of my function,
[03:10:37] because I'm taking the square root of x first,
and then adding 17. So I can write this as
[03:10:43] g of x plus 17.
[03:10:46] In the second example, I'm taking x and adding
12 first, then I take the square root of the
[03:10:53] whole thing. Since I'm adding the 12 to x
first that's on the inside of my function.
[03:10:58] So I write that as g of the quantity x plus
12.
[03:11:04] Remember that this notation means I plug in
the entire x plus 12, into the square root
[03:11:09] sign, which gives me exactly square root of
x plus 12. And this next example, undoing
[03:11:14] the square root first and then multiplying
by negative 36. So i minus 36, multiplications,
[03:11:20] outside my function, I can rewrite this as
minus 36 times g of x. Finally, in this last
[03:11:27] example, I take x multiplied by a fourth and
then apply the square root of x. So that's
[03:11:33] the same thing as g of 1/4 X, my 1/4 X is
on the inside of my function. In other words,
[03:11:40] it's inside the parentheses when I use function
notation.
[03:11:44] Let's graph the square root of x and two transformations
have this function,
[03:11:50] y equals the square root of x goes to the
point 0011. And for two, since the square
[03:11:56] root of four is two, it looks something like
this.
[03:12:01] In order to graph y equals the square root
of x minus two, notice that the minus two
[03:12:06] is on the outside of the function, that means
we're going to take the square root of x first
[03:12:10] and then subtract two. So for example, if
we start with the x value of zero, and compute
[03:12:17] the square root of zero, that's zero, then
we subtract two to give us a y value of negative
[03:12:21] two,
[03:12:23] an x value of one, which under the square
root function had a y value of one now has
[03:12:29] a y value that's decreased by two, one minus
two is negative one. And finally, an x value
[03:12:36] of four, which under the square root function
had a y value of two now has a y value of
[03:12:42] two minus two or zero, its y value is also
decreased by two. If I plot these new points,
[03:12:49] zero goes with negative two, one goes with
negative one, and four goes with zero, I have
[03:12:56] my transformed graph.
[03:13:00] Because I subtracted two on the outside of
my function, my y values were decreased by
[03:13:04] two, which brought my graph down by two units.
Next, let's look at y equals the square root
[03:13:12] of quantity x minus two. Now we're subtracting
two on the inside of our function, which means
[03:13:17] we subtract two from x first and then take
the square root. In order to get the same
[03:13:21] y value of zero as we had in our blue graph,
we need our x minus two to be zero, so we
[03:13:28] need our original x to be two.
[03:13:30] In order to get the y value of one that we
had in our blue graph, we need to be taking
[03:13:36] the square root of one, so we need x minus
two to be one, which means that we need to
[03:13:41] start with an x value of three.
[03:13:43] And in order to reproduce our y value of two
from our original graph, we need the square
[03:13:50] root of x minus two to be two, which means
we need to start out by taking the square
[03:13:55] root of four, which means our x minus two
is four, so our x should be up there at six.
[03:14:04] If I plot my x values, with my corresponding
y values of square root of x minus two, I
[03:14:10] get the following graph. Notice that the graph
has moved horizontally to the right by two
[03:14:18] units.
[03:14:19] To me, moving down by two units, makes sense
because we're subtracting two, so we're decreasing
[03:14:25] y's by two units, but the minus two on the
inside that kind of does the opposite of what
[03:14:30] I expect, I might expected to to move the
graph left, I might expect the x values to
[03:14:35] be going down by two units, but instead, it
moves the graph to the right,
[03:14:41] because the x units have to go up by two units
in order to get the right square root when
[03:14:46] I then subtract two units again, the observations
we made for these transformations of functions
[03:14:52] hold in general, according to the following
rules. First of all, numbers on the outside
[03:14:58] of the function like
[03:14:59] In our example, y equals a squared of x minus
two, those numbers affect the y values. And
[03:15:06] a result in vertical motions, like we saw,
these motions are in the direction you expect.
[03:15:12] So subtracting two was just down by two. If
we were adding two instead, that would move
[03:15:18] us up by two
[03:15:19] numbers on the inside of the function. That's
like our example, y equals the square root
[03:15:24] of quantity x minus two, those affect the
x values and results in a horizontal motion,
[03:15:31] these motions go in the opposite direction
from what you expect. Remember, the minus
[03:15:35] two on the inside actually shifted our graph
to the right, if it had had a plus two on
[03:15:40] the inside, that would actually shift our
graph to the left.
[03:15:45] Adding results in a shift those are called
translations, but multiplying like something
[03:15:51] like y equals three times the square root
of x, that would result in a stretch or shrink.
[03:15:59] In other words, if I start with the square
root of x,
[03:16:03] and then when I graph y equals three times
the square root of x, that stretches my graph
[03:16:08] vertically by a factor of three.
[03:16:12] Like this,
[03:16:13] if I want to graph y equals 1/3, times the
square root of x, that shrinks my graph vertically
[03:16:20] by a factor of 1/3.
[03:16:22] Finally, a negative sign results in reflection.
For example, if I start with the graph of
[03:16:29] y equals the square root of x, and then when
I graph y equals the square root of negative
[03:16:34] x, that's going to do a reflection in the
horizontal direction because the negative
[03:16:39] is on the inside of the square root sign.
[03:16:42] A reflection in the horizontal direction means
a reflection across the y axis.
[03:16:49] If instead, I want to graph y equals negative
A squared of x, that negative sign on the
[03:16:53] outside means a vertical reflection, a reflection
across the x axis.
[03:17:00] Pause the video for a moment and see if you
could describe what happens in these four
[03:17:06] transformations.
[03:17:07] In the first example, we're subtracting four
on the outside of the function. adding or
[03:17:12] subtracting means a translation or shift.
And since we're on the outside of the function
[03:17:18] affects the y value, so that's moving us vertically.
So this transformation should take the square
[03:17:24] root of graph and move it down by four units,
that would look something like this.
[03:17:31] In the next example, we're adding 12 on the
inside, that's still a translation. But now
[03:17:39] we're moving horizontally. And so since we
go the opposite direction, we expect we are
[03:17:44] going to go to the left by 12 units, that's
going to look something like this.
[03:17:51] And the next example, we're multiplying by
three and introducing a negative sign both
[03:17:55] on the outside of our function outside our
function means we're affecting the y values.
[03:18:00] So in multiplication means we're stretching
by a factor of three, the negative sign means
[03:18:06] we reflect in the vertical direction,
[03:18:10] here's stretching by a factor of three vertically,
before I apply the minus sign, and now the
[03:18:15] minus sign reflects in the vertical direction.
[03:18:19] Finally, in this last example, we're multiplying
by one quarter on the inside of our function,
[03:18:26] we know that multiplication means stretch
or shrink. And since we're on the inside,
[03:18:31] it's a horizontal motion, and it does the
opposite of what we expect. So instead of
[03:18:36] shrinking by a factor of 1/4, horizontally,
it's actually going to stretch by the reciprocal,
[03:18:41] a factor of four horizontally.
[03:18:44] that'll look something like this. Notice that
stretching horizontally by a factor of four
[03:18:50] looks kind of like shrinking vertically by
a factor of one half.
[03:18:55] And that's actually borne out by the algebra,
because the square root of 1/4 x is the same
[03:19:01] thing as the square root of 1/4 times the
square root of x, which is the same thing
[03:19:05] as one half times the square root of x. And
so now we can see algebraically that vertical
[03:19:12] shrink by a factor of one half is the same
as a horizontal stretch by a factor of four,
[03:19:18] at least for this function, the square root
function.
[03:19:22] This video gives some rules for transformations
of functions, which I'll repeat below. numbers
[03:19:27] on the outside correspond to changes in the
y values, or vertical motions.
[03:19:37] numbers on the inside of the function, affect
the x values, and result in horizontal motions.
[03:19:44] Adding and subtracting
[03:19:47] corresponds to translations or shifts.
[03:19:52] multiplying and dividing by numbers corresponds
to stretches and shrinks
[03:19:57] and putting in a negative sign.
[03:20:01] corresponds to a reflection,
[03:20:04] horizontal reflection, if the negative sign
is on the inside, and a vertical reflection,
[03:20:10] if the negative sign is on the outside,
[03:20:13] knowing these basic rules about transformations
empowers you to be able to sketch graphs of
[03:20:17] much more complicated functions, like y equals
three times the square root of x plus two,
[03:20:24] by simply considering the transformations,
one at a time.
[03:20:29] a quadratic function is a function that can
be written in the form f of x equals A x squared
[03:20:35] plus bx plus c, where a, b and c are real
numbers. And a is not equal to zero. The reason
[03:20:46] we require that A is not equal to zero is
because if a were equal to zero, we'd have
[03:20:53] f of x is equal to b x plus c, which is called
a linear function. So by making sure that
[03:21:01] a is not zero, we make sure there's really
an x squared term which is the hallmark of
[03:21:06] a quadratic function.
[03:21:08] Please pause the video for a moment and decide
which of these equations represent quadratic
[03:21:13] functions.
[03:21:15] The first function can definitely be written
in the form g of x equals A x squared plus
[03:21:23] bx plus c. fact it's already written in that
form, where a is negative five, B is 10, and
[03:21:30] C is three.
[03:21:32] The second equation is also a quadratic function,
because we can think of it as f of x equals
[03:21:39] one times x squared plus zero times x plus
zero.
[03:21:46] So it is in the right form, where A is one,
B is zero, and C is zero.
[03:21:53] It's perfectly fine for the coefficient of
x and the constant term to be zero for a quadratic
[03:21:59] function, we just need the coefficient of
x squared to be nonzero, so the x squared
[03:22:03] term is preserved.
[03:22:06] The third equation is not a quadratic function.
It's a linear function, because there's no
[03:22:13] x squared term.
[03:22:17] The fourth function might not look like a
quadratic function. But if we rewrite it by
[03:22:21] expanding out the X minus three squared, let's
see what happens. We get y equals two times
[03:22:28] x minus three times x minus three plus four.
So that's two times x squared minus 3x minus
[03:22:36] 3x, plus nine plus four, continuing, I get
2x squared,
[03:22:45] minus 12 access, plus 18 plus four, in other
words, y equals 2x squared minus 12x plus
[03:22:53] 22. So in fact, our function can be written
in the right form, and it is a quadratic function.
[03:23:02] A function that is already written in the
form y equals A x squared plus bx plus c is
[03:23:08] said to be in standard form.
[03:23:12] So our first example g of x is in standard
form
[03:23:17] a function that's written in the format of
the last function that is in the form of y
[03:23:23] equals a times x minus h squared plus k for
some numbers, a, h, and K. That said to be
[03:23:33] in vertex form,
[03:23:35] I'll talk more about standard form and vertex
form in another video.
[03:23:40] In this video, we identified some quadratic
functions, and talked about the difference
[03:23:45] between standard form
[03:23:49] and vertex form.
[03:23:54] This video is about graphing quadratic functions.
a quadratic function, which is typically written
[03:23:59] in standard form, like this, or sometimes
in vertex form, like this, always has the
[03:24:08] graph that looks like a parabola.
[03:24:11] This video will show how to tell whether the
problem is pointing up or down. How to find
[03:24:17] its x intercepts, and how to find its vertex.
[03:24:21] The bare bones basic quadratic function is
f of x equals x squared, it goes to the origin,
[03:24:30] since f of zero is zero squared, which is
zero, and it is a parabola pointing upwards
[03:24:38] like this.
[03:24:40] The vertex of a parabola is its lowest point
if it's pointing upwards, and its highest
[03:24:46] point if it's pointing downwards. So in this
case, the vertex is at 00.
[03:24:53] The x intercepts are where the graph crosses
the x axis. In other words, where y is zero
[03:24:58] In this function, y equals zero means that
x squared is zero, which happens only when
[03:25:05] x is zero. So the x intercept, there's only
one of them is also zero.
[03:25:11] The second function, y equals negative 3x
squared also goes to the origin. Since the
[03:25:17] functions value when x is zero, is y equals
zero. But in this case, the parabola is pointing
[03:25:25] downwards.
[03:25:27] That's because thinking about transformations
of functions, a negative sign on the outside
[03:25:32] reflects the function vertically over the
x axis, making the problem instead of pointing
[03:25:38] upwards, reflecting the point downwards.
[03:25:43] The number three on the outside stretches
the graph vertically by a factor of three.
[03:25:49] So it makes it kind of long and skinny like
this.
[03:25:55] In general, a negative coefficient to the
x squared term means the problem will be pointing
[03:26:00] down. Whereas a positive coefficient, like
here, the coefficients, one means the parabola
[03:26:05] is pointing up.
[03:26:06] Alright, that roll over here.
[03:26:09] So if a is bigger than zero, the parabola
opens up.
[03:26:16] And if the value of the coefficient a is less
than zero, then the parabola opens down.
[03:26:22] In this second example, we can see again that
the vertex is at 00. And the x intercept is
[03:26:27] x equals zero.
[03:26:30] Let's look at this third example.
[03:26:33] If we multiplied our expression out, we'd
see that the coefficient of x squared would
[03:26:38] be to a positive number. So that means our
parabola is going to be opening up.
[03:26:43] But the vertex of this parabola will no longer
be at the origin. In fact, we can find the
[03:26:50] parabolas vertex by thinking about transformations
of functions.
[03:26:55] Our function is related to the function y
equals to x squared, by moving it to the right
[03:27:02] by three and up by four. Since y equals 2x
squared has a vertex at 00. If we move that
[03:27:13] whole parabola including the vertex, right
by three, and up by four, the vertex will
[03:27:21] end up at the point three, four.
[03:27:26] So a parabola will look something like this.
[03:27:30] Notice how easy it was to just read off the
vertex when our quadratic function is written
[03:27:36] in this form. In fact, any parabola any quadratic
function written in the form a times x minus
[03:27:44] h squared plus k has a vertex at h k. By the
same reasoning, we're moving the parabola
[03:27:53] with a vertex at the origin to the right by
H, and by K.
[03:27:59] That's why this form of a quadratic function
is called the vertex form.
[03:28:06] Notice that this parabola has no x intercepts
because it does not cross the x axis.
[03:28:11] For our final function, we have g of x equals
5x squared plus 10x plus three,
[03:28:19] we know the graph of this function will be
a parabola pointing upwards, because the coefficient
[03:28:23] of x squared is positive.
[03:28:26] To find the x intercepts, we can set y equals
zero, since the x intercepts is where our
[03:28:32] graph crosses the x axis, and that's where
the y value is zero.
[03:28:37] So zero equals 5x squared plus 10x plus three,
I'm going to use the quadratic formula to
[03:28:43] solve that. So x is negative 10 plus or minus
the square root of 10 squared minus four times
[03:28:52] five times three, all over two times five.
That simplifies to x equals negative 10 plus
[03:28:59] or minus the square root of 40 over 10,
[03:29:03] which simplifies further to x equals negative
10 over 10, plus or minus the square root
[03:29:08] of 40 over 10, which is negative one plus
or minus two square root of 10 over 10, or
[03:29:16] negative one plus or minus square root of
10 over five.
[03:29:21] Since the square root of 10 is just a little
bit bigger than three, this works out to approximately
[03:29:27] about negative two fifths and negative eight
foot or so somewhere around right here. So
[03:29:34] our parabola is going to look something like
this. Notice that it goes through crosses
[03:29:40] the x axis at y equals three, that's because
when we plug in x equals zero, and two here
[03:29:48] we get y equal three, so the y intercept is
at three. Finally, we can find the vertex.
[03:29:56] Since this function is written in standard
form,
[03:30:00] On the Y equals a x plus a squared plus bx
plus c form, and not in vertex form, we can't
[03:30:10] just read off the vertex like we couldn't
before. But there's a trick called the vertex
[03:30:15] formula, which says that whenever you have
a function in a quadratic function in standard
[03:30:20] form, the vertex has an x coordinate
[03:30:24] of negative B over two A. So in this case,
that's an x coordinate of negative 10 over
[03:30:34] two times five or negative one, which is kind
of like where I put it on the graph. To find
[03:30:39] the y coordinate of that vertex, I can just
plug in negative one into my equation for
[03:30:47] x, which gives me at y equals five times negative
one squared, plus 10, times negative one plus
[03:30:54] three, which is negative two.
[03:30:59] So I think I better redraw my graph a little
bit to put that vertex down at a y coordinate
[03:31:04] of negative two where it's supposed to be.
[03:31:10] Let's summarize the steps we use to graph
these quadratic functions. First of all, the
[03:31:15] graph of a quadratic function has the shape
of a parabola.
[03:31:19] The parabola opens up, if the coefficient
of x squared, which we'll call a is greater
[03:31:27] than zero and down if a is less than zero.
To find the x intercepts, we set y equal to
[03:31:34] zero, or in other words, f of x equal to zero
and solve for x, the find the vertex, we can
[03:31:44] either read it off as h k, if our function
is in vertex form,
[03:31:53] or we can use the vertex formula
[03:31:57] and get the x coordinate
[03:32:01] of the vertex to be negative B over to a if
our function is in standard form.
[03:32:10] To find the y coordinate of the vertex in
this case, we just plug in the x coordinate
[03:32:19] and figure out what y is.
[03:32:21] Finally, we can always find additional points
on the graph by plugging in values of x.
[03:32:29] In this video, we learned some tricks for
graphing quadratic functions. In particular,
[03:32:34] we saw that the vertex can be read off as
h k, if our function is written in vertex
[03:32:43] form, and the x coordinate of the vertex can
be calculated
[03:32:50] as negative B over two A, if our function
is in standard form. For an explanation of
[03:32:57] why this vertex formula works, please see
the my other video.
[03:33:02] a quadratic and standard form looks like y
equals A x squared plus bx plus c, where a,
[03:33:09] b and c are real numbers, and a is not zero.
a quadratic function in vertex form looks
[03:33:16] like y equals a times x minus h squared plus
k, where a h and k are real numbers, and a
[03:33:24] is not zero. When a functions in vertex form,
it's easy to read off the vertex is just the
[03:33:30] ordered pair, H, okay.
[03:33:34] This video explains how to get from vertex
form two standard form and vice versa.
[03:33:40] Let's start by converting this quadratic function
from vertex form to standard form. That's
[03:33:46] pretty straightforward, we just have to distribute
out.
[03:33:50] So if I multiply out the X minus three squared,
[03:33:58] I get minus four times x squared minus 6x
plus nine plus one, distributing the negative
[03:34:07] four, I get negative 4x squared plus 24x minus
36 plus one. So that works out to minus 4x
[03:34:16] squared plus 24x minus 35. And I have my quadratic
function now in standard form.
[03:34:26] Now let's go the other direction and convert
a quadratic function that's already in standard
[03:34:31] form into vertex form. That is, we want to
put it in the form of g of x equals a times
[03:34:40] x minus h squared plus k, where the vertex
is at h k. To do this, it's handy to use the
[03:34:51] vertex formula.
[03:34:53] The vertex formula says that the x coordinate
of the vertex is given by negative
[03:35:00] over to a, where A is the coefficient of x
squared, and B is the coefficient of x. So
[03:35:08] in this case, we get an x coordinate of negative
eight over two times two or negative two.
[03:35:16] To find the y coordinate of the vertex, we
just plug in the x coordinate into our formula
[03:35:24] for a g of x. So that's g of negative two,
which is two times negative two squared plus
[03:35:30] eight times negative two plus six. And that
works out to be negative two by coincidence.
[03:35:37] So the vertex for our quadratic function has
coordinates negative two, negative two. And
[03:35:44] if I want to write g of x in vertex form,
it's going to be a times x minus negative
[03:35:50] two squared plus minus two. That's because
remember, we subtract H and we add K. So that
[03:36:00] simplifies to g of x equals a times x plus
two squared minus two. And finally, we just
[03:36:08] need to figure out what this leading coefficient
as. But notice, if we were to multiply, distribute
[03:36:14] this out, then the coefficient of x squared
would end up being a. So therefore, the coefficient
[03:36:23] of x squared here, which is a has to be the
same as the coefficient of x squared here,
[03:36:27] which we conveniently also called a, in other
words, are a down here needs to be two. So
[03:36:33] I'm going to write that as g of x equals two
times x plus two squared minus two, lots of
[03:36:39] twos in this problem. And that's our quadratic
function in vertex form. If I want to check
[03:36:46] my answer, of course, I could just distribute
out again, I'd get two times x squared plus
[03:36:53] 4x, plus two, minus two, in other words, 2x
squared plus 8x plus six, which checks out
[03:37:03] to exactly what I started with. This video
showed how to get from vertex form to standard
[03:37:09] form by distributing out and how to get from
standard form to vertex form by finding the
[03:37:14] vertex using the vertex formula.
[03:37:17] Suppose you have a quadratic function in the
form y equals x squared plus bx plus c, and
[03:37:26] you want to find where the vertex is, when
you graph it.
[03:37:31] The vertex formula says that the x coordinate
of this vertex is that negative B over two
[03:37:40] A,
[03:37:42] this video gives a justification for where
that formula comes from.
[03:37:47] Let's start with a specific example. Suppose
I wanted to find the x intercepts and the
[03:37:53] vertex for this quadratic function. To find
the x intercepts, I would set y equal to zero
[03:38:00] and solve for x. So that's zero equals 3x
squared plus 7x minus five. And to solve for
[03:38:07] x, I use the quadratic formula. So x is going
to be negative seven plus or minus the square
[03:38:13] root of seven squared minus four times three
times negative five, all over two times three.
[03:38:22] That simplifies to negative seven plus or
minus a squared of 109 over six, we could,
[03:38:29] I could also write this as negative seven,
six, plus the square root of 109 over six,
[03:38:35] or x equals negative seven, six minus the
square root of 109 over six, since the square
[03:38:41] root of 109 is just a little bit bigger than
10, this can be approximated by negative seven
[03:38:47] six plus 10, six, and negative seven, six
minus 10, six.
[03:38:54] So pretty close to, I guess about what half
over here and pretty close to about negative
[03:39:00] three over here, I'm just going to estimate
so that I can draw a picture of the function.
[03:39:06] Since the leading coefficient three is positive,
I know my parabola is going to be opening
[03:39:11] up and the intercepts are somewhere around
here and here. So roughly speaking, it's gonna
[03:39:20] look something like this.
[03:39:25] Now the vertex is going to be somewhere in
between the 2x intercepts, in fact, it's going
[03:39:30] to be by symmetry, it'll be exactly halfway
in between the 2x intercepts. Since the x
[03:39:37] intercepts are negative seven, six plus and
minus 100 squared of 109 over six, the number
[03:39:43] halfway in between those is going to be exactly
negative seven, six,
[03:39:47] right, because on the one hand, I have negative
seven six plus something. And on the other
[03:39:56] hand, I have negative seven six minus that
same thing.
[03:39:59] So negative seven, six will be exactly in
the middle. So my x coordinate of my vertex
[03:40:08] will be at negative seven sex. Notice that
I got that number from the quadratic formula.
[03:40:16] More generally, if I want to find the x intercepts
for any quadratic function, I set y equal
[03:40:24] to zero
[03:40:27] and solve for x using the quadratic formula,
negative b plus or minus the square root of
[03:40:33] b squared minus four AC Oliver to a, b, x
intercepts will be at these two values, but
[03:40:40] the x coordinate of the vertex, which is exactly
halfway in between the 2x intercepts will
[03:40:47] be at negative B over two A. That's where
the vertex formula comes from.
[03:40:53] And it turns out that this formula works even
when there are no x intercepts, even when
[03:40:58] the quadratic formula gives us no solutions.
That vertex still has the x coordinate, negative
[03:41:04] B over two A.
[03:41:06] And that's the justification of the vertex
formula.
[03:41:11] This video is about polynomials and their
graphs.
[03:41:14] We call that a polynomial is a function like
this one, for example.
[03:41:20] His terms are numbers times powers of x. I'll
start with some definitions. The degree of
[03:41:28] a polynomial is the largest exponent. For
example, for this polynomial, the degree is
[03:41:35] for
[03:41:37] the leading term is the term with the largest
exponent. In the same example, the leading
[03:41:44] term is 5x to the fourth, it's conventional
to write the polynomial in descending order
[03:41:51] of powers of x. So the leading term is first.
But the leading term doesn't have to be the
[03:41:56] first term. If I wrote the same polynomial
as y equals, say, two minus 17x minus 21x
[03:42:04] cubed plus 5x. Fourth, the leading term would
still be the 5x to the fourth, even though
[03:42:11] it was last.
[03:42:13] The leading coefficient is the number in the
leading term. In this example, the leading
[03:42:20] coefficient is five.
[03:42:21] Finally, the constant term is the term with
no x's in it. In our same example, the constant
[03:42:28] term is to
[03:42:29] please pause the video for a moment and take
a look at this next example of polynomial
[03:42:34] figure out what's the degree the leading term,
the leading coefficient and the constant term.
[03:42:40] The degree is again, four, since that's the
highest power we have, and the leading term
[03:42:48] is negative 7x. to the fourth, the leading
coefficient is negative seven, and the constant
[03:42:56] term is 18.
[03:42:58] In the graph of the polynomial Shown here
are the three marks points are called turning
[03:43:04] points, because the polynomial turns around
and changes direction at those three points,
[03:43:12] those same points can also be called local
extreme points, or they can be called local
[03:43:18] maximum and minimum points. For this polynomial,
the degree is four, and the number of turning
[03:43:25] points is three.
[03:43:28] Let's compare the degree and the number of
turning points for these next three examples.
[03:43:33] For the first one, the degree is to and there's
one turning point.
[03:43:39] For the second example, that agree, is three,
and there's two turning points.
[03:43:50] And for this last example, the degree is four,
and there's one turning point.
[03:43:56] For this first example, and the next two,
the number of turning points is exactly one
[03:44:01] less than the degree. So you might conjecture
that this is always true. But in fact, this
[03:44:07] is not always true. In this last example,
the degree is four, but the number of turning
[03:44:13] points is one, not three.
[03:44:16] In fact, it turns out that while the number
of turning points is not always equal to a
[03:44:21] minus one, it is always less than or equal
to the degree minus one. That's a useful factor.
[03:44:28] Remember, when you're sketching graphs are
recognizing graphs of polynomials.
[03:44:33] The end behavior of a function is how the
ends of the function look, as x gets bigger
[03:44:39] and bigger heads towards infinity, or x gets
goes through larger and larger negative numbers
[03:44:45] towards negative infinity.
[03:44:47] In this first example, the graph of the function
goes down as x gets towards infinity as x
[03:44:54] goes towards negative infinity. I can draw
this with two little arrows pointing
[03:44:59] Down on either side, or I can say in words,
that the function is falling as we had left
[03:45:05] and falling also as we had right.
[03:45:09] In the second example, the graph rises to
the left and rises to the right. In the third
[03:45:16] example, the graph falls to the left, but
rises to the right. And in the fourth example,
[03:45:23] it rises to the left and falls to the right.
If you study these examples, and others, you
[03:45:29] might notice there's a relationship between
the degree of the polynomial the leading coefficients
[03:45:36] of the polynomials and the end behavior. Specifically,
these four types of end behavior are determined
[03:45:44] by whether the degree is even or odd. And
by whether the leading coefficient is positive
[03:45:51] or negative.
[03:45:54] When the degree is even, and the leading coefficient
is positive, like in this example, where the
[03:46:00] leading coefficient is one, we have this sorts
of n behavior rising on both sides.
[03:46:08] When the degree is even at the leading coefficient
is negative, like in this example, we have
[03:46:14] the end behavior that's falling on both sides
[03:46:18] when the degree is odd, and the leading coefficient
is positive, that's like this example, with
[03:46:23] the degree three and the leading coefficient
three, then we have this sort of end behavior.
[03:46:30] And finally, when the degree is odd, and the
leading coefficient is negative, like in this
[03:46:35] example, we have this sort of NBA havior.
[03:46:40] I like to remember this chart just by thinking
of the most simple examples, y equals x squared,
[03:46:48] y equals negative x squared, y equals x cubed,
and y equals negative x cubed. Because I know
[03:46:55] by heart what those four examples look like,
[03:47:00] then I just have to remember that any polynomial
with a even degree and positive leading coefficient
[03:47:08] has the same end behavior as x squared.
[03:47:11] And similarly, any polynomial with even degree
and negative leading coefficient has the same
[03:47:18] end behavior as negative x squared. And similar
statements for x cubed and negative x cubed.
[03:47:25] We can use facts about turning points and
behavior, to say something about the equation
[03:47:30] of a polynomial just by looking at this graph.
[03:47:34] In this example, because of the end behavior,
we know that the degree is odd,
[03:47:42] we know that the leading coefficient
[03:47:46] must be positive.
[03:47:49] And finally, since there are 1234, turning
points, we know that the degree is greater
[03:47:58] than or equal to five.
[03:48:01] That's because the number of turning points
is less than or equal to the degree minus
[03:48:06] one. And in this case, the number of turning
points we said was four.
[03:48:10] And so solving that inequality, we get the
degree is bigger than equal to five.
[03:48:18] Put in some of that information together,
we see the degree could be
[03:48:23] five, or seven, or nine, or any odd number
greater than or equal to five. But it couldn't
[03:48:31] be for example, three or six. Because even
numbers and and also numbers less than five
[03:48:38] are right out.
[03:48:41] This video gave a lot of definitions, including
the definition of degree
[03:48:46] leading coefficients,
[03:48:49] turning points,
[03:48:51] and and behavior.
[03:48:54] We saw that knowing the degree and the leading
coefficient can help you make predictions
[03:48:59] about the number of turning points and the
end behavior, as well as vice versa.
[03:49:06] This video is about exponential functions
and their graphs.
[03:49:11] an exponential function is a function that
can be written in the form f of x equals a
[03:49:16] times b to the x, where a and b are any real
numbers. As long as a is not zero, and B is
[03:49:24] positive.
[03:49:26] It's important to notice that for an exponential
function, the variable x is in the exponent.
[03:49:33] This is different from many other functions
we've seen, for example, quite a quadratic
[03:49:37] function like f of x equals 3x squared has
the variable in the base, so it's not an exponential
[03:49:44] function.
[03:49:45] For exponential functions, f of x equals a
times b dx. We require that A is not equal
[03:49:53] to zero, because otherwise, we would have
f of x equals zero times b dx.
[03:50:00] Which just means that f of x equals zero.
And this is called a constant function, not
[03:50:04] an exponential function. Because f of x is
always equal to zero
[03:50:11] in an exponential function, but we require
that B is bigger than zero, because otherwise,
[03:50:17] for example, if b is equal to negative one,
say, we'd have f of x equals a times negative
[03:50:27] one to the x. Now, this would make sense for
a lot of values of x. But if we try to do
[03:50:34] something like f of one half, with our Bs,
negative one, that would be the same thing
[03:50:40] as a times the square root of negative one,
which is not a real number,
[03:50:45] we'd get the same problem for other values
of b, if the values of b were negative. And
[03:50:52] if we tried b equals zero, we'd get a kind
of ridiculous thing a times zero to the x,
[03:50:59] which again is always zero. So that wouldn't
count as an exponential either. So we can't
[03:51:03] use any negative basis, and we can't use zero
basis, if we want an exponential function.
[03:51:09] The number A in the expression f of x equals
a times b to the x is called the initial value.
[03:51:18] And the number B is called the base.
[03:51:21] The phrase initial value comes from the fact
that if we plug in x equals zero, we get a
[03:51:28] times b to the zero, well, anything to the
zero is just one. So this is equal to A. In
[03:51:34] other words, f of zero equals a. So if we
think of starting out,
[03:51:40] when x equals zero, we get the y value
[03:51:44] of a, that's why it's called the initial value.
[03:51:51] Let's start out with this example, where y
equals a times b to the x next special function,
[03:51:59] and we've set a equal to one and B equals
a two. Notice that the y intercept, the value
[03:52:06] when x is zero, is going to be one.
[03:52:09] If I change my a value, my initial value,
the y intercept changes, the function is stretched
[03:52:20] out.
[03:52:21] If I make the value of a go to zero, and then
negative,
[03:52:27] then my initial value becomes negative, and
my graph flips across the x axis. Let's go
[03:52:35] back to an a value of say one, and see what
happens when we change B. Right now, the B
[03:52:41] value the basis two, if I increase B, my y
intercept sticks at one, but my graph becomes
[03:52:52] steeper and steeper. If I put B back down
close to one, my graph becomes more flat at
[03:53:03] exactly one, my graph is just a constant.
[03:53:05] As B gets into fractional territory, point
8.7 point six, my graph starts to slope the
[03:53:14] other way, it's decreasing now instead of
increasing, but notice that the y intercept
[03:53:20] still hasn't changed, I can get it more and
more steep as my B gets farther and farther
[03:53:26] away from one of course, when B goes to negative
territory, my graph doesn't make any sense.
[03:53:33] So a changes the y intercept, and B changes
the steepness of the graph. So that it's added
[03:53:44] whether it's increasing for B values bigger
than one, and decreasing for values of the
[03:53:50] less than one.
[03:53:55] we'll summarize all these observations on
the next slide.
[03:53:58] We've seen that for an exponential function,
y equals a times b to the x, the parameter
[03:54:04] or number a gives the y intercept, the parameter
B tells us how the graph is increasing or
[03:54:13] decreasing. Specifically, if b is greater
than one, the graph is increasing.
[03:54:19] And if b is less than one, the graph is decreasing.
[03:54:23] The closer B is to the number one, the flatter
the graph.
[03:54:29] So for example, if I were to graph y equals
point two five to the x and y equals point
[03:54:37] four to the x, they would both be decreasing
graphs, since the base for both of them is
[03:54:44] less than one. But point two five is farther
away from one and point four is closer to
[03:54:52] one. So point four is going to be flatter.
And point two five is going to be more steep.
[03:54:58] So in this picture,
This red graph would correspond to point two
[03:55:03] five to the x, and the blue graph would correspond
to point four to the x. For all these exponential
[03:55:11] functions, whether the graphs are decreasing
or increasing, they all have a horizontal
[03:55:17] asymptote along the x axis. In other words,
at the line at y equals zero, the domain is
[03:55:25] always from negative infinity to infinity,
and the range is always from zero to infinity,
[03:55:34] because the range is always positive y values.
Actually, that's true. If a is greater than
[03:55:43] zero, if a is less than zero, then our graph
flips over the x axis, our domain stays the
[03:55:52] same, but our range becomes negative infinity
to zero. The most famous exponential function
[03:55:59] is f of x equals e to the x. This function
is also sometimes written as f of x equals
[03:56:04] x of x. The number E is Oilers number as approximately
2.7. This function has important applications
[03:56:13] to calculus and to some compound interest
problems. In this video, we looked at exponential
[03:56:21] functions, functions of the form a times b
to the x, where the variable is in the exponent,
[03:56:28] we saw that they all have the same general
shape, either increasing like this, or decreasing
[03:56:37] like this, unless a is negative, in which
case they flip over the x axis. They all have
[03:56:45] a horizontal asymptote at y equals zero, the
x axis. In this video, we'll use exponential
[03:56:55] functions to model real world examples. Let's
suppose you're hired for a job. The starting
[03:57:02] salary is $40,000. With a guaranteed annual
raise of 3% each year, how much will your
[03:57:10] salary be after one year, two years, five
years. And in general after two years, let
[03:57:16] me chart out the information. The left column
will be the number of years since you're hired.
[03:57:23] And the right column will be your salary.
When you start work, at zero years after you're
[03:57:30] hired, your salary will be $40,000. After
one year, you'll have gotten a 3% raise. So
[03:57:40] your salary will be the original 40,000 plus
3% of 40,000.03 times 40,000. I can think
[03:57:52] of this first number as one at times 40,000.
And I can factor out the 40,000 from both
[03:58:00] terms, to get 40,000 times one plus 0.03.
we rewrite this as 40,000 times 1.03. This
[03:58:15] is your original salary multiplied by a growth
factor of 1.03. After two years, you'll get
[03:58:26] a 3% raise from your previous year salary,
your previous year salary was 40,000 times
[03:58:33] 1.03.
[03:58:34] But you'll add
[03:58:36] 3% of that, again, I can think of the first
number is one times 40,000 times 1.03. And
[03:58:45] I can factor out the common factor of 40,000
times 1.03 from both terms, to get 40,000
[03:58:53] times 1.03 times one plus 0.03. Let me rewrite
that as 40,000 times 1.03 times 1.03 or 40,000
[03:59:08] times 1.3 squared. We can think of this as
your last year salary multiplied by the same
[03:59:19] growth factor of 1.03. After three years,
a similar computation will give you that your
[03:59:26] new salary is your previous year salary times
that growth factor of 1.03 that can be written
[03:59:37] as 40,000 times 1.03 cubed. And in general,
if you're noticing the pattern, after two
[03:59:45] years, your salary should be 40,000 times
1.03 to the t power. In other words, your
[03:59:53] salary after two years is your original salary
multiplied By the growth factor of 1.03, taken
[04:00:06] to the t power, let me write this as a formula
s of t, where S of t is your salary is equal
[04:00:15] to 40,000 times 1.03 to the T. This is an
exponential function, that is a function of
[04:00:24] the form a times b to the T, where your initial
value a is 40,000 and your base b is 1.03.
[04:00:38] Notice that your base is the amount that your
salary gets multiplied by each year. Given
[04:00:45] this formula, we can easily figure out what
your salary will be after, for example, five
[04:00:49] years by plugging in five for T. I worked
that out on my calculator to be $46,370.96
[04:01:01] to the nearest cent. exponential functions
are also useful in modeling population growth.
[04:01:08] The United Nations estimated that the world
population in 2010 was 6.7 9 billion growing
[04:01:15] at a rate of 1.1% per year. Assuming that
the growth rate stayed the same and will continue
[04:01:20] to stay the same, we'll write an equation
for the population at t years after the year
[04:01:26] 2010 1.1%, written as a decimal is 0.011.
So if we work out a chart as before, we see
[04:01:39] that after zero years since 2010, we have
our initial population 6.7 9 billion, after
[04:01:46] one year, we'll take that 6.7 9 billion and
add 1.1% of it. That is point 011 times 6.79.
[04:01:59] This works out to 6.79 times one plus point
011 or 6.79 times 1.011. Here we have our
[04:02:15] initial population of 6.7 9 billion, and our
growth factor of 1.011. That's how much the
[04:02:27] population got multiplied by in one year.
As before, we can work out that after two
[04:02:35] years, our population becomes 6.79 times 1.011
squared, since they got multiplied by 1.011,
[04:02:45] twice, and after two years, it'll be 6.79
times 1.011 to the t power. So our function
[04:02:56] that models population is going to be 6.79
times 1.011 to the T. Here, t represents time
[04:03:09] in years, since 2010. Just for fun, I'll plug
in t equals 40. That's 40 years since 2010.
[04:03:21] So that's the year 2050. And I get 6.79 times
1.011 to the 40th power, which works out to
[04:03:31] 10 point 5 billion. That's the prediction
based on this exponential model.
[04:03:40] The previous two examples were examples of
exponential growth. This last example is an
[04:03:45] example of exponential decay. The drugs Seroquel
is metabolized and eliminated from the body
[04:03:51] at a rate of 11% per hour. If 400 milligrams
are given how much remains in the body. 24
[04:03:58] hours later, I'll chart out my information
where the left column will be time in hours
[04:04:09] since the dose was given, and the right column
will be the number of milligrams of Seroquel
[04:04:15] still on the body. zero hours after the dose
is given, we have the full 400 milligrams
[04:04:22] in the body. One hour later, we have the formula
100 milligrams minus 11% of it, that's minus
[04:04:30] point one one times 400. If I factor out the
400 from both terms, I get 400 times one minus
[04:04:40] 0.11 or 400 times point eight nine. The 400
represents the initial amount. The point eight
[04:04:53] nine I'll call the growth factor, even though
in this case the quantity is decreasing, not
[04:04:58] growing. So really it's kind of a shrink factor.
Let's see what happens after two hours. Now
[04:05:08] I'll have 400 times 0.89 my previous amount,
it'll again get multiplied by point eight,
[04:05:15] nine, so that's going to be 400 times 0.89
squared. And in general, after t hours, I'll
[04:05:22] have 400 multiplied by this growth or shrinkage
factor of point eight nine, raised to the
[04:05:29] t power. Since each hour, the amount of Seroquel
gets multiplied by point eight, nine, that
[04:05:36] number less than one. All right, my exponential
decay function as f of t equals 400 times
[04:05:44] 0.89 to the T, where f of t represents the
number of milligrams of Seroquel in the body.
[04:05:54] And t represents the number of hours since
the dose was given. To find out how much is
[04:06:01] in the body after 24 hours, I just plug in
24 for t. This works out to about 24.4 milligrams,
[04:06:13] I hope you notice the common form for the
functions used to model all three of these
[04:06:17] examples. The functions are always in the
form f of t equals a times b to the T, where
[04:06:26] a represented the initial amount, and B represented
the growth factor. To find the growth factor
[04:06:36] B, we started with the percent increase or
decrease. We wrote it as a decimal. And then
[04:06:45] we either added or subtracted it from one,
depending on whether the quantity was increasing
[04:06:50] or decreasing. Let me show you that as a couple
examples. In the first example, we had a percent
[04:06:58] increase of 3% on the race as a decimal, I'll
call this r, this was point O three. And to
[04:07:09] get the growth factor, we added that to one
to get 1.03. In the world population example,
[04:07:16] we had a 1.1% increase, we wrote this as point
011 and added it to one to get 1.011. In the
[04:07:29] drug example, we had a decrease of 11%. We
wrote that as a decimal. And we subtracted
[04:07:42] the point one one from one to get 0.89. In
fact, you can always write the growth factor
[04:07:50] B as one plus the percent change written as
a decimal. If you're careful to make that
[04:07:58] percent change, negative when the quantity
is decreasing and positive when the quantity
[04:08:04] is increasing. Since here one plus negative
point 11
[04:08:09] gives us the correct growth factor of point
eight nine which is less than one as a good
[04:08:14] check. Remember that if your quantity is increasing,
the base b should be bigger than one. And
[04:08:23] if the quantity is decreasing, then B should
be less than one.
[04:08:30] exponential functions can also be used to
model bank accounts and loans with compound
[04:08:35] interest, as we'll see in another video. This
video is about interpreting exponential functions.
[04:08:42] An antique car is worth $50,000 now, and its
value increases by 7%. Each year, let's write
[04:08:49] an equation to model its value x years from
now. After one year, it's value is the 50,000
[04:09:00] plus 0.07 times the 50,000. That's because
its value has grown by 7% or point oh seven
[04:09:10] times 50,000. this can be written as 50,000
times one plus point O seven. Notice that
[04:09:20] adding 7% to the original value is the same
as multiplying the original value by one plus
[04:09:28] point oh seven or by 1.07. After two years,
the value will be 50,000 times one plus 0.07
[04:09:41] squared, or 50,000 times 1.07 squared. That's
because the previous year's value is multiplied
[04:09:51] again by 1.7. In general, after x years We
have x, the antique cars value will be 50,000
[04:10:06] times 1.07 to the x. That's because the original
value of 50,000 gets multiplied by 1.07x times
[04:10:16] one time for each year. If we dissect this
equation, we see that the number of 50,000
[04:10:24] comes from the original value of the car.
The one point is seven, which I call the growth
[04:10:29] factor comes from one plus point 707. The
point O seven being the, the percent increase
[04:10:41] written as a decimal. So the form of this
equation is an exponential equation, V of
[04:10:50] x equals a times b to the x, where A is the
initial value, and B is the growth factor.
[04:10:57] But we could also write this as a times one
plus r to the x, where A is still the initial
[04:11:06] value,
[04:11:07] but
[04:11:08] R is the percent increase written as a decimal.
This same equation will come up in the next
[04:11:15] example, here, my Toyota Prius is worth only
3000. Now, and its value is decreasing by
[04:11:22] 5% each year. So after one year, its value
will be 3000 minus 0.05 times 3000. That is
[04:11:34] 3000 times one minus 0.05. I can also write
that as 3000 times 0.95. Decreasing the value
[04:11:48] by 5% is like multiplying the value by one
minus point oh five, or by point nine, five.
[04:12:00] After two years, the value will be multiplied
by point nine five again. So the value will
[04:12:08] be 3000 times point nine, five squared. And
after x years, the value will be 3000 times
[04:12:17] point nine five to the x. So my equation for
the value is 3000 times point nine five to
[04:12:26] the x. This is again, an equation of the form
V of x equals a times b to the x, where here,
[04:12:36] a is 3000, the initial value, and B is point
nine, five, I'll still call that the growth
[04:12:47] factor, even though we're actually declining
value not growing. Now remember where this
[04:12:53] point nine five came from, it came from taking
one and subtracting point 05, because of the
[04:13:01] 5%, decrease in value, so I can again, write
my equation in the form a times well, this
[04:13:09] time times one minus r to the x, where R is
our point 05. That's our percent decrease
[04:13:19] written as a decimal. Please take a moment
to study this equation and the previous one.
[04:13:26] They say that when you have an exponential
function, the number here is the initial value.
[04:13:35] If it's written in this form, B is your growth
factor. But you can think of B as being either
[04:13:42] one minus r, where r is the percent decrease,
or one plus r, where r is the percent increase.
[04:13:50] In this example, we're given a function f
of x to model the number of bacteria in a
[04:13:55] petri dish x hours after 12 o'clock noon,
we want to know what was the number of bacteria
[04:14:03] at noon, and by what percent, the number of
bacteria is increasing every hour, we can
[04:14:09] see from the equation that the number of bacteria
is increasing and not decreasing, because
[04:14:14] the base of the exponential function 1.45
is bigger than one. Notice that our equation
[04:14:21] 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
[04:14:30] 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
[04:14:43] this familiar form, we can recognize that
the initial amount of bacteria is going to
[04:14:49] be 12 12,000. Since those are our units and
this value 1.45 is our growth factor. What
[04:15:00] the number of bacteria is multiplied by each
hour? Well are the point four five is the
[04:15:08] the rate of increase, in other words, a 45%
increase each hour. So our answers to the
[04:15:18] questions are 12,045%. In this example, the
population of salamanders is modeled by this
[04:15:29] exponential function, where x is the number
of years since 2015. Notice that the number
[04:15:37] of salamanders is decreasing, because the
base of our exponential function point seven,
[04:15:42] eight is less than one. So if we recognize
the form of our exponential function, a times
[04:15:50] b to the x, or we can think of this as a times
one minus r to the x, where A is our initial
[04:15:59] value, and r is our percent decrease written
as a decimal.
[04:16:08] Our initial value is 3000. So that's the number
of salamanders zero years after 2015. Our
[04:16:18] growth factor B is 0.78. But if I write that
as one minus r, I see that R has to be one
[04:16:27] minus 0.78, or 0.22. In other words, our population
is decreasing by 22% each year. In this video,
[04:16:40] we saw that exponential functions can be written
in the form f of x equals a times b to the
[04:16:47] x, where A is the initial value. And B is
the growth factor. We also saw that they can
[04:16:58] be written in the form a times one plus r
to the x when the amount is increasing. And
[04:17:05] as a times one minus r to the x when the amount
is decreasing. In this format, R is the percent
[04:17:16] increase or the percent decrease written as
a decimal. So 15% increase will be an R value
[04:17:28] of 0.15 and a growth factor B of 1.15. Whereas
a 12% decrease will be an R value of point
[04:17:43] one, two, and a B value of one minus point
one, two, or 0.7. Sorry, 0.88. These observations
[04:17:55] help us quickly interpret exponential functions.
For example, here, we have an initial value
[04:18:03] of 100 and a 15% increase. And here, we have
an initial value of 50 and a 40%. decrease.
[04:18:17] exponential functions can be used to model
compound interest for loans and bank accounts.
[04:18:23] Suppose you invest $200 in a bank account
that earns 3% interest every year. If you
[04:18:30] make no deposits or withdrawals, how much
money will you have accumulated after 10 years,
[04:18:37] because 3% of the money that's in the bank
is getting added each year, the money in the
[04:18:41] bank gets multiplied by 1.03 each year. So
after one year, the amount will be 200 times
[04:18:51] 1.3, after two years 200 times one point O
three squared, and and after t years 200 times
[04:19:02] one point O three to the t power. So the function
modeling the amount of money, I'll call it
[04:19:08] P of t is given by 200 times 1.03 to the T.
More generally, if you invest a dollars
[04:19:20] at an annual interest rate of our for t years.
In the end, you'll have P of t equals a times
[04:19:32] one plus r to the T
[04:19:37] here r needs to be written as a decimal, so
0.03 In our example, for the 3% annual interest
[04:19:45] rate. Going back to our specific example,
After 10 years, the amount of money is going
[04:19:51] to be P of 10 which is 200 times 1.03 to the
10 which works out to 206 $68.78 to the nearest
[04:20:03] cent. In this problem, we've assumed that
the interest accumulates once per year. But
[04:20:10] in the next few examples, we'll see what happens
when the interest rate accumulates more frequently,
[04:20:14] twice a year, or every month. For example,
let's deposit $300 in an account that earns
[04:20:23] 4.5% annual interest compounded semi annually,
this means two times a year or every six months.
[04:20:31] A 4.5% annual interest rate compounded two
times a year means that we're actually getting
[04:20:40] 4.5 over 2% interest, every time the interest
is compounded. That is every six months. That's
[04:20:53] 2.25% interest every half a year. Note that
2.25% is the same as 0.02 to five as a decimal.
[04:21:05] So every time we earn interest, our money
gets multiplied by 1.02 to five. Let me make
[04:21:17] a chart of what happens. After zero years,
which is also zero half years, we have our
[04:21:25] original $300. for half a year, that's one
half year, our money gets earns interest one
[04:21:33] time, so we multiply the 300 by 1.02 to five,
after one year, that's two half years, our
[04:21:42] money earns interest two times. So we multiply
300 by 1.0225 squared. continuing this way,
[04:21:52] after 1.5 years, that's three half years,
we have 300 times 1.02 to five cubed. And
[04:22:01] after two years or four half years, we have
300 times 1.02 to five to the fourth. In general,
[04:22:09] after t years, which is to T half years, our
money will grow to 300 times 1.02 to five
[04:22:20] to the two t power. Because we've compounded
interest to tee times our formula for our
[04:22:28] amount of money is P of t equals 300 times
one point O two to five to the two t where
[04:22:35] t is the number of years. To finish the problem,
after seven years, we'll have P of $7, which
[04:22:45] is 300 times 1.02 to five to the two times
seven or 14 power. And that works out to $409.65
[04:22:59] to the nearest cent. In this next example,
we're going to take out a loan for $1,200
[04:23:06] in annual interest rate of 6% compounded monthly.
Although loans and bank accounts might feel
[04:23:14] different, they're mathematically the same.
It's like from the bank's point of view, they're
[04:23:19] investing money in you and getting interest
on their money from you. So we can work them
[04:23:25] out with the same kind of math 6% annual interest
rate compounded monthly means you're compounding
[04:23:36] 12 times a year. So each time you compound
interest, you're just going to get six over
[04:23:42] 12% interest. That's point 5% interest. And
as a decimal, that's 0.005. Let's try it out
[04:23:56] again, what happens. Time is zero, of course,
you'll have the original loan amount of 1200.
[04:24:05] After one year, that's 12 months, your loan
has had interest added to it 12 times so it
[04:24:15] gets multiplied by 1.05 to the 12th.
[04:24:22] After two years, that's 24 months, it's had
interest added to it 24 times, so it gets
[04:24:30] multiplied by 1.05 to the 24th power. Similarly,
after three years or 36 months, your loan
[04:24:39] amount will be 1200 times 1.05 to the 36th
power. And in general, after two years, that's
[04:24:49] 12 t months. So the interest will be compounded
12 t times and so we have to raise the 1.05
[04:24:59] To the 12 t power. This gives us the general
formula for the money owed is P of t equals
[04:25:08] 1200 times 1.05 to the 12 T, where T is the
number of years. In particular, after three
[04:25:20] years, we'll have to pay back a total of 1200
times 1.05 to the 12 times three or 36 power,
[04:25:31] which works out to $1,436.02 to the nearest
cent. These last two problems follow a general
[04:25:42] pattern. If A is the initial amount of the
loan or bank account, and r is the annual
[04:25:51] interest rate, compounded n times per year,
then our formula for the amount of money is
[04:26:03] going to be a times one plus r over n to the
n t. This formula exactly matches what we
[04:26:13] did in this problem. First, we took the interest
rate our which was 6%, and divided it by the
[04:26:20] number of compounding periods each year 12.
We wrote that as a decimal and added one to
[04:26:25] it. That's where we got the 1.05 from. We
raised this to not to the number of years,
[04:26:35] but to 12 times the number of years. That's
the number of compounding periods per year,
[04:26:39] times the number of years. And we multiplied
all that by the initial amount of money, which
[04:26:46] was 1200. This formula for compound interest
is a good one to memorize. But it's also important
[04:26:52] to be able to reason your way through it,
like we did in this chart. There's one more
[04:26:57] type of compound interest. And that's interest
compounded continuously. You can think of
[04:27:02] continuous compounding as the limit of compounding
more and more frequently 10 times a year,
[04:27:09] 100 times a year 1000 times a year a million
times a year. In the limit, you get continuous
[04:27:14] compounding. The formula for continuous compounding
is P of t equals a times e to the RT, where
[04:27:24] p of t is the amount of money, t is the time
in years. A is the initial amount of money.
[04:27:37] And R is the annual interest rate written
as a decimal. So 0.025 in this problem from
[04:27:49] the 2.5% annual interest rate. He represents
the famous constant Oilers constant, which
[04:27:57] is about 2.718. So in this problem, we have
P of t is 4000 times e to the 0.025 T. And
[04:28:12] after five years, we'll have P of five, which
is 4000 times e to the 0.0 to five times five,
[04:28:21] which works out to $4,532.59 to the nearest
cent. To summarize, if R represents the annual
[04:28:32] interest rate written as a decimal, that is
2% would be 0.02. And t represents the number
[04:28:43] of years and a represents the initial amount
of money, then for just simple annual interest
[04:28:50] compounded once a year. Our formula is P of
t is a times one plus r to the T for compound
[04:29:00] interest compounded n times per year. Our
formula is P of t is a times one plus r over
[04:29:08] n to the n T. And for compound interest compounded
continuously, we get P of t is a times e to
[04:29:16] the RT.
[04:29:18] In this video, we looked at three kinds of
compound interest problems, simple annual
[04:29:24] interest, interest compounded and times per
year, and continuously compounded interest.
[04:29:33] This video introduces logarithms. logarithms
are a way of writing exponents. The expression
[04:29:42] log base a of B equals c means that a to the
C equals b. In other words, log base a of
[04:29:52] B is the exponent that you raise a to to get
BE THE NUMBER A It is called the base of the
[04:30:02] logarithm. It's also called the base when
we write it in this exponential form. Some
[04:30:07] students find it helpful to remember this
relationship. log base a of B equals c means
[04:30:14] a to the C equals b, by drawing arrows,
[04:30:17] a to the C equals b.
[04:30:20] Other students like to think of it in terms
of asking a question, log base a of B, asks,
[04:30:30] What power do you raise a to in order to get
b? Let's look at some examples. log base two
[04:30:41] of eight is three, because two to the three
equals eight. In general, log base two of
[04:30:50] y is asking you the question, What power do
you have to raise to to to get y? So for example,
[04:30:59] log base two of 16 is four, because it's asking
you the question to what power equals 16?
[04:31:08] And the answer is four. Please pause the video
and try some of these other examples. log
[04:31:15] base two of two is asking, What power do you
raise to two to get two? And the answer is
[04:31:22] one. Two to the one equals two. log base two
of one half is asking two to what power gives
[04:31:34] you one half? Well, to get one half, you need
to raise two to a negative power. So that
[04:31:40] would be two to the negative one. So the answer
is negative one. log base two of 1/8 means
[04:31:49] what power do we raise to two in order to
get 1/8. Since 1/8, is one over two cubed,
[04:31:58] we have to raise two to the negative three
power to get one over two cubed. So our exponent
[04:32:06] is negative three. And that's our answer to
our log expression. Finally, log base two
[04:32:13] of one is asking to what power equals one.
Well, anything raised to the zero power gives
[04:32:21] us one, so this log expression evaluates to
zero. Notice that we can get positive negative
[04:32:28] and zero answers for our logarithm expressions.
Please pause the video and figure out what
[04:32:36] these logs evaluate to. to work out log base
10 of a million. Notice that a million is
[04:32:46] 10 to the sixth power. Now we're asking the
question, What power do we raise tend to to
[04:32:54] get a million? So that is what power do we
raise 10 to to get 10 to the six? Well, of
[04:32:59] course, the answer is going to be six. Similarly,
since point O one is 10 to the minus three,
[04:33:11] this log expression is the same thing as asking,
what's the log base 10 of 10 to the minus
[04:33:16] three? Well, what power do you have to raise
10? to to get 10 to the minus three? Of course,
[04:33:22] the answer is negative three. Log base 10
of zero is asking, What power do we raise
[04:33:31] 10 to to get zero. If you think about it,
there's no way to raise 10 to an exponent
[04:33:38] get zero. Raising 10 to a positive exponent
gets us really big positive numbers. Raising
[04:33:43] 10 to a negative exponent is like one over
10 to a power that's giving us tiny fractions,
[04:33:49] but they're still positive numbers, we're
never going to get zero. Even if we raise
[04:33:55] 10 to the zero power, we'll just get one.
So there's no way to get zero and the log
[04:33:59] base 10 of zero does not exist. If you try
it on your calculator using the log base 10
[04:34:05] button, you'll get an error message. Same
thing happens when we do log base 10 of negative
[04:34:11] 100. We're asking 10 to what power equals
negative 100. And there's no exponent that
[04:34:17] will work. And more generally, it's possible
to take the log of numbers that are greater
[04:34:23] than zero, but not for numbers that are less
than or equal to zero. In other words, the
[04:34:29] domain of the function log base a of x, no
matter what base you're using for a, the domain
[04:34:36] is going to be all positive numbers. A few
notes on notation. When you see ln of x, that's
[04:34:44] called natural log, and it means the log base
e of x where he is that famous number that's
[04:34:50] about 2.718. When you see log of x with no
base at all, by convention, that means log
[04:34:59] base 10 of x And it's called the common log.
Most scientific calculators have buttons for
[04:35:05] natural log, and for common log. Let's practice
rewriting expressions with logs in them. log
[04:35:13] base three of one nine is negative two, can
be rewritten as the expression three to the
[04:35:20] negative two equals 1/9.
[04:35:26] Log of 13 is shorthand for log base 10 of
13. So that can be rewritten as 10 to the
[04:35:34] 1.11394 equals 13. Finally, in this last expression,
ln means natural log, or log base e, so I
[04:35:45] can rewrite this equation as log base e of
whenever E equals negative one. Well, that
[04:35:52] means the same thing as e to the negative
one equals one over e, which is true. Now
[04:36:00] let's go the opposite direction. We'll start
with exponential equations and rewrite them
[04:36:05] as logs. Remember that log base a of B equals
c means the same thing as a to the C equals
[04:36:14] b, the base stays the same in both expressions.
So for this example, the base of three in
[04:36:23] the exponential equation, that's going to
be the same as the base in our log. Now I
[04:36:29] just have to figure out what's in the argument
of the log. And what goes on the other side
[04:36:33] of the equal sign. Remember that the answer
to a log is an exponent. So the thing that
[04:36:40] goes in this box should be my exponent for
my exponential equation. In other words, you.
[04:36:48] And I'll put the 9.78 as the argument of my
log. This works, because log base three of
[04:36:57] 9.78 equals view means the same thing as three
to the U equals 9.78, which is just what we
[04:37:06] started with. In the second example, the base
of my exponential equation is E. So the base
[04:37:13] of my log is going to be the answer to my
log is an exponent. In this case, the exponent
[04:37:22] 3x plus seven. And the other expression, the
four minus y becomes my argument of my log.
[04:37:32] Let me check, log base e of four minus y equals
3x plus seven means e to the 3x plus seven
[04:37:41] equals four minus y, which is just what I
started with. I can also rewrite log base
[04:37:47] e as natural log. This video introduced the
idea of logs, and the fact that log base a
[04:37:56] of B equal c means the same thing as a to
the C equals b. So log base a of B is asking
[04:38:07] you the question, What power exponent Do you
raise a to in order to get b. In this video,
[04:38:16] we'll work out the graph, so some log functions
and also talk about their domains. For this
[04:38:22] first example, let's graph a log function
by hand by plotting some points. The function
[04:38:29] we're working with is y equals log base two
of x, I'll make a chart of x and y values.
[04:38:36] Since we're working this out by hand, I want
to pick x values for which it's easy to compute
[04:38:41] log base two of x. So I'll start out with
the x value of one because log base two of
[04:38:47] one is zero, log base anything of one is 02
is another x value that's easy to compute
[04:38:56] log base two of two, that's asking, What power
do I raise to two to get to one? And the answer
[04:39:03] is one. Power other powers of two are easy
to work with. So for example, log base two
[04:39:11] of four that saying what power do I raise
to two to get four, so the answer is two.
[04:39:18] Similarly, log base two of eight is three
and log base two of 16 is four. Let me also
[04:39:26] work with some fractional values for X. If
x is one half, then log base two of one half
[04:39:33] that saying what power do I raise to two to
get one half? Well, that needs a power of
[04:39:39] negative one. It's also easy to compute by
hand, the log base two of 1/4 and 1/8. log
[04:39:48] base two of 1/4 is negative two since two
to the negative two is 1/4. And similarly,
[04:39:57] log base two of one eight is negative f3 I'll
put some tick marks on my x and y axes. Please
[04:40:04] pause the video and take a moment to plot
these points. Let's see, I have the point,
[04:40:10] one zero, that's here to one that's here,
for two,
[04:40:17] that is here, and then eight, three, which
is here. And the fractional x values, one
[04:40:26] half goes with negative one, and 1/4 with
negative two 1/8 with negative three. And
[04:40:33] if I connect the dots, I get a graph that
looks something like this. If I had smaller
[04:40:39] and smaller fractions, I would keep getting
more and more negative answers when I took
[04:40:44] log base two of them, so my graph is getting
more and more negative, my y values are getting
[04:40:51] more and more negative, as x is getting close
to zero. Now I didn't draw any parts of the
[04:40:56] graph over here with negative X values, I
didn't put any negative X values in my chart,
[04:41:02] that omission is no accident. Because if you
try to take the log base two or base anything
[04:41:08] of a negative number, like say negative four
or something, there's no answer. This doesn't
[04:41:15] exist because there's no power that you can
raise to to to get a negative number. So there
[04:41:24] are no points on the graph for negative X
values. And similarly, there are no points
[04:41:29] on the graph where x is zero, because you
can't take log base two of zero, there's no
[04:41:35] power you can raise to two to get zero. I
want to observe some key features of this
[04:41:42] graph. First of all, the domain is x values
greater than zero. In interval notation, I
[04:41:51] can write that as a round bracket because
I don't want to include zero to infinity,
[04:41:58] the range is going to be the y values, while
they go all the way down into the far reaches
[04:42:05] of the negative numbers. And the graph gradually
increases y value is getting bigger and bigger.
[04:42:11] So the range is actually all real numbers
are an interval notation negative infinity
[04:42:17] to infinity. Finally, I want to point out
that this graph has a vertical asymptote at
[04:42:25] the y axis, that is at the line x equals zero.
I'll draw that on my graph with a dotted line.
[04:42:35] A vertical asymptote is a line that our functions
graph gets closer and closer to. So this is
[04:42:42] the graph of y equals log base two of x. But
if I wanted to graph say, y equals log base
[04:42:50] 10 of x, it would look very similar, it would
still have a domain of X values greater than
[04:42:57] zero, a range of all real numbers and a vertical
asymptote at the y axis, it will still go
[04:43:02] through the point one zero, but it would go
through the point 10 one instead, because
[04:43:10] log base 10 of 10 is one, it would look pretty
much the same, just a lot flatter over here.
[04:43:20] But even though it doesn't look like it with
the way I've drawn it, it's still gradually
[04:43:24] goes up to n towards infinity. In fact, the
graph of y equals log base neaa of X for a
[04:43:34] bigger than one looks pretty much the same,
and has the same three properties. Now that
[04:43:42] we know what the basic log graph looks like,
we can plot at least rough graphs of other
[04:43:46] log functions without plotting points. Here
we have the graph of natural log of X plus
[04:43:52] five. And again, I'm just going to draw a
rough graph. If I did want to do a more accurate
[04:43:57] graph, I probably would want to plot some
points. But I know that roughly a log graph,
[04:44:02] if it was just like y equals ln of x, that
would look something like this, and it would
[04:44:09] go through the point one zero, with a vertical
asymptote along the y axis. Now if I want
[04:44:17] a graph, ln of x plus five, that just shifts
our graph by five units, it'll still have
[04:44:23] the same vertical asymptote. Since the vertical
line shifted up by five units, there's still
[04:44:27] a vertical line, but instead of going through
one zero, it'll go through the point one,
[04:44:33] five. So I'll draw a rough sketch here. Let's
compare our starting function y equals ln
[04:44:42] x and the transformed version y equals ln
x plus five in terms of the domain, the range
[04:44:51] and the vertical asymptote. Our original function
y equals ln x had a domain of zero to infinity
[04:44:59] Since adding five on the outside affects the
y values, and the domain is the x values,
[04:45:08] this transformation doesn't change the domain.
So the domain is still from zero to infinity.
[04:45:16] Now the range of our original y equals ln
x was from negative infinity to infinity.
[04:45:22] Shifting up by five does affect the y values,
and the range is talking about the y values.
[04:45:28] But since the original range was all real
numbers, if you add five to all set of all
[04:45:33] real numbers, you still get the set of all
real numbers. So in this case, the range doesn't
[04:45:38] change either. And finally, we already saw
that the original vertical asymptote of the
[04:45:43] y axis x equals zero, when we shift that up
by five units, it's still the vertical line
[04:45:49] x equals zero. In this next example, we're
starting with a log base 10 function. And
[04:45:57] since the plus two is on the inside, that
means we shift that graph left by two. So
[04:46:02] I'll draw our basic log function. Here's our
basic log function. So I'll think of that
[04:46:09] as y equals log of x going through the point
one, zero,
[04:46:15] here's its vertical asymptote. Now I need
to shift everything left by two. So my vertical
[04:46:23] asymptote shifts left, and now it's at the
line x equals negative two, instead of at
[04:46:28] x equals zero, and my graph, let's see my
point, one zero gets shifted to, let's see
[04:46:37] negative one zero, since I'm subtracting two
from the axis, and here's a rough sketch of
[04:46:45] the resulting graph. Let's compare the features
of the two graphs drawn here. We're talking
[04:46:54] about domains, the original had a domain of
from zero to infinity. But now I've shifted
[04:47:01] that left. So I've subtracted two from all
my x values. And here's my new domain, which
[04:47:07] I can also verify just by looking at the picture.
My range was originally from negative infinity
[04:47:15] to infinity, well, shifting left only affects
the x value. So it doesn't even affect the
[04:47:20] range. So my range is still negative infinity
to infinity. My vertical asymptote was originally
[04:47:26] at x equals zero. And since I subtract two
from all x values, that shifts it to x equals
[04:47:32] negative two. In this last problem, I'm not
going to worry about drawing this graph. I'll
[04:47:38] just use algebra to compute its domain. So
let's think about what's the issue, when you're
[04:47:44] taking the logs of things? Well, you can't
take the log of a negative number is zero.
[04:47:50] So whatever's inside the argument of the log
function, whatever is being fed into log had
[04:47:56] better be greater than zero. So I'll write
that down, we need to minus 3x to be greater
[04:48:03] than zero. Now it's a matter of solving an
inequality to it's got to be greater than
[04:48:08] 3x. So two thirds is greater than x. In other
words, x has to be less than two thirds. So
[04:48:15] our domain is all the x values from negative
infinity to two thirds, not including two
[04:48:23] thirds. It's a good idea to memorize the basic
shape of the graph of a log function. It looks
[04:48:31] something like this, go through the point
one zero, and has a vertical asymptote on
[04:48:36] the y axis. Also, if you remember that you
can't take the log of a negative number, or
[04:48:44] zero, then that helps you quickly compute
domains for log functions. Whatever's inside
[04:48:52] the log function, you set that greater than
zero, and solve. This video is about combining
[04:49:01] logs and exponents. Please pause the video
and take a moment to use your calculator to
[04:49:07] evaluate the following four expressions. Remember,
that log base 10 on your calculator is the
[04:49:15] log button. While log base e on your calculator
is the natural log button, you should find
[04:49:24] 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
[04:49:34] the log base 10 of 1000 is 1000. And eat the
log base e of 9.6 is 9.6. In each case, the
[04:49:45] log and the exponential function with the
same base undo each other and we're left with
[04:49:51] the exponent. In fact, it's true that for
any base a the log base a of a to the x is
[04:49:59] equal to x, the same sort of cancellation
happens if we do the exponential function
[04:50:05] in the log function with the same base in
the opposite order. For example, we take 10
[04:50:10] to the power of log base 10 of 1000, the 10
to the power and the log base 10 undo each
[04:50:15] other, and we're left with the 1000s. This
happens for any base, say, a data log base
[04:50:21] a of x is equal to x. We can describe this
by saying that an exponential function and
[04:50:31] a log function with the same base undo each
other. If you're familiar with the language
[04:50:40] of inverse functions, the exponential function
and log function are inverses. Let's see why
[04:50:46] these roles hold for the first log role. log
base a of a dx is asking the question, What
[04:50:55] power do we raise a two in order to get a
to the x? In other words, a two what power
[04:51:03] is a dx? Well, the answer is clearly x. And
that's why log base a of a to the x equals
[04:51:11] x. For the second log rule,
[04:51:15] notice that the log base a of x means the
power we raise a two to get x. But this expression
[04:51:25] is saying that we're supposed to raise a to
that power. If we raise a to the power, we
[04:51:30] need to raise a two to get x, then we'll certainly
get x. Now let's use these two rules. In some
[04:51:37] examples. If we want to find three to the
log base three of 1.43 to the power and log
[04:51:44] base three undo each other, so we're left
with 1.4. If we want to find ln of e iliacs.
[04:51:51] Remember that ln means log base e. So we're
taking log base e of e to the x, well, those
[04:51:59] functions undo each other, and we're left
with x. If we want to take 10 to the log of
[04:52:05] three z, remember that log without a base
written implies that the base is 10. So really,
[04:52:12] we want to take 10 to the log base 10 of three
z will tend to a power and log base 10 undo
[04:52:18] each other. So we're left with a three z.
Finally, this last statement hold is ln of
[04:52:26] 10 to the x equal to x, well, ln means log
base e. So we're taking log base e of 10 to
[04:52:36] the x, notice that the base of the log and
the base of the exponential function are not
[04:52:41] the same. So they don't undo each other. And
in fact, log base e of 10 to the x is not
[04:52:47] usually equal to x, we can check with one
example. Say if x equals one, then log base
[04:52:53] e of 10 to the one, that's log base e of 10.
And we can check on the calculator that's
[04:52:59] equal to 2.3. And some more decimals, which
is not the same thing as one. So this statement
[04:53:06] is false, it does not hold. We need the basis
to be the same for logs and exponents to undo
[04:53:13] each other. In this video, we saw that logs
and exponents with the same base undo each
[04:53:23] other. Specifically,
[04:53:26] 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
[04:53:33] x
[04:53:34] for any values of x and any base a. This video
is about rules or properties of logs. The
[04:53:42] log rules are closely related to the exponent
rules. So let's start by reviewing some of
[04:53:46] the exponent rules. To keep things simple,
we'll write everything down with a base of
[04:53:51] two. Even though the exponent rules hold for
any base. We know that if we raise two to
[04:53:57] the zero power, we get one, we have a product
rule for exponents, which says that two to
[04:54:04] the M times two to the n is equal to two to
the m plus n. In other words, if we multiply
[04:54:12] two numbers, then we add the exponents. We
also have a quotient rule that says that two
[04:54:18] to the M divided by n to the n is equal to
two to the m minus n. In words, that says
[04:54:26] that if we divide two numbers, then we subtract
the exponents. Finally, we have a power rule
[04:54:34] that says if we take a power to a power, then
we multiply the exponents. Each of these exponent
[04:54:43] rules can be rewritten as a log rule. The
first rule, two to the zero equals one can
[04:54:50] be rewritten in terms of logs as log base
two of one equals zero. That's because log
[04:54:57] base two of one equals zero mean To the zero
equals one. The second rule, the product rule
[04:55:05] can be rewritten in terms of logs by saying
log of x times y equals log of x plus log
[04:55:14] of y. I'll make these base two to agree with
my base that I'm using for my exponent rules.
[04:55:21] In words, that says the log of the product
is the sum of the logs. Since logs really
[04:55:28] represent exponent, this is saying that when
you multiply two numbers together, you add
[04:55:34] their exponents, which is just what we said
for the exponent version. The quotient rule
[04:55:41] for exponents can be rewritten in terms of
logs by saying the log of x divided by y is
[04:55:48] equal to the log of x minus the log of y.
In words, we can say that the log of the quotient
[04:55:58] is equal to the difference of the logs. Since
logs are really exponents, another way of
[04:56:04] saying the same thing is that when you divide
two numbers, you subtract their exponents.
[04:56:11] That's how we described the exponent rule
above. Finally, the power rule for exponents
[04:56:18] can be rewritten in terms of logs by saying
the log of x to the n is equal to n times
[04:56:27] log of x. Sometimes people describe this rule
by saying when you take the log of an expression
[04:56:34] with an exponent, you can bring down the exponent
and multiply. If we think of x as being some
[04:56:40] power of two, this is really saying when we
take a power to a power, we multiply their
[04:56:45] exponents. That's exactly how we described
the power rule above. It doesn't really matter
[04:56:50] if you multiply this exponent on the left
side, or on the right side. But it's more
[04:56:56] traditional to multiply it on the left side.
I've given these rules with the base of two,
[04:57:03] but they actually work for any base. To help
you remember them, please take a moment to
[04:57:07] write out the log roles using a base of a
you should get the following chart. Let's
[04:57:17] use the log rules to rewrite the following
expressions as a sum or difference of logs.
[04:57:23] In the first expression, we have a log base
10 of a quotient. So we can rewrite the log
[04:57:30] of the quotient as the difference of the logs.
Now we still have the log of a product, I
[04:57:39] can rewrite the log of a product as the sum
of the logs. So that is log of y plus log
[04:57:47] of z. When I put things together, I have to
be careful because here I'm subtracting the
[04:57:55] entire log expression. So I need to subtract
both terms of this song on the make sure I
[04:58:02] do that by putting them in parentheses. Now
I can simplify a little bit by distributing
[04:58:08] the negative sign. And here's my final answer.
In my next expression, I have the log of a
[04:58:16] product. So I can rewrite that as the sum
of two logs.
[04:58:21] I can also use my power rule to bring down
the exponent T and multiply it in the front.
[04:58:31] That gives me the final expression log of
five plus t times log of to one common mistake
[04:58:38] on this problem is to rewrite this expression
as t times log of five times two. In fact,
[04:58:47] those two expressions are not equal. Because
the T only applies to the two, not to the
[04:58:54] whole five times two, we can't just bring
it down in front using the power wall. After
[04:58:59] all, the power rule only applies to a single
expression raised to an exponent, and not
[04:59:08] to a product like this. And these next examples,
we're going to go the other direction. Here
[04:59:14] we're given sums and differences of logs.
And we want to wrap them up into a single
[04:59:18] log expression. By look at the first two pieces,
that's a difference of logs. So I know I can
[04:59:24] rewrite it as the log of a quotient. Now I
have the sum of two logs. So I can rewrite
[04:59:34] that as the log of a product. I'll clean that
up a little bit and rewrite it as log base
[04:59:43] five of a times c over B. In my second example,
I can rewrite the sum of my logs as the log
[04:59:54] of a product now, I will Like to rewrite this
difference of logs as the log of a quotient.
[05:00:03] But I can't do it yet, because of that factor
of two multiplied in front. But I can use
[05:00:09] the power rule backwards to put that two back
up in the exponent. So I'll do that first.
[05:00:15] So I will copy down the ln of x plus one times
x minus one, and rewrite this second term
[05:00:23] as ln of x squared minus one squared. Now
I have a straightforward difference of two
[05:00:31] logs, which I can rewrite as the log of a
quotient. I can actually simplify this some
[05:00:40] more. Since x plus one times x minus one is
the same thing as x squared minus one. I can
[05:00:50] cancel factors to get ln of one over x squared
minus one. In this video, we saw four rules
[05:01:00] for logs that are related to exponent rules.
First, we saw that the log with any base of
[05:01:07] one is equal to zero. Second, we saw the product
rule, the log of a product is equal to the
[05:01:15] sum of the logs. We saw the quotient rule,
the log of a quotient is the difference of
[05:01:23] the logs. And we saw the power rule. When
you take a log of an expression with an exponent
[05:01:30] in it, you can bring down the exponent and
multiply it. It's worth noticing that there's
[05:01:37] no log rule that helps you split up the log
of a psalm. In particular, the log of a psalm
[05:01:44] is not equal to the sum of the logs. If you
think about logs and exponent rules going
[05:01:52] together, this kind of makes sense, because
there's also no rule for rewriting the sum
[05:02:00] of two exponential expressions.
[05:02:04] Log rules will be super handy, as we start
to solve equations using logs. Anytime you
[05:02:10] have an equation like this one that has variables
in the exponent logs are the tool of choice
[05:02:17] for getting those variables down where you
can solve for them. In this video, I'll do
[05:02:23] a few examples of solving equations with variables
in the exponent. For our first example, let's
[05:02:29] solve for x the equation five times two to
the x plus one equals 17. As my first step,
[05:02:37] I'm going to isolate the difficult spot, the
part that has the variables and the exponent.
[05:02:42] In this example, I can do that by dividing
both sides by five. That gives me two to the
[05:02:47] x plus one equals 17 over five. Next, I'm
going to take the log of both sides, it's
[05:02:56] possible to take the log with any base, but
I prefer to take the log base 10, or the log
[05:03:00] base e for the simple reason that my calculator
has buttons for those logs. So in this example,
[05:03:06] I'll just take the log base 10. So I can omit
the base because it's a base 10 is implied
[05:03:12] here. And that gives me this expression. As
my next step, I'm going to use log roles to
[05:03:19] bring down my exponent and multiply it on
the front. It's important to use parentheses
[05:03:24] here because the entire x plus one needs to
get multiplied by log two. So that was my
[05:03:31] third step using the log roles. Now all my
variables are down from the exponent where
[05:03:37] I can work with them, but I still need to
solve for x. Right now x is trapped in the
[05:03:42] parentheses. So I'm going to free it from
the parentheses by distributing. So I get
[05:03:47] x log two plus log two equals log of 17 fifths.
Now I will try to isolate x by moving all
[05:03:55] my terms with x's in them to one side, and
all my terms without x's in them to the other
[05:04:01] side. Finally, I factor out my x. Well, it's
already kind of factored out, and I divide
[05:04:08] to isolate it. So I read out what I did. So
far I distributed. I moved all the x terms
[05:04:17] on one side, and the terms without x's on
the other side. And then I isolated the x
[05:04:24] by factoring out and dividing. We have an
exact solution for x. This is correct, but
[05:04:31] maybe not so useful if you want a decimal
answer. At this point, you can type in everything
[05:04:36] into your calculator using parentheses liberally
to get a decimal answer about 0.765. It's
[05:04:46] always a good idea to check your work by typing
in this decimal answer and seeing if the equation
[05:04:51] checks out. This next example is trickier
because there are variables in the exponent
[05:04:57] in two places with two different bases. First
step is normally to clean things up and simplify
[05:05:05] by isolating the tricky stuff. But in this
example, there's nothing really to clean up
[05:05:10] or simplifier, or no way to isolate anything
more than it already is. So we'll go ahead
[05:05:15] to the next step and take the log of both
sides. Again, I'll go ahead and use log base
[05:05:20] 10. But we couldn't use log base e instead.
Next, we can use log roles to bring down the
[05:05:27] exponents. This gives me 2x minus three in
parentheses, log two equals x minus two, log
[05:05:35] five. Now I'm going to distribute things out
to free the excess from the parentheses. That
[05:05:43] gives me 2x log two minus three log two equals
x log five minus two log five. Now I need
[05:05:52] to group the x terms on one side, and the
other terms that don't involve x on the other
[05:05:59] side. So I'll keep the 2x log two on the left,
put the minus log x log five on the left,
[05:06:08] and that gives me the minus two log five that
stays on the right and a plus three log two
[05:06:13] on the right. Finally, I need to isolate x
by factoring and dividing by factoring I just
[05:06:22] mean I factor out the x from all the terms
on the left, so that gives me x times the
[05:06:27] quantity to log two minus log five. And that
equals this mess on the right. Now I can divide
[05:06:37] the right side by the quantity on the left
side. When you type this into your calculator,
[05:06:48] I encourage you to type the whole thing in
rather than type bits and pieces in because
[05:06:53] if you round off, you'll get a less accurate
answer than if you type the whole thing in
[05:06:58] at once. In this example, when I type it in,
I get a final answer of about 5.106.
[05:07:06] In this equation, we have the variable t in
the exponent in two places. The letter E is
[05:07:12] not a variable, it represents the number e
whose decimal approximation is about 2.7.
[05:07:19] Because there's already an E and the expression,
it's going to be handy to use natural log
[05:07:24] in this problem instead of log base 10. But
before we take the log of both sides, let's
[05:07:29] clean things up. by isolating the tricky parts,
we could at least divide both sides by five.
[05:07:35] And that will give us either the negative
0.05 t is equal to three fifths it is 0.2
[05:07:42] T. One way to proceed would now be to clean
things out further by dividing both sides
[05:07:47] by either the point two t, but I'm going to
take a different approach and go ahead and
[05:07:51] take the natural log of both sides. That gives
me ln of e to the minus 0.05 t is equal to
[05:08:01] ln of three fifths e to the 0.2 t. Now on
the left side, I can immediately use my log
[05:08:08] rule to bring down my exponent and get minus
0.05 t, L and E. But on the right side, I
[05:08:18] can't bring down the exponent yet because
this e to the point two t is multiplied by
[05:08:23] three fifths. So before I can bring down the
exponent, I need to split up this product
[05:08:28] using the product rule. So I can rewrite this
as ln of three fifths plus ln e to the 0.2
[05:08:36] T. And now I can bring down the exponent.
So that third step was using log roles to
[05:08:47] ultimately bring down my exponents. Now ln
of E is a really nice expression, ln IV means
[05:08:56] log base e of E. So that's asking what power
do I raise E to in order to get e? And the
[05:09:02] answer is one. So anytime I have ln of E,
I can just replace that with one. That's why
[05:09:10] using natural log is a little bit handier
here than using log base 10. You can make
[05:09:14] that simplification. Next, I'm ready to solve
for t. So I don't need to distribute here.
[05:09:23] But I do need to bring my T terms to one side
of my terms without t to the other side. So
[05:09:29] let's see. I'll put my T terms on the left
and my team turns without T on the right.
[05:09:37] And finally I'm going to isolate t by factoring
and dividing. So by factoring I mean I factor
[05:09:45] out my T 
and now I can divide. Using my calculator,
[05:09:57] I can get a decimal answer of 2.04 Three,
three. This video gave some examples of solving
[05:10:04] equations with variables in the exponent.
And the key idea was to take the log of both
[05:10:10] sides and use the log properties to bring
the exponents down. This video, give some
[05:10:21] examples of equations with logs in them like
this one. In order to solve the equations
[05:10:26] like this one, we have to free the variable
from the log. And we'll do that using exponential
[05:10:32] functions. My first step in solving pretty
much any kind of equation is to simplify it
[05:10:38] and isolate the tricky part. In this case,
the tricky part is the part with the log in
[05:10:44] it. So I can isolate it by first adding three
to both sides. That gives me two ln 2x plus
[05:10:51] five equals four, and then I can divide both
sides by two. Now that I've isolated the tricky
[05:11:01] part, I still need to solve for x, but x is
trapped inside the log function. to free it,
[05:11:07] I need to somehow undo the log function. Well
log functions and exponential functions undo
[05:11:13] each other. And since this is a log base e,
I need to use the exponential function base
[05:11:18] e also. So I'm going to take e to the power
of both sides. In other words, I'll take e
[05:11:26] to the ln 2x plus five, and that's going to
equal e to the power of two.
[05:11:33] Now e to the ln of anything, I'll just write
a for anything. That means e to the log base
[05:11:40] e of a, that you the power and log base e
undo each other. So we just get a, I'm going
[05:11:46] to use that principle over here, e to the
ln of 2x plus five, the E and the log base
[05:11:53] e undo each other. And we're left with 2x
plus five on the left side. So 2x plus five
[05:11:59] is equal to E squared. And from there, it's
easy to finish the problem and solve for x
[05:12:04] by subtracting five from both sides and then
dividing by two. So I'll just write the third
[05:12:11] step here is to just say finish solving for
x. There's one last step we need to do when
[05:12:17] solving equations with logs in them. And that's
to check our answers. Because we may get extraneous
[05:12:25] solutions. an extraneous solution is a solution
that comes out of the solving process, but
[05:12:31] doesn't actually satisfy the original equation.
And those can happen for equations with logs
[05:12:36] in them, because we might get a solution that
makes the argument of the log negative or
[05:12:42] zero, and we can't take the log of a negative
numbers zero. So let's check and plug in the
[05:12:52] solution of E squared minus five over two.
We'll plug that in for x in our original equation
[05:13:01] and see if that works. So let's see the twos
cancel here. So I get two ln e squared minus
[05:13:09] five plus five, minus three, I want that to
equal one. And now my fives cancel. And so
[05:13:17] I have two ln e squared minus three that I
want to equal one. Well, I think this is going
[05:13:23] to work out because let's see, ln is log base
e. and log base e of E squared that's asking
[05:13:34] the question, What power do I raise e two
to get e squared? Well, I have to raise it
[05:13:38] to the power of two to get e squared. So this
becomes two times two minus three. And does
[05:13:46] that equal one, four minus three does equal
one. So that all checks out. And we didn't
[05:13:53] have any problem with taking the log of negative
numbers zero, we didn't get any extraneous
[05:13:58] solutions. So this is our solution. The second
equation is a little bit trickier, because
[05:14:04] there's a log into places. Now notice that
this is a log where there's no base written.
[05:14:10] So a base 10 is implied. So I'm already thinking
in order to undo a log base 10, I'm going
[05:14:16] to want to take a 10 to the power of both
sides. Of course, it's still a good idea to
[05:14:23] isolate the tricky part, but there's nothing
really to isolate here. So I'm going to say,
[05:14:29] can't do it here. So we'll just jump right
to straight step two, and take 10 to the power
[05:14:37] of both sides. Okay, so that's going to give
me 10 to the whole thing, log x plus three
[05:14:46] plus log x, that whole thing is in the exponent
equals 10 to the One Power. Well, I know what
[05:14:52] to do with the right side 10 to the one is
just 10. But what do I do with the 10 to these
[05:14:58] two things added
[05:14:59] up Up,
[05:15:00] while remembering my exponent rules, I know
that when you add up the exponent, that's
[05:15:06] what happens when you multiply two things.
So this is the same thing as 10 to the log
[05:15:12] x plus three times 10 to the log x, right,
because when you multiply two things, you
[05:15:18] add the exponent, so these are the same. Okay,
now we're in business, because 10 to the log
[05:15:24] base 10, those undo each other. And so this
whole expression here simplifies to x plus
[05:15:31] three. Similarly, 10 to the log base, 10 of
x is just x. So I'm multiplying x plus three
[05:15:38] by x, that's equal to 10. Now I have an equation
I can deal with, it's a quadratic, so I'm
[05:15:46] going to first multiply out to make it look
more like a quadratic, get everything to one
[05:15:52] side. So is equal to zero. And, and now I
can either factor or use the quadratic formula.
[05:15:59] I think this one factors, it looks like X
plus five times x minus two, so I'm going
[05:16:06] to get x is negative five, or x is two. So
that was all the third step to finish solving
[05:16:14] for x. Finally, we need to check our solutions
to make sure we haven't gotten some extraneous
[05:16:20] ones. So let's see if x equals negative five.
If I plug that into my original equation,
[05:16:27] that says, I'm checking that log of negative
five plus three, plus log of negative five,
[05:16:36] checking that's equal to one. Well, this is
giving me a queasy feeling. And I hope it's
[05:16:40] giving you a queasy feeling too, because log
of negative two does not exist, right, you
[05:16:48] can't take the log of a negative number. Same
thing with log of negative five. So x equals
[05:16:53] negative five is an extraneous solution, it
doesn't actually solve our original equation.
[05:17:00] Let's check the other solution, x equals two.
So now we're checking to see if log of two
[05:17:08] plus three plus log of two is equal to one.
Since there's no problem with taking logs
[05:17:15] of negative numbers, or zero here, this should
work out fine. And just to finish checking,
[05:17:20] we can see let's see this is log of five plus
log of two, we want that equal one. But using
[05:17:27] my log rules, the sum of two logs is the log
of the product. So log of five times two,
[05:17:34] we want that to equal one. And that's just
log base 10 of 10. And that definitely equals
[05:17:41] one because log base 10 of 10 says, What power
do I raise tend to to get 10. And that power
[05:17:46] is one. So the second solution x equals two
does check out. And that's our final answer.
[05:17:56] Before I leave this problem, I do want to
mention that some people have an alternative
[05:18:00] approach. Some people like to start with the
original equation, and then use log roles
[05:18:05] to combine everything into one log expression.
So since we have the sum of two logs, we know
[05:18:12] that's the same as the log of a product, right,
so we can rewrite the left side as log of
[05:18:18] x plus three times x, that equals one, then
we do the same trick of taking 10 to the power
[05:18:25] of both sides. And as before, the 10 to the
power and the log base 10 undo each other,
[05:18:32] and we get x plus three times x equals 10,
just like we did before. In the first solution,
[05:18:40] we ended up using exponent rules to rewrite
things. In the second alternative method,
[05:18:45] we use log rules to rewrite things. So the
methods are really pretty similar, pretty
[05:18:50] equivalent, and they certainly will get us
to the same answer. So we've seen a couple
[05:18:54] examples of equations with logs in them and
how to solve them. And the key step is to
[05:19:02] use exponential functions to undo the log.
In other words, take e to the power of both
[05:19:13] sides to undo natural log, and take 10 to
the power of both sides. To undo log base
[05:19:25] 10.
[05:19:27] In this video, we'll use exponential equations
in some real life applications, like population
[05:19:33] growth and radioactive decay. I'll also introduce
the ideas of half life and doubling time.
[05:19:40] In this first example, let's suppose we invest
$1,600 in a bank account that earns 6.5% annual
[05:19:47] interest compounded once a year. How many
years will it take until the account has $2,000
[05:19:52] in it if you don't make any further deposits
or withdrawals since our money is earning
[05:19:59] six point 5% interest each year, that means
that every year the money gets multiplied
[05:20:05] by 1.065. So after t years, my 1600 gets multiplied
by 1.065 to the t power. I'll write this in
[05:20:18] function notation as f of t equals 1600 times
1.065 to the T, where f of t is the amount
[05:20:30] of money after t years. Now we're trying to
figure out how long it will take to get $2,000
[05:20:41] $2,000 is a amount of money. So that's an
amount for f of t. And we're trying to solve
[05:20:48] for T the amount of time. So let me write
out my equation and make a note that I'm solving
[05:20:55] for t. Now to solve for t, I want to first
isolate the tricky part. So I'm going to the
[05:21:04] tricky part is the part with the exponential
in it. So I'm going to divide both sides by
[05:21:08] 1600. That gives me 2000 over 1600 equals
1.065. To the tee, I can simplify this a little
[05:21:18] bit further as five fourths. Now that I've
isolated the tricky part, my next step is
[05:21:24] going to be to take the log of both sides.
That's because I have a variable in the exponent.
[05:21:29] And I know that if I log take the log of both
sides, I can use log roles to bring that exponent
[05:21:33] down where I can solve for it. I think I'll
use log base e this time. So I have ln fi
[05:21:40] force equals ln of 1.065 to the T. Now by
the power rule for logs, on the right side,
[05:21:48] I can bring that exponent t down and multiply
it in the front. Now it's easy to isolate
[05:21:56] t just by dividing both sides by ln of 1.065.
Typing that into my calculator, I get that
[05:22:07] t is approximately 3.54 years. And the next
example, we have a population of bacteria
[05:22:17] that initially contains 1.5 million bacteria,
and it's growing by 12% per day, we want to
[05:22:24] find the doubling time, the doubling time
means the amount of time it takes for a quantity
[05:22:34] to double in size. For example, the amount
of time it takes to get from the initial 1.5
[05:22:41] million bacteria to 3 million bacteria would
be the doubling time. Let's start by writing
[05:22:47] an equation for the amount of bacteria. So
if say, P of t represents the number of bacteria
[05:22:56] in millions, then my equation and T represents
time in days, then P of t is going to be given
[05:23:08] by the initial amount of bacteria times the
growth factor 1.12 to the T. That's because
[05:23:19] my population of bacteria is growing by 12%
per day. That means the number of bacteria
[05:23:25] gets multiplied by one point 12. Since we're
looking for the doubling time, we're looking
[05:23:30] for the t value when P of D will be twice
as big. So I can set P of t to be three and
[05:23:39] solve for t. As before, I'll start by isolating
the tricky part, taking the log
[05:23:56] bringing the T down. And finally solving for
T. Let's see three over 1.5 is two so I can
[05:24:08] write this as ln two over ln 1.12. Using my
calculator, that's about 6.12 days. It's an
[05:24:19] interesting fact that doubling time only depends
on the growth rate, that 12% growth, not the
[05:24:32] initial population. In fact, I could have
figured out the doubling time without even
[05:24:37] knowing how many bacteria were in my initial
population. Let me show you how that would
[05:24:41] work. If I didn't know how many I started
with, I could still write P of t equals a
[05:24:48] times 1.12 to the t where a is our initial
population that I don't know what it is.
[05:24:55] Then if I want to figure out how long it takes
for my population, to double Well, if I start
[05:25:01] with a and double that I get to a. So I'll
set my population equal to two a. And I'll
[05:25:08] solve for t. Notice that my A's cancel. And
so when I take the log of both sides and bring
[05:25:19] the T down and solve for t, I get the exact
same thing as before, it didn't matter what
[05:25:32] the initial population was, I didn't even
have to know what it was. In this next example,
[05:25:38] we're told the initial population, and we're
told the doubling time, we're not told by
[05:25:42] what percent the population increases each
minute, or by what number, we're forced to
[05:25:49] multiply the population by each minute. So
we're gonna have to solve for that, I do know
[05:25:53] that I want to use an equation of the form
y equals a times b to the T, where T is going
[05:25:59] to be the number of minutes, and y is going
to be the number of bacteria. And I know that
[05:26:04] my initial amount a is 350. So I can really
write y equals 350 times b to the T. Now the
[05:26:11] doubling time tells me that when 15 minutes
have elapsed, my population is going to be
[05:26:19] twice as big, or 700. plugging that into my
equation, I have 700 equals 350 times b to
[05:26:28] the 15. Now I need to solve for b. Let me
clean this up a little bit by dividing both
[05:26:34] sides by 350. That gives me 700 over 350 equals
b to the 15th. In other words, two equals
[05:26:43] b to the 15th. To solve for B, I don't have
to actually use logs here, because my variables
[05:26:51] in the base not in the exponent, so I don't
need to bring that exponent down. Instead,
[05:26:57] the easiest way to solve this is just by taking
the 15th root of both sides or equivalently.
[05:27:03] The 1/15 power. That's because if I take B
to the 15th to the 1/15, I multiply by exponents,
[05:27:12] that gives me B to the one is equal to two
to the 1/15. In other words, B is two to the
[05:27:19] 1/15, which as a decimal is approximately
1.047294. I like to use a lot of decimals
[05:27:29] if I'm doing a decimal approximation and these
kind of problems to increase accuracy. But
[05:27:34] of course, the most accurate thing is just
to leave B as it is. And I'll do that and
[05:27:38] rewrite my equation is y equals 350 times
two to the 1/15 to the T. Now I'd like to
[05:27:50] work this problem one more time. And this
time, I'm going to use the form of the equation
[05:27:56] y equals a times e to the RT. This is called
a continuous growth model. It looks different,
[05:28:04] but it's actually an equivalent form to this
growth model over here. And I'll say more
[05:28:09] about why these two forms are equivalent at
the end, I can use the same general ideas
[05:28:14] to solve in this form. So I know that my initial
amount is 350. And I know again that when
[05:28:21] 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.
[05:28:33] Again, I'm going to simplify things by dividing
both sides by 350. That gives me two equals
[05:28:44] e to the r times 15. This time my variable
does end up being in the exponent. So I do
[05:28:51] want to take the log of both sides, I'm going
to use natural log, because I already have
[05:28:56] an E and my problem. So natural log and E
are kind of more harmonious the other than
[05:29:01] then common log with base 10 and E. So I take
the log of both sides. Now I can pull the
[05:29:09] exponent down. So that's our times 15 times
ln of E. Well Elena V is just one, right?
[05:29:14] Because Elena V is asking what power do I
raise E to to get e that's just one. And so
[05:29:19] I get 15 r equals ln two. So r is equal to
ln two divided by 15. Let me plug that back
[05:29:27] in to my original equation as e to the ln
two over 15 times t. Now I claimed that these
[05:29:35] two equations were actually the same thing
just looking different. And the way to see
[05:29:40] that is if I start with this equation, and
I rewrite as e to the ln two over 15 like
[05:29:49] that
[05:29:50] to the tee. Well, I claim that this quantity
right here is the same thing as my two to
[05:29:58] the 1/15. And in fact One way to see that
is e to the ln two already to the 115. The
[05:30:06] Right, that's the same, because every time
I take a power to a power, I multiply exponents.
[05:30:10] But what's Ed Oh, and two, E and ln undo each
other, so that's just 350 times two to the
[05:30:18] 1/15 to the T TA, the equations are really
the same. And these, you'll always be able
[05:30:24] to find two different versions of an exponential
equation, sort of the standard one, a times
[05:30:30] b to the T, or the continuous growth one,
a times e to the RT. In this last example,
[05:30:38] we're going to work with half life, half life
is pretty much like doubling time, it just
[05:30:43] means the amount of time that it takes for
a quantity to decrease to half as much as
[05:30:55] we originally started with, we're told that
the half life of radioactive carbon 14 is
[05:31:02] 5750 years. So that means it takes that long
for a quantity of radioactive carbon 14 to
[05:31:11] decay, so that you just have half as much
left and the rest is nonradioactive form.
[05:31:16] So we're told a sample of bone that originally
contained 200 grams of radioactive carbon
[05:31:21] 14 now contains only 40 grams, we're supposed
to find out how old the sample is, is called
[05:31:29] carbon dating. Let's use the continuous growth
model this time. So our final amount, so this
[05:31:36] is our amount of radioactive c 14 is going
to be the initial amount
[05:31:49] times e to the RT, we could have used the
other model to we could have used f of t equals
[05:31:54] a times b to the T, but I just want to use
the continuous model for practice. So we know
[05:31:59] that our half life is 5750. So what that means
is when t is 5750, our amount is going to
[05:32:13] be one half of what we started with. Let me
see if I can plug that into my equation and
[05:32:19] figure out use that to figure out what r is.
That's ours called the continuous growth rate.
[05:32:26] So I plug in one half a for the final amount,
a is still the initial amount, e to the r
[05:32:35] and I have 5750 I can cancel my A's. And now
I want to solve for r, r is in my exponent.
[05:32:46] So I do need to take the log of both sides
to solve for it. I'm going to use log base
[05:32:50] e since I already have an E and my problem,
log base e is more compatible with E then
[05:32:57] log base 10 is okay. Now, on the left side,
I still have log of one half and natural log
[05:33:04] of one half on the right side, ln n e to a
power those undo each other. So I'm left with
[05:33:09] R times 5750. Now I can solve for r, it's
ln one half over 5750 could work that as a
[05:33:20] decimal, but it's actually more accurate just
to keep it in exact form. So now I can rewrite
[05:33:25] my equation, I have f of t equals a times
e to the ln one half over five 750 t. Now
[05:33:40] I can use that to figure out my problem. And
my problem the bone originally contained 200
[05:33:46] grams, that's my a, I want to figure out when
it's going to contain only 40 grams, that's
[05:33:53] my final amount. And so I need to solve for
t. I'll clean things up by dividing both sides
[05:34:00] by 200. Let's say 40 over 200 is 1/5. Now
I'm going to take the ln of both sides. And
[05:34:10] ln and e to the power undo each other. So
I'm left with Elena 1/5 equals ln of one half
[05:34:19] divided by five 750 T. And finally I can solve
for t but super messy but careful use of my
[05:34:27] calculator gives me an answer of 13,351 years
approximately.
[05:34:35] That kind of makes sense in terms of the half
life because to get from 200 to 40 you have
[05:34:41] to decrease by half a little more than two
times right or increasing by half once would
[05:34:45] get you to 100 decreasing to half again would
get you to 50 a little more than 40 and two
[05:34:50] half lives is getting pretty close to 13,000
years. This video introduced a lot of new
[05:34:57] things it introduced continue Less growth
model, which is another equivalent way of
[05:35:04] writing an exponential function. The relationship
is that the B in this example, is the same
[05:35:16] thing as EDR. In that version, it also introduced
the ideas of doubling time. And HalfLife the
[05:35:27] amount of time it takes a quantity to double
or decreased to half in an exponential growth
[05:35:32] model. Recall that a linear equation is equation
like for example, 2x minus y equals one. It's
[05:35:42] an equation without any x squared or y squared
in it, something that could be rewritten in
[05:35:47] the form y equals mx plus b, the equation
for a line a system of linear equations is
[05:35:55] a collection of two or more linear equations.
For example, I could have these two equations.
[05:36:03] A solution to a system of equations is that
an x value and a y value that satisfy both
[05:36:10] of the equations. For example, the ordered
pair two three, that means x equals two, y
[05:36:18] equals three is a solution to this system.
Because if I plug in x equals two, and y equals
[05:36:26] three into the first equation, it checks out
since two times two minus three is equal to
[05:36:33] one. And if I plug in x equals two and y equals
three into the second equation, it also checks
[05:36:40] out two plus three equals five. However, the
ordered pair one for that is x equals one,
[05:36:49] y equals four is not a solution to the system.
Even though this X and Y value work in the
[05:36:55] second equation, since one plus four does
equal five, it doesn't work in the first equation,
[05:37:02] because two times one minus four is not equal
to one. In this video, we'll use systematic
[05:37:10] methods to find the solutions to systems of
linear equations. In this first example, we
[05:37:17] want to solve this system of equations, there
are two main methods we could use, we could
[05:37:22] use the method of substitution, or we could
use the method of elimination. If we use the
[05:37:29] method of substitution, the main idea is to
isolate one variable in one equation, and
[05:37:38] then substitute it in to the other equation.
For example, we can start with the first equation
[05:37:44] 3x minus two y equals four, and isolate the
x by adding two y to both sides, and then
[05:37:53] dividing both sides by three. Think I'll rewrite
that a little bit by breaking up the fraction
[05:38:01] into two fractions four thirds plus two thirds
y. Now, I'm going to copy down the second
[05:38:07] equation 5x plus six y equals two. And I'm
going to substitute in my expression for x.
[05:38:16] That gives me five times four thirds plus
two thirds y plus six y equals two. And now
[05:38:24] I've got an equation with only one variable
in it y. So I can solve for y as a number.
[05:38:30] First, I'm going to distribute the five so
that gives me 20 thirds plus 10 thirds y plus
[05:38:38] six y equals two. And now I'm going to keep
all my y terms, my terms with y's in them
[05:38:45] on the left side, but I'll move all my terms
without y's in them to the right side. At
[05:38:51] this point, I could just add up all my fractions
and solve for y. But since I don't really
[05:38:56] like working with fractions, I think I'll
do the trick of clearing the denominators
[05:39:00] here. So I'm going to actually multiply both
sides by my common denominator of three just
[05:39:05] to get rid of the denominators and not have
to work with fractions. So let me write that
[05:39:10] down. Distributing the three, I get 10 y plus
eight t and y equals six minus 20. Now add
[05:39:20] things together. So that's 28 y equals negative
14. So that means that y is going to be negative
[05:39:28] 14 over 28, which is negative one half.
[05:39:32] So I've solved for y. And now I can go back
and plug y into either of my equations to
[05:39:38] solve for x, I plug it into my first equation.
So I'm plugging in negative one half for y.
[05:39:47] That gives me 3x plus one equals four. So
3x equals three, which means that x is equal
[05:39:55] to one. I've solved my system of equations
and gotten x equals one On y equals minus
[05:40:01] one half, I can also write that as an ordered
pair, one, negative one half for my solution.
[05:40:09] Now let's go back and solve the same system,
but use a different method, the method of
[05:40:14] elimination, the key idea to the method of
elimination is to multiply each equation by
[05:40:23] a constant to make the coefficients of one
variable match. Let me start by copying down
[05:40:34] my two equations. Say I'm trying to make the
coefficient of x match. One way to do that
[05:40:42] is to multiply the first equation by five,
and the second equation by three. That way,
[05:40:48] the coefficient of x will be 15 for both equations,
so let me do that. So for the first equation,
[05:40:55] I'm going to multiply both sides by five.
And for the second equation, I'm going to
[05:41:00] multiply both sides by three. That gives me
for the first equation 15x minus 10, y equals
[05:41:09] 20. And for the second equation, 15x plus
18, y is equal to six. Notice that the equate
[05:41:19] the coefficients of x match. So if I subtract
the second equation from the first, the x
[05:41:25] term will completely go away, it'll be zero
times x, and I'll be left with let's see negative
[05:41:32] 10 y minus 18, y is going to give me minus
28 y. And if I do 20 minus six, that's going
[05:41:44] to give me 14. solving for y, I get y is 14
over minus 28, which is minus one half just
[05:41:52] like before. Now we can continue, like we
did in the previous solution, and substitute
[05:41:59] that value of y into either one of the equations.
I'll put it again in here. And my solution
[05:42:06] proceeds as before. So once again, I get the
solution that x equals one and y equals minus
[05:42:13] one half. Before we go on to the next problem,
let me show you graphically what this means.
[05:42:20] Here I've graphed the equations 3x minus two,
y equals four, and 5x plus six y equals two.
[05:42:29] And we can see that these two lines intersect
in the point with coordinates one negative
[05:42:36] one half, just like we predicted by solving
equations algebraically. Let's take a look
[05:42:42] at another system of equations. I'm going
to rewrite the first equation, so the x term
[05:42:49] is on the left side with the y term, and the
constant term stays on the right. And I'll
[05:42:55] rewrite or copy down the second equation.
Since the coefficient of x in the first equation
[05:43:02] is minus four, and the second equation is
three, I'm going to try using the method of
[05:43:08] elimination and multiply the first equation
by three and the second equation by four.
[05:43:14] That'll give me a coefficient of x of negative
12. And the first equation, and 12. And the
[05:43:19] second equation, those are equal and opposite,
right, so I'll be able to add together my
[05:43:23] equations to cancel out access. So let's do
that. My first equation becomes negative 12x
[05:43:31] plus 24, y equals three, and I'll put everything
by three. And my second equation, I'll multiply
[05:43:39] everything by four. So that's 12x minus 24,
y equals eight. Now something kind of funny
[05:43:46] has happened here, not only do the x coefficients
match has in with opposite signs, but the
[05:43:53] Y coefficients do also. So if I add together
my two equations, in order to cancel out the
[05:44:00] x term, I'm also going to cancel out the y
term, and I'll just get zero plus zero is
[05:44:06] equal to three plus eight is 11. Well, that's
a contradiction, we can't have zero equal
[05:44:12] to 11. And that shows that these two equations
actually have no solution.
[05:44:19] Let's look at this situation graphically.
If we graph our two equations, we see that
[05:44:25] they're parallel lines with the same slope.
This might be more clear, if I isolate y in
[05:44:31] each equation, the first equation, I get y
equals, let's see, dividing by eight that's
[05:44:36] the same thing as four eighths or one half
x plus one eight. And the second equation,
[05:44:43] if I isolate y, let's say minus six y equals
minus 3x plus two divided by minus six, that's
[05:44:50] y equals one half x minus 1/3. So indeed,
they have the same slope. And so they're parallel
[05:44:59] with different In intercepts, and so they
can have no intersection. And so it makes
[05:45:04] sense that we have no solution to our system
of linear equations. This kind of system that
[05:45:10] has no solution is called an inconsistent
system. In this third example, yet a third
[05:45:18] behavior happens. This time, I think I'm going
to solve by substitution because I already
[05:45:24] have X with a coefficient of one. So it's
really easy to just isolate X in the first
[05:45:30] equation, and then plug in to the second equation
to get three times six minus five y plus 15,
[05:45:39] y equals 18. If I distribute out, I get the
strange phenomenon that the 15 y's cancel
[05:45:46] and I just get 18 equals 18, which is always
true. This is what's called a dependent system
[05:45:54] of linear equations. If you look more closely,
you can see that the second equation is really
[05:46:00] just a constant multiple, the first equation
is just three times every term is three times
[05:46:06] as big as the corresponding term and the first
equation. So there's no new information in
[05:46:10] the second equation, anything, any x and y
values that satisfy the first one will satisfy
[05:46:16] the second one. So this system of equations
has infinitely many solutions. Any ordered
[05:46:23] pair x y, where X plus five y equals six,
or in other words, X's minus five y plus six
[05:46:32] will satisfy this system of equations. That
would include a y value of zero corresponding
[05:46:40] x value of six or a y value of one corresponding
to an x value of one, or a y value of 1/3.
[05:46:49] Corresponding to an x value of 13 thirds just
by plugging into this equation will work.
[05:46:55] Graphically, if I graph both of these equations,
the lines will just be on top of each other,
[05:47:00] so I'll just see one line. In this video,
we've solved systems of linear equations,
[05:47:06] using the method of substitution and the method
of elimination. We've seen that systems of
[05:47:13] linear equations can have one solution. When
the lines that the equations represent intersect
[05:47:19] in one point, they can be inconsistent, and
have no solutions that corresponds to parallel
[05:47:26] lines, or they can be dependent and have infinitely
many solutions that corresponds to the lines
[05:47:33] lying on top of each other. In this video,
I'll work through a problem involving distance
[05:47:39] rate and time. The key relationship to keep
in mind is that the rate of travel is the
[05:47:46] distance traveled divided by the time it takes
to travel again. For example, if you're driving
[05:47:51] 60 miles an hour, that's your rate. And that's
because you're going a distance of 60 miles
[05:47:57] in one hour. Sometimes it's handy to rewrite
that relationship by multiplying both sides
[05:48:02] by T time. And that gives us that R times
T is equal to D. In other words, distance
[05:48:10] is equal to rate times time. There's one more
important principle to keep in mind. And that's
[05:48:15] the idea that rates add. For example, if you
normally walk at three miles per hour, but
[05:48:23] you're walking on a moving sidewalk, that's
going at a rate of two miles per hour, then
[05:48:29] your total speed of travel with respect to
you know something stationary is going to
[05:48:34] be three plus two, or five miles per hour.
All right, that is a formula as our one or
[05:48:43] the first rate per second rate is equal to
the total rate. Let's use those two key ideas.
[05:48:51] distance equals rate times time, and rates
add in the following problem. else's boat
[05:48:59] has a top speed of six miles per hour and
still water. While traveling on a river at
[05:49:04] top speed. She went 10 miles upstream in the
same amount of time she went 30 miles downstream,
[05:49:10] we're supposed to find the rate of the river
currents. I'm going to organize the information
[05:49:15] in this problem into a chart.
[05:49:17] During the course of Elsa stay, there were
two situations we need to keep in mind. For
[05:49:23] one period of time she was going upstream.
And for another period of time she was going
[05:49:27] downstream. For each of those, I'm going to
chart out the distance you traveled. The rate
[05:49:35] she went at and the time it took when she
was going upstream. She went a total distance
[05:49:42] of 10 miles. When she was going downstream
she went a longer distance of 30 miles. But
[05:49:49] the times to travel those two distances were
the same. Since I don't know what that time
[05:49:55] was, I'll just give it a variable I'll call
it T now Think about her rate of travel, the
[05:50:02] rate she traveled upstream was slower because
she was going against the current and faster
[05:50:07] when she was going downstream with the current.
We don't know what the speed of the current
[05:50:11] is, that's what we're trying to figure out.
So maybe I'll give it a variable R. But we
[05:50:17] do know that in still water also can go six
miles per hour. And when she's going downstream,
[05:50:23] since she's going with the direction of the
current rates should add, and her rate downstream
[05:50:29] should be six plus R, that's her rate, and
still water plus the rate of the current.
[05:50:40] On the other hand, when she's going upstream,
then she's going against the current, so her
[05:50:46] rate of six miles per hour, we need to subtract
the rate of the current from that. Now that
[05:50:53] we've charted out our information, we can
turn it into equations using the fact that
[05:50:57] distance equals rate times time, we actually
have two equations 10 equals six minus R times
[05:51:04] T, and 30 is equal to six plus R times T.
Now that we've converted our situation into
[05:51:12] a system of equations, our next job is to
solve the system of equations. In this example,
[05:51:17] I think the easiest way to proceed is to isolate
t in each of the two equations. So in the
[05:51:23] first equation, I'll divide both sides by
six minus r, and a second equation R divided
[05:51:28] by six plus R. That gives me 10 over six minus
r equals T, and 30 over six plus r equals
[05:51:36] t. Now if I set my T variables equal to each
other, I get, I get 10 over six minus r is
[05:51:45] equal to 30 over six plus R. I'm making progress
because now I have a single equation, the
[05:51:52] single variable that I need to solve. Since
the variable R is trapped in the denominator,
[05:51:58] I'm going to proceed by clearing the denominator.
So I'm going to multiply both sides by the
[05:52:03] least common denominator, that is six minus
r times six plus R. Once I cancel things out,
[05:52:14] I get that the six plus r times 10 is equal
to 30 times six minus r, if I distribute,
[05:52:23] I'm going to get 60 plus xR equals 180 minus
30 ar, which I can now solve, let's see, that's
[05:52:36] going to be 40 r is equal to 120. So our,
the speed of my current is going to be three
[05:52:43] miles per hour. This is all that the problem
asked for the speed of the current. If I also
[05:52:51] wanted to solve for the other unknown time,
I could do so by plugging in R into one of
[05:52:57] my equations and solving for T. In this video,
I saw the distance rate and time problem by
[05:53:04] charting out my information for the two situations
and my problem using the fact that rates add
[05:53:12] to fill in some of my boxes, and then using
the formula distance equals rate times time
[05:53:17] to build a system of equations. In this video,
I'll do a standard mixture problem in which
[05:53:24] we have to figure out what quantity of two
solutions to mix together. household bleach
[05:53:30] contains 6% sodium hypochlorite. The other
94% is water. How much household bleach should
[05:53:39] be combined with 70 liters of a weaker 1%
hypochlorite solution in order to form a solution,
[05:53:50] that's 2.5% sodium hypochlorite. I want to
turn this problem into a system of equations.
[05:53:59] So I'm asking myself what quantities are going
to be equal to each other? Well, the total
[05:54:05] amount of sodium hypochlorite that has symbol
and a co o before mixing,
[05:54:14] it should equal the total amount of sodium
hypochlorite after mixing. Also, the total
[05:54:23] amount of water before mixing should equal
the total amount of water after. And finally,
[05:54:30] there's just the total amount of solution.
In my two jugs, sodium hypochlorite together
[05:54:37] with water should equal the total amount of
solution after that gives me a hint for what
[05:54:47] I'm looking for. But before I start reading
out equations, I find it very helpful to chart
[05:54:51] out my quantities. So I've got the 6% solution.
The household leech, I've got the 1% solution.
[05:55:05] And I've got my Desired Ending 2.5% solution.
Now on each of those solutions, I've got a
[05:55:16] certain volume of sodium hypochlorite. I've
also got a volume of water. And I've got a
[05:55:26] total volume of solution. Let me see which
of these boxes I can actually fill in, I know
[05:55:37] that I'm adding 70 liters of the 1% solution.
So I can put a 70 in the total volume of solution
[05:55:45] here. I don't know what volume of the household
bleach, I want to add, that's what I'm trying
[05:55:53] to find out. So I'm going to just call that
volume x. Now since my 2.5% solution is made
[05:56:02] by combining my other two solutions, I know
its volume is going to be the sum of these
[05:56:08] two volumes, so I'll write 70 plus x in this
box. Now, the 6% solution means that whatever
[05:56:16] the volume of solution is, 6% of that is the
sodium hypochlorite. So the volume of the
[05:56:22] sodium hypochlorite is going to be 0.06 times
x, the volume of water and that solution is
[05:56:31] whatever's left, so that's going to be x minus
0.06x, or point nine four times X with the
[05:56:40] following the same reasoning for the 1% solution
1% of the 70 liters is a sodium hypochlorite.
[05:56:46] So that's going to be 0.01 times 70. Or point
seven, the volume of water and that solution
[05:56:56] is going to be 99% or point nine, nine times
seven day. That works out to 69.3. Finally,
[05:57:05] for the 2.5% solution, the volume of the sodium
hypochlorite is going to be 0.0 to five times
[05:57:15] the volume of solution 70 plus x and the volume
of water is going to be the remainder. So
[05:57:22] that's 0.975 times 70 plus x. Now I've already
used the fact that the volume of solutions
[05:57:32] before added up is the volume of solution
after in writing a 70 plus x in this box.
[05:57:39] But I haven't yet used the fact that the volume
of the sodium hypochlorite is preserved before
[05:57:43] and after. So I can write that down as an
equation. So that means 0.06x plus 0.7 is
[05:57:53] equal to 0.025 times 70 plus x. Now I've got
an equation with a variable. I'll try to solve
[05:58:01] it. Since I don't like all these decimals,
I'm going to multiply both sides of my equation
[05:58:06] by let's see, 1000 should get rid of all the
decimals. After distributing, I get 60x plus
[05:58:15] 700 equals 25 times 70 plus x. Distributing
some more, I get 60x plus 700 is equal to
[05:58:27] 1750 plus 25x. So let's see 60 minus 25 is
35x is equal to 1050. Which gives me that
[05:58:42] x equals 30 liters of the household bleach.
[05:58:49] Notice that I never actually had to use the
fact that the water quantity of water before
[05:58:54] mixing is equal to the quantity of water after
that is I never use the information in this
[05:59:00] column. In fact, that information is redundant.
Once I know that the quantities of sodium
[05:59:08] hypochlorite add up, and the total volume
of solutions add up. The fact that that volume
[05:59:15] of waters add up is just redundant information.
The techniques to use to solve this equation
[05:59:21] involving solutions can be used to solve many
many other equations involving mixtures of
[05:59:28] items. My favorite method is to first make
a chart involving the types of mixtures and
[05:59:36] the types of items in your mixture. Fill in
as many boxes size I can and then use the
[05:59:43] fact that the quantities add. This video is
about rational functions and their graphs.
[05:59:53] Recall that a rational function is a function
that can be written as a ratio or quotient
[05:59:59] of two power. No Here's an example. The simpler
function, f of x equals one over x is also
[06:00:07] considered a rational function, you can think
of one and x as very simple polynomials. The
[06:00:14] graph of this rational function is shown here.
This graph looks different from the graph
[06:00:20] of a polynomial. For one thing, its end behavior
is different. The end behavior of a function
[06:00:27] is the way the graph looks when x goes through
really large positive, or really large negative
[06:00:33] numbers, we've seen that the end behavior
of a polynomial always looks like one of these
[06:00:38] cases. That is why marches off to infinity
or maybe negative infinity, as x gets really
[06:00:44] big or really negative. But this rational
function has a different type of end behavior.
[06:00:50] Notice, as x gets really big, the y values
are leveling off
[06:00:54] at about a y value of three. And similarly,
as x values get really negative, our graph
[06:01:00] is leveling off near the line y equals three,
I'll draw that line, y equals three on my
[06:01:07] graph, that line is called a horizontal asymptote.
A horizontal asymptote is a horizontal line
[06:01:16] that our graph gets closer and closer to as
x goes to infinity, or as X goes to negative
[06:01:21] infinity, or both. There's something else
that's different about this graph from a polynomial
[06:01:26] graph, look at what happens as x gets close
to negative five. As we approach negative
[06:01:32] five with x values on the right, our Y values
are going down towards negative infinity.
[06:01:37] And as we approach the x value of negative
five from the left, our Y values are going
[06:01:42] up towards positive infinity. We say that
this graph has a vertical asymptote at x equals
[06:01:49] negative five. A vertical asymptote is a vertical
line that the graph gets closer and closer
[06:01:56] to. Finally, there's something really weird
going on at x equals two, there's a little
[06:02:02] open circle there, like the value at x equals
two is dug out. That's called a hole. A hole
[06:02:10] is a place along the curve of the graph where
the function doesn't exist. Now that we've
[06:02:16] identified some of the features of our rational
functions graph,
[06:02:19] I want to look back at the equation and see
how we could have predicted those features
[06:02:24] just by looking at the equation. To find horizontal
asymptotes, we need to look at what our function
[06:02:31] is doing when x goes through really big positive
or really big negative numbers. Looking at
[06:02:36] our equation for our function, the numerator
is going to be dominated by the 3x squared
[06:02:42] term when x is really big, right, because
three times x squared is going to be absolutely
[06:02:47] enormous compared to this negative 12. If
x is a big positive or negative number, in
[06:02:53] the denominator, the denominator will be dominated
by the x squared term. Again, if x is a really
[06:02:59] big positive or negative number, like a million,
a million squared will be much, much bigger
[06:03:04] than three times a million or negative 10.
For that reason, to find the end behavior,
[06:03:10] or the horizontal asymptote, for our function,
we just need to look at the term on the numerator
[06:03:18] and the term on the denominator that have
the highest exponent, those are the ones that
[06:03:22] dominate the expression in size. So as x gets
really big, our functions y values are going
[06:03:29] to be approximately 3x squared over x squared,
which is three. That's why we have a horizontal
[06:03:37] asymptote at y equals three. Now our vertical
asymptotes, those tend to occur where our
[06:03:45] denominator of our function is zero. That's
because the function doesn't exist when our
[06:03:51] denominator is zero. And when we get close
to that place where our denominator is zero,
[06:03:56] we're going to be dividing by tiny, tiny numbers,
which will make our Y values really big in
[06:04:01] magnitude. So to check where our denominators
zero, let's factor our function. In fact,
[06:04:06] I'm going to go ahead and factor the numerator
and the denominator. So the numerator factors,
[06:04:11] let's see, pull out the three, I get x squared
minus four, factor in the denominator, that
[06:04:18] factors into X plus five times x minus two,
I can factor a little the numerator a little
[06:04:25] further, that's three times x minus two times
x plus two over x plus 5x minus two. Now,
[06:04:34] when x is equal to negative five, my denominator
will be zero, but my numerator will not be
[06:04:41] zero. That's what gives me the vertical asymptote
at x equals negative five. Notice that when
[06:04:49] x equals two, the denominator is zero, but
the numerator is also zero. In fact, if I
[06:04:57] cancelled the x minus two factor from the
numerator, and in nominator, I get a simplified
[06:05:01] form for my function that agrees with my original
function as long as x is not equal to two.
[06:05:11] That's because when x equals two, the simplified
function exists, but the original function
[06:05:16] does not, it's zero over zero, it's undefined.
But for every other x value, including x values
[06:05:23] near x equals to our original functions just
the same as this function. And that's why
[06:05:29] our function only has a vertical asymptote
at x equals negative five, not one at x equals
[06:05:36] two, because the x minus two factor is no
longer in the function after simplifying,
[06:05:41] it does have a hole at x equals two, because
the original function is not defined there,
[06:05:45] even though the simplified version is if we
want to find the y value of our hole, we can
[06:05:52] just plug in x equals two into our simplified
version of our function, that gives a y value
[06:05:59] of three times two plus two over two plus
seven, or 12 ninths, which simplifies to four
[06:06:06] thirds. So our whole is that to four thirds.
Now that we've been through one example in
[06:06:15] detail, let's summarize our findings. We find
the vertical asymptotes and the holes by looking
[06:06:22] where the denominator is zero. The holes happen
where the denominator and numerator are both
[06:06:29] zero and those factors cancel out. The vertical
asymptotes are all other x values where the
[06:06:35] denominator is zero, we find the horizontal
asymptotes. By considering the highest power
[06:06:42] term on the numerator and the denominator,
I'll explain this process in more detail in
[06:06:47] three examples. In the first example, if we
circle the highest power terms, that simplifies
[06:06:55] to 5x over 3x squared, which is five over
3x. As x gets really big, the denominator
[06:07:03] is going to be huge. So I'm going to be dividing
five by a huge, huge number, that's going
[06:07:09] to be going very close to zero. And therefore
we have a horizontal asymptote
[06:07:15] at y equals zero. In the second example, the
highest power terms, 2x cubed, over 3x cubed
[06:07:24] simplifies to two thirds. So as x gets really
big, we're going to be heading towards two
[06:07:30] thirds, and we have a horizontal asymptote
at y equals two thirds. In the third example,
[06:07:36] the highest power terms, x squared over 2x
simplifies to x over two. As x gets really
[06:07:45] big, x over two is getting really big. And
therefore, we don't have a horizontal asymptote
[06:07:51] at all. This is going to infinity, when x
gets through go through big positive numbers,
[06:07:59] and is going to negative infinity when x goes
through a big negative numbers. So in this
[06:08:07] case, the end behavior is kind of like that
of a polynomial, and there's no horizontal
[06:08:12] asymptote. In general, when the degree of
the numerator is smaller than the degree of
[06:08:18] the denominator, we're in this first case
where the denominator gets really big compared
[06:08:23] to the numerator and we go to zero. In the
second case, where the degree of the numerator
[06:08:29] and the degree of the dominant are equal,
things cancel out, and so we get a horizontal
[06:08:34] asymptote at the y value, that's equal to
the ratio of the leading coefficients. Finally,
[06:08:42] in the third case, when the degree of the
numerator is bigger than the degree of the
[06:08:46] denominator, then the numerator is getting
really big compared to the denominator, so
[06:08:52] we end up with no horizontal asymptote. Final
Finally, let's apply all these observations
[06:08:57] to one more example. Please pause the video
and take a moment to find the vertical asymptotes,
[06:09:03] horizontal asymptotes and holes for this rational
function. To find the vertical asymptotes
[06:09:09] and holes, we need to look at where the denominator
is zero. In fact, it's going to be handy to
[06:09:15] factor both the numerator and the denominator.
Since there if there are any common factors,
[06:09:20] we might have a whole instead of a vertical
asymptote. The numerator is pretty easy to
[06:09:25] factor. Let's see that's 3x times x plus one
for the denominator or first factor out an
[06:09:31] x. And then I'll factor some more using a
guess and check method. I know that I'll need
[06:09:40] a 2x and an X to multiply together to the
2x squared and I'll need a three and a minus
[06:09:47] one or alpha minus three and a one. Let's
see if that works. If I multiply out 2x minus
[06:09:56] one times x plus three that does get me back
to x squared plus 5x minus three, so that
[06:10:02] checks out. Now I noticed that I have a common
factor of x in both the numerator and the
[06:10:08] denominator. So that's telling me I'm going
to have a hole at x equals zero. In fact,
[06:10:14] I could rewrite my rational function by cancelling
out that common factor. And that's equivalent,
[06:10:21] as long as x is not equal to zero. So the
y value of my whole is what I get when I plug
[06:10:28] zero into my simplified version, that would
be three times zero plus one over two times
[06:10:35] zero minus one times zero plus three, which
is three over negative three or minus one.
[06:10:42] So my whole is at zero minus one. Now all
the remaining places in my denominator that
[06:10:49] make my denominator zero will get me vertical
asymptotes. So I'll have a vertical asymptote,
[06:10:55] when 2x minus one times x plus three equals
zero, that is, when 2x minus one is zero,
[06:11:04] or x plus three is zero. In other words, when
x is one half, or x equals negative three.
[06:11:13] Finally, to find my horizontal asymptotes,
I just need to consider the highest power
[06:11:20] term in the numerator and the denominator.
That simplifies to three over 2x, which is
[06:11:28] bottom heavy, right? When x gets really big,
this expression is going to zero. And that
[06:11:35] means that we have a horizontal asymptote
at y equals zero. So we found the major features
[06:11:41] of our graph, the whole, the vertical asymptotes
and the horizontal asymptotes. Together, this
[06:11:48] would give us a framework for what the graph
of our function looks like. horizontal asymptote
[06:11:54] at y equals zero, vertical asymptotes at x
equals one half, and x equals minus three
[06:12:02] at a hole at the point zero minus one. plotting
a few more points, or using a graphing calculator
[06:12:09] of graphing program, we can see that our actual
function will look something like this.
[06:12:17] Notice that the x intercept when x is negative
one, corresponds to where the numerator of
[06:12:24] our rational function or reduced rational
function is equal to zero. That's because
[06:12:29] a zero on the numerator that doesn't make
the denominator zero makes the whole function
[06:12:33] zero. And an X intercept is where the y value
of the whole function is zero. In this video,
[06:12:40] we learned how to find horizontal asymptotes
have rational functions. By looking at the
[06:12:44] highest power terms, we learned to find the
vertical asymptotes and holes. By looking
[06:12:50] at the factored version of the functions.
The holes correspond to the x values that
[06:12:56] make the numerator and denominator zero, his
corresponding factors cancel. The vertical
[06:13:03] asymptotes correspond to the x values that
make the denominator zero, even after factoring
[06:13:09] any any common and in common factors in the
numerator denominator.
[06:13:13] This video is about combining functions by
adding them subtracting them multiplying and
[06:13:19] dividing them. Suppose we have two functions,
f of x equals x plus one and g of x equals
[06:13:26] x squared. One way to combine them is by adding
them together. This notation, f plus g of
[06:13:34] x means the function defined by taking f of
x and adding it to g of x. So for our functions,
[06:13:43] that means we take x plus one and add x squared,
I can rearrange that as the function x squared
[06:13:50] plus x plus one. So f plus g evaluated on
x means x squared plus x plus one. And if
[06:14:01] I wanted to evaluate f plus g, on the number
two, that would be two squared plus two plus
[06:14:08] one, or seven. Similarly, the notation f minus
g of x means the function we get by taking
[06:14:18] f of x and subtracting g of x. So that would
be x plus one minus x squared. And if I wanted
[06:14:25] to take f minus g evaluated at one, that would
be one plus one minus one squared, or one,
[06:14:35] the notation F dot g of x, which is sometimes
also written just as f g of x. That means
[06:14:44] we take f of x times g of x. In other words,
x plus one times x squared, which could be
[06:14:52] simplified as x cubed plus x squared. The
notation f divided by g of x means I take
[06:15:02] f of x and divided by g of x. So that would
be x plus one divided by x squared. In this
[06:15:11] figure, the blue graph represents h of x.
And the red graph represents the function
[06:15:19] p of x, we're asked to find h minus p of zero.
[06:15:26] We don't have any equations to work with,
but that's okay. We know that for any x, h
[06:15:32] minus p of x is defined as h of x minus p
of x. So for x equals zero, H minus p of zero
[06:15:40] is going to be h of zero minus p of zero.
Using the graph, we can find h of zero by
[06:15:48] finding the value of zero on the x axis, and
finding the corresponding y value for the
[06:15:55] function h of x. So that's about 1.8. Now
P of zero, we can find similarly by looking
[06:16:04] for zero on the x axis, and finding the corresponding
y value for the function p of x, and that's
[06:16:10] a y value of one 1.8 minus one is 0.8. So
that's our approximate value for H minus p
[06:16:18] of zero. If we want to find P times h of negative
three, again, we can rewrite that as P of
[06:16:28] negative three times h of negative three.
And using the graphs, we see that for an x
[06:16:34] value of negative three, the y value for P
is two. And the x value of negative three
[06:16:45] corresponds to a y value of negative two for
H.
[06:16:50] Two times negative two is negative four. So
that's our value for P times h of negative
[06:16:55] three.
[06:16:56] In this video, we saw how to add two functions,
subtract two functions, multiply two functions,
[06:17:03] and divide two functions in the following
way. When you compose two functions, you apply
[06:17:13] the first function, and then you apply the
second function to the output of the first
[06:17:19] function. For example, the first function
might compute population size from time in
[06:17:27] years. So its input would be time in years,
since a certain date, as output would be number
[06:17:35] of people in the population. The second function
g, might compute health care costs as a function
[06:17:45] of population size. So it will take population
size as input, and its output will be healthcare
[06:17:53] costs. If you put these functions together,
that is compose them, then you'll go all the
[06:18:00] way from time in years to healthcare costs.
This is your composition, g composed with
[06:18:07] F. The composition of two functions, written
g with a little circle, f of x is defined
[06:18:16] as follows. g composed with f of x is G evaluated
on f of x, we can think of it schematically
[06:18:26] and diagram f x on a number x and produces
a number f of x, then g takes that output
[06:18:36] f of x and produces a new number, g of f of
x. Our composition of functions g composed
[06:18:45] with F is the function that goes all the way
from X to g of f of x. Let's work out some
[06:18:51] examples where our functions are defined by
tables of values. If we want to find g composed
[06:18:58] with F of four, by definition, this means
g of f of four. To evaluate this expression,
[06:19:08] we always work from the inside out. So we
start with the x value of four, and we find
[06:19:15] f of four, using the table of values for f
of x, when x equals four, f of x is seven,
[06:19:23] so we can replace F of four with the number
seven. Now we need to evaluate g of seven,
[06:19:33] seven becomes our new x value in our table
of values for G, the x value of seven corresponds
[06:19:39] to the G of X value of 10. So g of seven is
equal to 10. We found that g composed with
[06:19:47] F of four is equal to 10. If instead we want
to find f composed with g of four, well, we
[06:19:56] can rewrite that as f of g of four Again work
from the inside out. Now we're trying to find
[06:20:04] g of four. So four is our x value. And we
use our table of values for G to see that
[06:20:10] g of four is one. So we replaced you a four
by one. And now we need to evaluate f of one.
[06:20:20] Using our table for F values, f of one is
eight. Notice that when we've computed g of
[06:20:29] f of four, we got a different answer than
when we computed f of g of four. And in general,
[06:20:36] g composed with F is not the same thing as
f composed with g. Please pause the video
[06:20:43] and take a moment to compute the next two
examples. We can replace f composed with F
[06:20:49] of two by the equivalent expression, f of
f of two. Working from the inside out, we
[06:20:56] know that f of two is three, and f of three
is six. If we want to find f composed with
[06:21:06] g of six, rewrite that as f of g of six, using
the table for g, g of six is eight. But F
[06:21:17] of eight, eight is not on the table as an
x value for the for the f function. And so
[06:21:25] there is no F of eight, this does not exist,
we can say that six is not in the domain,
[06:21:35] for F composed with g. Even though it was
in the domain of g, we couldn't follow all
[06:21:42] the way through and get a value for F composed
with g of six. Next, let's turn our attention
[06:21:49] to the composition of functions that are given
by equations.
[06:21:54] 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.
[06:22:03] As usual, I can rewrite this as Q of P of
one and work from the inside out. P of one
[06:22:14] is one squared plus one, so that's two. So
this is the same thing as Q of two. But Q
[06:22:21] of two is negative two times two or negative
four. So this evaluates to negative four.
[06:22:28] In this next example, we want to find q composed
with P of some arbitrary x, or rewrite it
[06:22:35] as usual as Q of p of x and work from the
inside out. Well, p of x, we know the formula
[06:22:44] for that. That's x squared plus x. So I can
replace my P of x with that expression. Now,
[06:22:52] I'm stuck with evaluating q on x squared plus
x. Well, Q of anything is negative two times
[06:22:59] that thing. So q of x squared plus x is going
to be negative two times the quantity x squared
[06:23:08] plus x, what I've done is I've substituted
in the whole expression x squared plus x,
[06:23:13] where I saw the X in this formula for q of
x, it's important to use the parentheses here.
[06:23:20] So that will be multiplying negative two by
the whole expression and not just by the first
[06:23:24] piece, I can simplify this a bit as negative
2x squared minus 2x. And that's my expression
[06:23:32] for Q composed with p of x. Notice that if
I wanted to compute q composed with P of one,
[06:23:40] which I already did in the first problem,
I could just use this expression now, negative
[06:23:45] two times one squared minus two and I get
negative four, just like I did before. Let's
[06:23:52] try another one. Let's try p composed with
q of x. First I read write this P of q of
[06:24:00] x. Working from the inside out, I can replace
q of x with negative 2x. So I need to compute
[06:24:07] P of negative 2x. Here's my formula for P.
to compute P of this expression, I need to
[06:24:15] plug in this expression everywhere I see an
x in the formula for P. So that means negative
[06:24:23] 2x squared plus negative 2x. Again, being
careful to use parentheses to make sure I
[06:24:29] plug in the entire expression in forex. let
me simplify. This is 4x squared minus 2x.
[06:24:40] Notice that I got different expressions for
Q of p of x, and for P of q of x. Once again,
[06:24:50] we see that q composed with P is not necessarily
equal to P composed with Q. Please pause the
[06:24:57] video and try this last example yourself.
rewriting. And working from the inside out,
[06:25:06] we're going to replace p of x with its expression
x squared plus x. And then we need to evaluate
[06:25:13] p on x squared plus x. That means we plug
in x squared plus x, everywhere we see an
[06:25:22] x in this formula, so that's x squared plus
x quantity squared plus x squared plus x.
[06:25:29] Once again, I can simplify by distributing
out, that gives me x to the fourth plus 2x
[06:25:36] cubed plus x squared plus x squared plus x,
or x to the fourth plus 2x cubed plus 2x squared
[06:25:45] plus x. In this last set of examples, we're
asked to go backwards, we're given a formula
[06:25:52] for a function of h of x. But we're supposed
to rewrite h of x as a composition of two
[06:25:57] functions, F and G. Let's think for a minute,
which of these two functions gets applied
[06:26:04] first, f composed with g of x, let's see,
that means f of g of x. And since we evaluate
[06:26:13] these expressions from the inside out, we
must be applying g first, and then F. In order
[06:26:21] to figure out what what f and g could be,
I like to draw a box around some thing inside
[06:26:27] my expression for H, so I'm going to draw
a box around x squared plus seven, then whatever's
[06:26:32] inside the box, that'll be my function, g
of x, the first function that gets applied,
[06:26:38] whatever happens to the box, in this case,
taking the square root sign, that becomes
[06:26:43] my outside function, my second function f.
[06:26:46] So here, we're gonna say g of x is equal to
x squared plus seven, and f of x is equal
[06:26:54] to the square root of x, let's just check
and make sure that this works. So I need to
[06:27:00] check that when I take the composition, f
composed with g, I need to get the same thing
[06:27:08] as my original h. So let's see, if I do f
composed with g of x, well, by definition,
[06:27:17] that's f of g of x, working from the inside
out, I can replace g of x with its formula
[06:27:23] x squared plus seven. So I need to evaluate
f of x squared plus seven. That means I plug
[06:27:30] in x squared plus seven, into the formula
for for F. So that becomes the square root
[06:27:37] of x squared plus seven to the it works because
it matches my original equation. So we found
[06:27:44] a correct answer a correct way of breaking
h down as a composition of two functions.
[06:27:48] But I do want to point out, this is not the
only correct answer. I'll write down my formula
[06:27:54] for H of X again, and this time, I'll put
the box in a different place, I'll just box
[06:27:59] the x squared. If I did that, then my inside
function of my first function, g of x would
[06:28:08] be x squared. And my second function is what
happens to the box. So my f of x is what happens
[06:28:17] to the box, and the box gets added seven to
it, and taking the square root. So in other
[06:28:23] words, f of x is going to be the square root
of x plus seven. Again, I can check that this
[06:28:32] 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
[06:28:39] I'm taking f of x squared. When I plug in
x squared for x, I do in fact, get the square
[06:28:46] root of x squared plus seven. So this is that
alternative, correct solution. In this video,
[06:28:53] we learn to evaluate the composition of functions.
by rewriting it and working from the inside
[06:29:00] out. We also learn to break apart a complicated
function into a composition of two functions
[06:29:10] by boxing one piece of the function and letting
the first function applied in the composition.
[06:29:16] Let that be the inside of the box, and the
second function applied in the composition
[06:29:21] be whatever happens to the box.
[06:29:33] The inverse of a function undoes what the
function does. So the inverse of tying your
[06:29:38] shoes would be to untie them. And the inverse
of the function that adds two to a number
[06:29:46] would be the function that subtracts two from
a number. This video introduces inverses and
[06:29:53] their properties. Suppose f of x is a function
defined by this chart. In other words, Have
[06:30:00] two is three, f of three is five, f of four
is six, and f of five is one, the inverse
[06:30:08] function for F written f superscript. Negative
1x undoes what f does. Since f takes two to
[06:30:17] three, F inverse takes three, back to two.
So we write this f superscript, negative one
[06:30:26] of three is to. Similarly, since f takes three
to five, F inverse takes five to three. And
[06:30:38] since f takes four to six, f inverse of six
is four. And since f takes five to one, f
[06:30:46] inverse of one is five. I'll use these numbers
to fill in the chart. Notice that the chart
[06:30:54] of values when y equals f of x and the chart
of values when y equals f inverse of x are
[06:31:00] closely related. They share the same numbers,
but the x values for f of x correspond to
[06:31:07] the y values for f inverse of x, and the y
values for f of x correspond to the x values
[06:31:14] for f inverse of x. That leads us to the first
key fact inverse functions reverse the roles
[06:31:21] of y and x. I'm going to plot the points for
y equals f of x in blue. Next, I'll plot the
[06:31:29] points for y equals f inverse of x in red.
Pause the video for a moment and see what
[06:31:35] kind of symmetry you observe in this graph.
How are the blue points related to the red
[06:31:40] points, you might have noticed that the blue
points and the red points are mirror images
[06:31:46] over the mirror line, y equals x. So our second
key fact is that the graph of y equals f inverse
[06:31:55] of x can be obtained from the graph of y equals
f
[06:31:58] of x
[06:31:59] by reflecting over the line y equals x. This
makes sense, because inverses, reverse the
[06:32:06] roles of war annex. In the same example, let's
compute f inverse of f of two, this open circle
[06:32:15] means composition. In other words, we're computing
f inverse of f of two, we compute this from
[06:32:23] the inside out. So that's f inverse of three.
Since F of two is three, and f inverse of
[06:32:33] three, we see as to similarly, we can compute
f of f inverse of three. And that means we
[06:32:44] take f of f inverse of three. Since f inverse
of three is two, that's the same thing as
[06:32:52] computing F of two, which is three. Please
pause the video for a moment and compute these
[06:33:02] other compositions. You should have found
that in every case, if you take f inverse
[06:33:11] of f of a number, you get back to the very
same number you started with. And similarly,
[06:33:16] if you take f of f inverse of any number,
you get back to the same number you started
[06:33:20] with. So in general, f inverse of f of x is
equal to x, and f of f inverse of x is also
[06:33:29] equal to x. This is the mathematical way of
saying that F and n f inverse undo each other.
[06:33:37] Let's look at a different example. Suppose
that f of x is x cubed. Pause the video for
[06:33:43] a moment, and guess what the inverse of f
should be. Remember, F inverse undoes the
[06:33:48] work that F does. You might have guessed that
f inverse of x is going to be the cube root
[06:33:57] function, we can check that this is true by
looking at f of f inverse of x, that's F of
[06:34:05] the cube root of function, which means the
cube root function
[06:34:09] cubed, which gets us back to x. Similarly,
if we compute f inverse of f of x, that's
[06:34:16] the cube root of x cubed.
[06:34:19] And we get back to excellence again. So the
cube root function really is the inverse of
[06:34:25] the cubing function. When we compose the two
functions, we get back to the number that
[06:34:30] we started with. It'd be nice to have a more
systematic way of finding inverses of functions
[06:34:37] besides guessing and checking. One method
uses the fact that inverses reverse the roles
[06:34:44] of y and x. So if we want to find the inverse
of the function, f of x equals five minus
[06:34:50] x over 3x. We can write it as y equals five
minus x over 3x. Reverse the roles of y and
[06:34:59] x To get x equals five minus y over three
y, and then solve for y. To solve for y, let's
[06:35:09] multiply both sides by three y. Bring all
terms with y's in them to the left side, and
[06:35:19] alternate without wizened them to the right
side, factor out the why. And divide to isolate
[06:35:28] why this gives us f inverse of x as five over
3x plus one. Notice that our original function
[06:35:40] f and our inverse function, f inverse are
both rational functions, but they're not the
[06:35:46] reciprocals of each other. And in general,
f inverse of x is not usually equal to one
[06:35:53] over f of x. This can be confusing, because
when we write two to the minus one, that does
[06:36:01] mean one of our two, but f to the minus one
of x means the inverse function and not the
[06:36:08] reciprocal. It's natural to ask us all functions
have inverse functions. That is for any function
[06:36:16] you might encounter. Is there always a function
that its is its inverse? In fact, the answer
[06:36:23] is no. See, if you can come up with an example
of a function that does not have an inverse
[06:36:30] function. The word function here is key. Remember
that a function is a relationship between
[06:36:37] x values and y values, such that for each
x value in the domain, there's only one corresponding
[06:36:47] y value. One example of a function that does
not have an inverse function is the function
[06:36:57] f of x equals x squared.
[06:37:01] To see that,
[06:37:02] the inverse of this function is not a function.
Note that for the x squared function, the
[06:37:09] number two and the number negative two, both
go to number four. So if I had an inverse,
[06:37:17] you would have to send four to both two and
negative two, the inverse would not be a function,
[06:37:27] it might be easier to understand the problem,
when you look at a graph of y equals x squared.
[06:37:34] Recall that inverse functions reverse the
roles of y and x and flip the graph over the
[06:37:40] line y equals x. But when I flipped the green
graph over the line y equals x, I get this
[06:37:47] red graph. This red graph is not the graph
of a function because it violates the vertical
[06:37:53] line test. The reason that violates the vertical
line test is because the original green function
[06:37:59] violates the horizontal line test, and has
2x values with the same y value. In general,
[06:38:10] a function f has an inverse function if and
only if the graph of f satisfies the horizontal
[06:38:14] line test, ie every horizontal line intersects
the graph. In it most one point, pause the
[06:38:21] video for a moment and see which of these
four graphs satisfy the horizontal line test.
[06:38:26] In other words, which of the four corresponding
functions would have an inverse function?
[06:38:34] You may have found that graphs A and B violate
the horizontal line test. So their functions
[06:38:42] would not have inverse functions. But graph
C and D satisfy the horizontal line test.
[06:38:48] So these graphs represent functions that do
have inverses. functions that satisfy the
[06:38:54] horizontal line test are sometimes called
One to One functions. Equivalently a function
[06:39:02] is one to one, if for any two different x
values, x one and x two, the y value is f
[06:39:10] of x one and f of x two are different numbers.
Sometimes, as I said, f is one to one, if,
[06:39:18] whenever f of x one is equal to f of x two,
then x one has to equal x two. As our last
[06:39:27] example, let's try to find P inverse of x,
where p of x is the square root of x minus
[06:39:32] two drawn here. If we graph P inverse on the
same axis as p of x, we get the following
[06:39:40] graph simply by flipping over the line y equals
x. If we try to solve the problem algebraically
[06:39:49] we can write y equal to a squared of x minus
two, reverse the roles of y and x and solve
[06:39:56] for y by squaring both sides adding two. Now
if we were to graph y equals x squared plus
[06:40:07] two, that would look like a parabola, it would
look like the red graph we've already drawn
[06:40:13] together with another arm on the left side.
But we know that our actual inverse function
[06:40:22] consists only of this right arm, we can specify
this algebraically by making the restriction
[06:40:31] that x has to be bigger than or equal to zero.
This corresponds to the fact that on the original
[06:40:40] graph for the square root of x, y was only
greater than or equal to zero. Looking more
[06:40:47] closely at the domain and range of P and P
inverse, we know that the domain of P is all
[06:40:55] values of x such that x minus two is greater
than or equal to zero. Since we can't take
[06:41:01] the square root of a negative number. This
corresponds to x values being greater than
[06:41:07] or equal to two, or an interval notation,
the interval from two to infinity. The range
[06:41:14] of P, we can see from the graph is all y value
is greater than or equal to zero, or the interval
[06:41:20] from zero to infinity. Similarly, based on
the graph, we see the domain of P inverse
[06:41:29] is x values greater than or equal to zero,
the interval from zero to infinity. And the
[06:41:34] range of P inverse is Y values greater than
or equal to two, or the interval from two
[06:41:40] to infinity. If you look closely at these
domains and ranges, you'll notice that the
[06:41:45] domain of P corresponds exactly to the range
of P inverse, and the range of P corresponds
[06:41:53] to the domain of P inverse.
[06:41:57] This makes sense, because inverse functions
reverse the roles of y and x. The domain of
[06:42:04] f inverse of x is the x values for F inverse,
which corresponds to the y values or the range
[06:42:10] of F. The range of f inverse is the y values
for F inverse, which correspond to the x values
[06:42:18] or the domain of f. In this video, we discussed
five key properties of inverse functions.
[06:42:28] inverse functions, reverse the roles of y
and x. The graph of y equals f inverse of
[06:42:36] x is the graph of y equals f of x reflected
over the line y equals x. When we compose
[06:42:47] F with F inverse, we get the identity function
y equals x. And similarly, when we compose
[06:42:56] f inverse with F, that brings x to x. In other
words, F and F inverse undo each other. The
[06:43:04] function f of x has an inverse function if
and only if the graph of y equals f of x satisfies
[06:43:18] the horizontal line test. And finally, the
domain of f is the range of f inverse and
[06:43:31] the range of f is the domain of f inverse.
These properties of inverse functions will
[06:43:39] be important when we study exponential functions
and their inverses logarithmic functions.