College Algebra - Full Course — Full Transcript

2:01

of times as multiplied total.

2:04

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

2:10

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

2:17

same base,

2:19

then I can subtract their exponents.

2:23

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

2:31

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

2:37

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

2:45

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

2:50

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

2:54

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

3:01

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

3:08

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

3:17

I get to multiply the exponents.

3:21

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

3:29

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

3:34

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

3:40

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

3:49

again.

3:51

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

4:03

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

4:09

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

4:14

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

4:20

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

4:27

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

4:35

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

4:40

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

4:47

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

4:53

it work with the quotient rule.

4:55

And the same argument shows that anything to the zero power

5:00

has to be equal to one.

5:02

What happens when we take something to a negative power,

5:07

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

5:15

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

5:20

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

5:26

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

5:33

and we just said that that is equal to one.

5:35

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

5:41

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

5:47

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

5:52

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

5:58

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

6:03

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

6:14

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

6:20

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

6:26

Now, the square root of nine is just three.

6:29

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

6:39

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

6:46

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

6:55

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

7:01

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

7:07

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

7:12

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

7:18

to x to the n times y to the N.

7:22

For example, five times seven,

7:27

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

7:33

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

7:39

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

7:46

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

7:51

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

7:56

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

8:02

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

8:09

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

8:15

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

8:23

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

8:31

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

8:37

fifth, over seven to the fifth, as wanted.

8:43

We've seen that we can distribute an exponent

8:47

over multiplication, and division.

8:51

But be careful, because we cannot distribute next bowknot.

8:56

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

9:06

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

9:14

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

9:19

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

9:28

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

9:36

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

9:42

rule,

9:43

the quotient rule, the power rule,

9:46

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

9:55

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

1:00:07

is 30, we know that two times three will divide it, and three times five will also divide

1:00:14

it. And we won't be able to get a denominator any smaller, because we need the factors two,

1:00:19

three, and five, in order to ensure both these numbers divided. Once we have our least common

1:00:24

denominator, we can rewrite each of our fractions in terms of that denominator. So seven, six,

1:00:31

I need to get a 30 in the denominator, so I'm going to multiply that by five over five,

1:00:38

and multiply by the factors that are missing from the current denominator in order to get

1:00:45

my least common denominator of 30. For the second fraction, for 15th 15 times two is

1:00:54

30. So I'm going to multiply by two over two,

1:00:57

I can rewrite this as 3530 s minus 8/30. And now that I have a common denominator, I can

1:01:08

just subtract my two numerators. And I get 27/30.

1:01:15

If I factor, I can reduce this

1:01:20

to three squared over two times five, which is nine tenths. The process for finding the

1:01:28

sum of two rational expressions with variables in them follows the exact same process. First,

1:01:33

we have to find the least common denominator, I'll do that by factoring the two denominators.

1:01:39

So 2x plus two factors as two times x plus 1x squared minus one, that's a difference

1:01:45

of two squares. So that's x plus one times x minus one. Now for the least common denominator,

1:01:51

I'm going to take all the factors, I need to get an expression that each of these divides

1:01:57

into, so I need the factor two, I need the factor x plus one, and I need the factor x

1:02:02

minus one, I don't have to repeat the factor x plus one I just need to have at one time.

1:02:08

And so I will get my least common denominator two times x plus one times x minus one, I'm

1:02:14

not going to bother multiplying this out, it's actually better to leave it in factored

1:02:17

form to help me simplify later. Now I can rewrite each of my two rational expressions

1:02:24

by multiplying by whatever's missing from the denominator in terms of the least common

1:02:29

denominator. So what I mean is, I can rewrite three over 2x plus two, I'll write the 2x

1:02:36

plus two is two times x plus one, I'll write it in factored form. And then I noticed that

1:02:41

compared to the least common denominator, I'm missing the factor of x minus one. So

1:02:45

I multiply the numerator and the denominator by x minus one, I just need it in the denominator.

1:02:52

But I can't get away with just multiplying by the denominator without changing my expression,

1:02:56

I have to multiply by it on the numerator and the denominator. So I'm just multiplying

1:03:00

by one and a fancy form and not changing the value here. So now I do the same thing for

1:03:06

the second rational expression, I'll I'll write the denominator in factored form to

1:03:10

make it easier to see what's missing from the denominator. What's missing in this denominator,

1:03:16

compared to my least common denominator is just the factor two, so multiply the numerator

1:03:21

and the denominator by two. Now I can rewrite everything. So the first rational expression

1:03:27

becomes three times x minus one over two times x plus 1x minus one, and the second one becomes

1:03:36

five times two over two times x plus 1x minus one, notice that I now have a common denominator.

1:03:44

So I can just add together my numerators. So I get three times x minus one plus 10 over

1:03:51

two times x plus 1x minus one. I'd like to simplify this. And the best way to do that

1:03:58

is to leave the denominator in factored form. But I do have to multiply out the numerator

1:04:04

so that I can add things together. So I get 3x minus three plus 10 over two times x plus

1:04:12

1x minus one, or 3x plus seven over two times x plus 1x minus one. Now 3x plus seven doesn't

1:04:22

factor. And there's therefore no factors that I can cancel out. So this is already reduced.

1:04:29

As much as it can be. This is my final answer.

1:04:33

In this video, we saw how to simplify rational expressions to lowest terms by factoring and

1:04:40

canceling common factors.

1:04:42

We also saw how to multiply rational expressions by multiplying the numerator and multiplying

1:04:47

the denominator, how to divide rational expressions by flipping and multiplying and how to add

1:04:53

and subtract rational expressions by writing them in terms of the least common denominator

1:05:00

This video is about solving quadratic equations. a quadratic equation is an equation that contains

1:05:07

the square of the variable, say x squared, but no higher powers of x.

1:05:11

The standard form for a quadratic equation is the form a x squared plus b x plus c equals

1:05:20

zero, where A, B and C

1:05:25

represent real numbers. And a is not zero so that we actually have an x squared term.

1:05:31

Let me give you an example. 3x squared plus 7x minus two equals zero is a quadratic equation

1:05:40

in standard form, here a is three, B is seven, and C is minus two. The equation 3x squared

1:05:49

equals minus 7x plus two is also a quadratic equation, it's just not in standard form.

1:05:56

The key steps to solving quadratic equations are usually to write the equation in standard

1:06:02

form, and then either factor it

1:06:07

or use the quadratic formula, which I'll show you later in this video.

1:06:14

Let's start with the example y squared equals 18 minus seven y. Here our variable is y,

1:06:21

and we need to rewrite this quadratic equation in standard form, we can do this by subtracting

1:06:26

18 from both sides and adding seven y to both sides. That gives us the equation y squared

1:06:33

minus 18 plus seven y equals zero. And I can rearrange a little bit to get y squared plus

1:06:40

seven y minus 18 equals zero. Now I've got my equation in standard form. Next, I'm going

1:06:48

to try to factor it. So I need to look for two numbers that multiply to negative 18 and

1:06:54

add to seven.

1:06:57

two numbers that work are nine and negative two, so I can factor my expression on the

1:07:02

left as y plus nine times y minus two equals zero. Now, anytime you have two quantities

1:07:12

that multiply together to give you zero, either the first quantity has to be zero, or the

1:07:17

second quantity has to be zero, or I suppose they both could be zero. In some situations,

1:07:22

this is really handy, because that means that I know that either y plus nine equals zero,

1:07:27

or y minus two equals zero. So I can as my next step, set my factors equal to zero. So

1:07:36

y plus nine equals zero, or y minus two equals zero, which means that y equals negative nine,

1:07:43

or y equals two.

1:07:46

It's not a bad idea to check that those answers actually work by plugging them into the original

1:07:51

equation, negative nine squared, does that equal 18 minus seven times negative nine,

1:07:56

and you can work out that it does. And similarly, two squared equals 18 minus seven times two.

1:08:04

In the next example, let's find solutions of the equation w squared equals 121. This

1:08:11

is a quadratic equation, because it's got a square of my variable W, I can rewrite it

1:08:17

in standard form by subtracting 121 from both sides.

1:08:24

Notice that A is equal to one b is equal to zero because there's no w term, and c is equal

1:08:31

to negative 121 in the standard form, a W squared plus BW plus c equals zero. Next step,

1:08:39

I'm going to try to factor this expression. So since 121, is 11 squared, this is a difference

1:08:47

of two squares, and it factors as w plus 11, times w minus 11 is equal to zero. If I set

1:08:56

the factors equal to zero,

1:08:59

I get w plus 11 equals zero or w minus 11 equals zero. So w equals minus 11, or w equals

1:09:07

11. In this example, I could have solved the equation more simply, I could have instead

1:09:14

said that if W squared is 121, and then w has to equal plus or minus the square root

1:09:20

of 121. In other words, W is plus or minus 11.

1:09:26

If you saw the equation this way, it's important to remember the plus or minus since minus

1:09:30

11 squared equals 121, just like 11 squared does.

1:09:36

Now let's find the solutions for the equation. x times x plus two equals seven. Some people

1:09:41

might be tempted to say that, oh, if two numbers multiply to equal seven, then one of them

1:09:46

better equal one and the other equals seven or maybe negative one and negative seven.

1:09:51

But that's faulty reasoning in this case, because x and x plus two don't have to be

1:09:57

whole numbers. They could be crazy.

1:10:00

fractions or even irrational numbers. So instead, let's rewrite this equation in standard form.

1:10:08

To do that, I'm first going to multiply out. So x times x is x squared x times two is 2x.

1:10:16

That equals seven, and I'll subtract the seven from both sides to get x squared plus 2x minus

1:10:23

seven is zero. Now I'm looking to factor it. So I need two numbers that multiply to negative

1:10:28

seven and add to two, since the only way to factor negative seven is as negative one times

1:10:34

seven or seven times negative one, it's easy to see that there are no whole numbers that

1:10:39

will do will work. So there's no way to factor this expression over the integers. Instead,

1:10:46

let's use the quadratic equation. So we have our leading coefficient of x squared is one.

1:10:53

So A is one, B is two, and C is minus seven. And we're going to plug that into the equation

1:11:00

quadratic equation, which goes x equals negative b plus or minus the square root of b squared

1:11:06

minus four, I see all over two

1:11:10

different people have different ways of remembering this formula, I'd like to remember it by seeing

1:11:15

it x equals negative b plus or minus the square root of b squared minus four, I see oh over

1:11:22

to a, but you can use any pneumonic you like. Anyway, plugging in here, we have x equals

1:11:28

negative two plus or minus the square root of two squared minus four times one times

1:11:34

negative seven support and remember the negative seven there to all over two times one.

1:11:40

Now two squared is four, and four times one times negative seven is negative 28. So this

1:11:48

whole quantity under the square root sign becomes four minus negative 28, or 32. So

1:11:55

I can rewrite this as x equals negative two plus or minus the square root of 32, all over

1:12:00

two.

1:12:01

Since 32, is 16 times two and 16 is a perfect square, I can rewrite this as negative two

1:12:11

plus or minus the square root of 16 times the square root of two over two, which is

1:12:15

negative two plus or minus four times the square root of two over two.

1:12:19

Next, I'm going to split out my fraction

1:12:26

as negative two over two plus or minus four square root of two over two, and then simplify

1:12:32

those fractions. This becomes negative one plus or minus two square root of two. So my

1:12:38

answers are negative one plus two square root of two and negative one minus two square root

1:12:45

of two. And if I need a decimal answer for any reason, I could work this out on my calculator.

1:12:52

As our final example, let's find all real solutions for the equation, one half y squared

1:12:57

equals 1/3 y minus two. I'll start as usual by putting it in standard form. So that gives

1:13:04

me one half y squared minus 1/3 y plus two equals zero, I could go ahead and start trying

1:13:12

to factor or use the quadratic formula right now. But I find fractional coefficients kind

1:13:17

of annoying. So I'd like to get rid of them. By doing what I call clearing the denominator,

1:13:22

that means I'm going to multiply the whole entire equation by the least common denominator.

1:13:27

In this case, the least common denominator is two times three or six. So I'll multiply

1:13:33

the whole equation by six have to make sure I multiplied both sides of the equation, but

1:13:37

in this case, six times zero is just zero. And when I distribute the six, I get three

1:13:42

y squared minus two y plus 12 equals zero.

1:13:47

Now, I could try to factor this, but I think it's easier probably just to plunge in and

1:13:51

use the quadratic formula.

1:13:54

So I get x equals negative B, that's negative negative two or two, plus or minus the square

1:14:01

root of b squared, minus four times a times c, all over to a.

1:14:12

Working out the stuff in a square root sign, negative two squared is four. And here we

1:14:17

have, let's see 144.

1:14:21

So this simplifies to x equals to plus or minus the square root of four minus 144. That's

1:14:28

negative 140. All of our sex. Well, if you're concerned about that negative number under

1:14:35

the square root sign, you should be we can't take the square root of a negative number

1:14:40

and get an N get a real number is our answer. There's no real number whose square is a negative

1:14:45

number. And therefore, our conclusion is we have no real solutions to this quadratic equation.

1:14:54

In this video, we solve some quadratic equations by first writing them in standard form and

1:14:59

then

1:15:00

Either factoring or using the quadratic formula.

1:15:03

In some examples, factoring doesn't work, it's not possible to factor the equation.

1:15:08

But in fact, using the quadratic formula will always work even if it's also possible to

1:15:13

solve it by factoring. So you can't really lose by using the quadratic formula. It's

1:15:19

just sometimes it'll be faster to factor instead.

1:15:23

This video is about solving rational equations. A rational equation, like this one is an equation

1:15:29

that has rational expressions in that, in other words, an equation that has some variables

1:15:34

in the denominator.

1:15:35

There are several different approaches for solving a rational equation, but they all

1:15:41

start by finding the least common denominator. In this example, the denominators are x plus

1:15:46

three and x, we can think of one as just having a denominator of one.

1:15:52

Since the denominators don't have any factors in common, I can find the least common denominator

1:15:57

just by multiplying them together.

1:16:00

My next step is going to be clearing the denominator.

1:16:04

By this, I mean that I multiply both sides of my equation by this least common denominator,

1:16:12

x plus three times x, I multiply on the left side of the equation, and I multiply by the

1:16:19

same thing on the right side of the equation.

1:16:22

Since I'm doing the same thing to both sides of the equation, I don't change the the value

1:16:28

of the equation. Multiplying the least common denominator on both sides of the equation

1:16:33

is equivalent to multiplying it by all three terms in the equation, I can see this when

1:16:38

I multiply out,

1:16:41

I'll rewrite the left side the same as before, pretty much. And then I'll distribute the

1:16:47

right side to get x plus three times x times one plus x plus three times x times one over

1:16:56

x. So I've actually multiplied the least common denominator by all three terms of my equation.

1:17:01

Now I can have a blast canceling things. The x plus three cancels with the x plus three

1:17:07

on the denominator.

1:17:10

The here are nothing cancels out because there's no denominator, and here are the x in the

1:17:15

numerator cancels with the x in the denominator.

1:17:18

So I can rewrite my expression as x squared equals x plus three times x times one plus

1:17:26

x plus three. Now I'm going to simplify.

1:17:29

So I'll leave the x squared alone on this side, I'll distribute out x squared plus 3x

1:17:36

plus x plus three, hey, look, the x squared is cancel on both sides. And so I get zero

1:17:44

equals 4x plus three, so 4x is negative three, and x is negative three fourths. Finally,

1:17:51

I'm going to plug in my answer to check. This is a good idea for any kind of equation. But

1:17:57

it's especially important for a rational equation because occasionally for rational equations,

1:18:01

you'll get what's called extraneous solution solutions that don't actually work in your

1:18:05

original equation because they make the denominator zero. Now, in this example, I don't think

1:18:09

we're going to get any extraneous equations because negative three fourths is not going

1:18:16

to make any of these denominators zero, so it should work out fine when I plug in. If

1:18:20

I plug in,

1:18:22

I get this, I can simplify

1:18:29

the denominator here, negative three fourths plus three, three is 12 fourths, as becomes

1:18:34

nine fourths. And this is one I'll flip and multiply to get minus four thirds. So here,

1:18:42

I can simplify my complex fraction, it ends up being negative three nights, and one minus

1:18:49

four thirds is negative 1/3. So that all seems to check out.

1:18:55

And so my final answer is x equals negative three fourths.

1:19:00

This next example looks a little trickier. And it is, but the same approach will work.

1:19:05

First off, find the least common denominator. So here, my denominators are c minus five,

1:19:13

c plus one, and C squared minus four c minus five, I'm going to factor that as C minus

1:19:22

five times c plus one. Now, my least common denominator needs to have just enough factors

1:19:29

to that each of these denominators divided into it. So I need the factor c minus five,

1:19:35

I need the factor c plus one. And now I've already got all the factors I need for this

1:19:39

denominator. So here is my least common denominator. Next step is to clear the denominators.

1:19:47

So I do this by multiplying both sides of the equation by my least common denominator.

1:19:53

In fact, I can just multiply each of the three terms by this least common denominator

1:20:01

I went ahead and wrote my third denominator in factored form to make it easier to see

1:20:06

what cancels. Now canceling time dies, this dies. And both of those factors die. cancelling

1:20:16

out the denominator is the whole point of multiplying by the least common denominator,

1:20:21

you're multiplying by something that's big enough to kill every single denominator, so

1:20:25

you don't have to deal with denominators anymore.

1:20:28

Now I'm going to simplify by multiplying out.

1:20:32

So I get, let's see, c plus one times four c, that's four c squared plus four c, now

1:20:41

I get minus just c minus five, and then over here, I get three c squared plus three,

1:20:51

I can rewrite the minus quantity c minus five is minus c plus five.

1:21:00

And now I can subtract the three c squared from both sides to get just a C squared over

1:21:06

here, and the four c minus c that becomes a three C.

1:21:15

And finally, I can subtract the three from both sides to get c squared plus three c plus

1:21:21

two equals zero. got myself a quadratic equation that looks like a nice one that factors. So

1:21:28

this factors to C plus one times c plus two equals zero. So either c plus one is zero,

1:21:36

or C plus two is zero. So C equals negative one, or C equals negative two.

1:21:44

Now let's see, we need to still check our answers.

1:21:49

Without even going to the trouble of calculating anything, I can see that C equals negative

1:21:54

one is not going to work, because if I plug it in to this denominator here, I get a denominator

1:22:01

zero, which doesn't make sense. So C equals minus one is an extraneous solution, it doesn't

1:22:09

actually satisfy my original equation. And so I can just cross it right out, C equals

1:22:16

negative two.

1:22:19

I can go if I go ahead, and that doesn't make any of my denominators zero. So if I haven't

1:22:23

made any mistakes, it should satisfy my original equation, but, but I'll just plug it in to

1:22:30

be sure.

1:22:33

And after some simplifying,

1:22:36

I get a true statement.

1:22:39

So my final answer is C equals negative two. In this video, we saw the couple of rational

1:22:46

equations using the method of finding the least common denominator and then clearing

1:22:53

the denominator,

1:22:55

we cleared the denominator by multiplying both sides of the equation by the least common

1:22:59

denominator or equivalently. multiplying each of the terms by that denominator.

1:23:04

There's another equivalent method that some people prefer, it still starts out the same,

1:23:10

we find the least common denominator, but then we write all the fractions

1:23:16

over that least common denominator. So in this example, we'd still use the least common

1:23:22

denominator of x plus three times x. But our next step would be to write each of these

1:23:27

rational expressions over that common denominator by multiplying the top and the bottom by the

1:23:32

appropriate things. So one, in order to get the common denominator of x plus 3x, I need

1:23:38

to multiply the top and the bottom by x plus three times x, one over x, I need to multiply

1:23:44

the top and the bottom just by x plus three since that's what's missing from the denominator

1:23:49

x. Now, if I simplify a little bit,

1:23:53

let's say this is x squared over that common denominator, and

1:24:00

here I have just x plus three times x over that denominator, and here I have x plus three

1:24:09

over that common denominator. Now add together my fractions on the right side, so they have

1:24:17

a common denominator.

1:24:18

So this is x plus three times x plus x plus three. And now I have two fractions that have

1:24:27

that are equal that have the same denominator, therefore their numerators have to be equal

1:24:32

also. So the next step is to set the numerators equal.

1:24:37

So I get x squared is x plus three times x plus x plus three. And if you look back at

1:24:46

the previous way, we solve this equation, you'll recognize this equation. And so from

1:24:52

here on we just continue as before.

1:24:57

When choosing between these two methods, I personally tend

1:25:00

prefer the clear the denominators method, because it's a little bit less writing, you

1:25:03

don't have to get rid of those denominators earlier, you don't have to write them as many

1:25:07

times. But some people find this one a little bit easier to remember, a little easier to

1:25:12

understand either of these methods is fine.

1:25:15

One last caution, don't forget at the end, to check your solutions and eliminate any

1:25:21

extraneous solutions.

1:25:23

These will be solutions that make the denominators of your original equations go to zero.

1:25:30

This video is about solving radical equations, that is equations like this one that have

1:25:36

square root signs in them, or cube roots or any other kind of radical.

1:25:40

When I see an equation with a square root in it, I really want to get rid of the square

1:25:45

root. But it'll be easiest to get rid of the square root. If I first isolate the square

1:25:51

root. In other words, I want to get the term with the square root and that on one side

1:25:56

of the equation by itself, and everything else on the other side of the equation. If

1:26:01

I start with my original equation, x plus the square root of x equals 12. And I subtract

1:26:07

x from both sides, then that does isolate the square root term on the left side with

1:26:12

everything else on the right. Once I've isolated the term with the square root, I want to get

1:26:19

rid of the square root. And I'll do that by squaring both sides

1:26:27

of my equation. So I'll take the square root of x equals 12 minus x and square both sides.

1:26:36

Now the square root of x squared is just x, taking the square root and then squaring those

1:26:42

operations undo each other

1:26:45

to work out 12 minus x squared, write it out and distribute 12 times 12 is 144 12 times

1:26:54

minus x is minus 12x. I get another minus 12x from here.

1:27:02

And finally minus x times minus x is positive x squared. So I can combine my minus 12 x's,

1:27:09

that's minus 24x. And now I can subtract x from both sides to get zero equals 144 minus

1:27:18

25x plus x squared. That's a quadratic equation, I'll rewrite it in a little bit more standard

1:27:26

form here.

1:27:27

So now I've got a familiar quadratic equation with no radical signs left in my equation,

1:27:34

I'll just proceed to solve it like I usually do for a quadratic, I'll try to factor it.

1:27:38

So I'm going to look for two numbers that multiply to 144 and add to minus 25. I know

1:27:46

I'm going to need negative numbers to get to negative 25. And in fact, I'll need two

1:27:53

negative numbers. So they still multiply to a positive number. So I'll start listing some

1:27:57

factors of 144, I could have negative one and a negative 144, negative two, and negative

1:28:04

72, negative four, negative 36, and so on. Once I've listed out the possible factors,

1:28:12

it's not hard to find the two that add to negative 25. So that's negative nine and negative

1:28:17

16. So now I can factor in my quadratic equation as x minus nine times x minus 16 equals zero,

1:28:28

that means that x minus nine is zero or x minus 16 is zero. So x equals nine or x equals

1:28:34

16. I'm almost done. But there's one last very important step. And that's to check the

1:28:41

Solutions so that we can eliminate any extraneous solutions and extraneous solution as a solution

1:28:49

that we get that does not actually satisfy our original equation and extraneous solutions

1:28:54

can happen when you're solving equations with radicals in them. So let's first check x equals

1:29:00

nine. If we plug in to our original equation, we get nine plus a squared of nine and we

1:29:06

want that to equal 12. Well, the square root of nine is three, and nine plus three does

1:29:11

indeed equal 12. So that solution checks out. Now, let's try x equals 16. Plugging in, we

1:29:18

get 16 plus a squared of 16. And that's supposed to equal 12. Well, that says 16 plus four

1:29:25

is supposed to equal 12. But that most definitely is not true. And so x equals 16. Turns out

1:29:31

to be extraneous solution, and our only solution is x equals nine

1:29:47

This next equation might not look like an equation involving radicals. But in fact,

1:30:26

we can think of a fractional exponent as being a radical in disguise. Let's start by doing

1:30:32

the same thing we did on the previous problem by isolating This time, we'll isolate the

1:30:38

part of the equation

1:30:41

that involves the fractional exponent. So I'll start with the original equation, two

1:30:46

times P to the four fifths equals 1/8. And I'll divide both sides by two or equivalently,

1:30:53

I can multiply both sides by one half, that gives me p to the four fifths equals 1/16.

1:31:00

And I've effectively isolated the part of the equation with the fractional exponent

1:31:05

as much as possible. Now, in the previous example, the next step was to get rid of the

1:31:12

radical. In this example, we're going to get rid of the fractional exponent. And I'm going

1:31:18

to actually do this in two stages. First, I'm going to raise both sides to the fifth

1:31:24

power.

1:31:26

That's because when I take an exponent to an exponent, I'm multiply my exponents. And

1:31:34

so that becomes just p to the fourth equals 1/16 to the fifth power. Now, I'm going to

1:31:41

get rid of the fourth power by raising both sides to the 1/4 power, or by taking the fourth

1:31:48

root, there's something that you need to be careful about though, when taking

1:31:53

an even root, or the one over an even number power, you always have to consider plus or

1:32:01

minus your answer.

1:32:04

It's kind of like when you write x squared equals four, and you take the square root

1:32:10

of both sides, x could equal plus or minus the square root of four, right, x could equal

1:32:16

plus or minus two, since minus two squared is four, just as well as two squared. So that's

1:32:23

why when you take an even root, or a one over an even power of both sides, you always need

1:32:32

to include the plus or minus sign, when it's an odd root or one over an odd power, you

1:32:39

don't need to do that. If you had something like x cubed equals negative eight, then x

1:32:46

equals the cube root of negative eight, which is negative two would be your only solution,

1:32:51

you don't need to do the plus or minus because positive two wouldn't work.

1:32:55

So that aside, explains why we need this plus or minus power. And now p to the four to the

1:33:02

1/4. When I raise a power to a power, I multiply by exponents, so that's just p to the one,

1:33:09

which is equal to plus or minus 1/16 to the fifth power to the 1/4 power.

1:33:17

Now I just need to simplify this expression, I don't really want to raise 1/16 to the fifth

1:33:23

power, because 16 to the fifth power is like a really huge number. So I think I'm actually

1:33:28

going to rewrite this first

1:33:32

as P equals plus or minus 1/16. I'll write it back as the 5/4 power again.

1:33:42

And as I continue to solve using my exponent rules,

1:33:49

I'm going to prefer to write this as 1/16 to the fourth root to the fifth power, because

1:33:58

it's going to be easier to take the fourth root, let's see the fourth root of 1/16 is

1:34:03

the same thing as the fourth root of one over the fourth root of 16. Raise all that to the

1:34:08

fifth power. fourth root of one is just one and the fourth root of 16 is to raise that

1:34:14

to the fifth power, that's just going to be one to the fifth over two to the fifth, which

1:34:19

is plus or minus 130 seconds.

1:34:24

The last step is to check answers.

1:34:28

So I have the two answers p equals 130 seconds, and P equals minus 130 seconds. And if I plug

1:34:35

those both in

1:34:36

1/32 to the fourth fifth power.

1:34:43

That gives me two times one to the fourth power over 32 to the fourth power, which is

1:34:51

two times one over 32/5 routed to the fourth power. fifth root of 32 is two

1:34:59

Raisa to the fourth power, I get 16. So this is two times 1/16, which is 1/8, just as we

1:35:07

wanted in the original equation up here. Similarly, we can check that the P equals negative 1/32

1:35:16

actually does satisfy the equation, I'll leave that step to the viewer.

1:35:24

So our two solutions are p equals one over 32, and P equals minus one over 32. I do want

1:35:30

to point out an alternate approach to getting rid of the fractional exponent, we could have

1:35:35

gotten rid of it all in one fell swoop by raising both sides of our equation to the

1:35:41

five fourths power.

1:35:45

five fourths is the reciprocal of four fifths. So when I use my exponent rules, and say that

1:35:51

when I raise the power to the power, I multiply my exponents, that gives me p to the four

1:35:56

fifths times five fourths is plus or minus 1/16 to the five fourths, in other words,

1:36:03

P to the One Power, which is just P is plus or minus 1/16, to the five fourths, so that's

1:36:10

an alternate and possibly faster way to get the solution. Once again, the plus or minus

1:36:16

comes from the fact that when we take the 5/4 power, we're really taking an even root

1:36:22

a fourth root, and so we need to consider both positive and negative answers.

1:36:27

This video is about solving radical equations, that is equations like this one that have

1:36:33

square root signs in them, or cube roots or any other kind of radical.

1:36:39

In this video, we solved radical equations by first isolating the radical sign, or the

1:36:45

fractional exponent,

1:36:48

and then removing the radical sine or the fractional exponent

1:36:54

by either squaring both sides or taking the reciprocal power of both sides.

1:37:01

This video is about solving equations with absolute values in them.

1:37:06

Recall that the absolute value of a positive number is just the number, but the absolute

1:37:12

value of a negative number is its opposite. In general, I think of the absolute value

1:37:17

of a number as representing its distance from zero on the number line,

1:37:23

the number four,

1:37:25

and the number of negative four are both at distance for from zero, and so the absolute

1:37:32

value of both of them is four.

1:37:35

Similarly, if I write the equation, the absolute value of x is three, that means that x has

1:37:44

to be three units away from zero on the number line.

1:37:50

And so X would have to be either negative three, or three. Let's start with the equation

1:37:56

three times the absolute value of x plus two equals four.

1:38:00

I'd like to isolate the absolute value part of the equation, I can do this by starting

1:38:08

with my original equation,

1:38:14

subtracting two from both sides

1:38:17

and dividing both sides by three.

1:38:23

Now I'll think in terms of distance on a number line, the absolute value of x is two thirds

1:38:31

means that x is two thirds away from zero. So x could be here for here at negative two

1:38:40

thirds or two thirds.

1:38:43

And the answer to my equation is x is negative two thirds, or two thirds,

1:38:50

I can check my answers by plugging in

1:38:52

three times the added value of negative two thirds plus two, I need to check it that equals

1:39:00

four. Well, the absolute value of negative two thirds is just two thirds. So this is

1:39:04

three times two thirds plus two, which works out to four.

1:39:10

Similarly, if I plug in positive two thirds, it also works out to give me the correct answer.

1:39:20

The second example is a little different, because the opposite value sign is around

1:39:24

a more complicated expression, not just around the X.

1:39:29

I would start by isolating the absolute value part.

1:39:35

But it's already isolated. So I'll just go ahead and jump to thinking about distance

1:39:41

on the number line. So on my number line, the whole expression 3x plus two is supposed

1:39:49

to be at a distance of four from zero.

1:39:53

So that means that 3x plus two is here at four or 3x plus two

1:39:59

though is it negative four, all right, those as equations

1:40:03

3x plus two equals four, or 3x plus two is minus four. And then I can solve.

1:40:12

So this becomes 3x equals two, or x equals two thirds. And over here, I get 3x equals

1:40:20

minus six, or x equals minus two. Finally, I'll check my answers.

1:40:29

I'll leave it to you to verify that they both work.

1:40:32

A common mistake on absolute value equations is to get rid of the absolute value signs

1:40:38

like we did here, and then just solve for one answer, instead of solving for both answers.

1:40:44

Another mistake sometimes people make is, once they get the first answer, they just

1:40:48

assume that the negative of that works also.

1:40:52

But that doesn't always work. In the first example, our two answers were both the negatives

1:40:57

of each other. But in our second examples, or two answers, were not just the opposites

1:41:02

of each other one was two thirds and the other was negative two.

1:41:06

In this third example, let's again, isolate the absolute value part of the equation.

1:41:12

So starting with our original equation,

1:41:16

we can subtract 16 from both sides,

1:41:22

and divide both sides by five or equivalently, multiply by 1/5.

1:41:31

Now let's think about distance on the number line,

1:41:35

we have an absolute value needs to equal negative three. So that means whatever is inside the

1:41:41

absolute value sign needs to be at distance negative three from zero, well, you can't

1:41:48

be at distance negative three from zero. Another way of thinking about this is you can't have

1:41:52

the absolute value of something and end up with a negative number if the value is always

1:41:57

positive, or zero. So this equation doesn't actually make sense, and there are no solutions

1:42:05

to this equation.

1:42:07

In this video, we solved absolute value equations. In many cases, an absolute value equation

1:42:15

will have two solutions. But in some cases, it'll have no solutions. And occasionally,

1:42:20

it'll have just one solution.

1:42:25

This video is about interval notation, and easy and well known way to record inequalities.

1:42:30

Before dealing with interval notation, it is important to know how to deal with inequalities.

1:42:39

Our first example of an inequality is written here, all numbers between one and three, not

1:42:45

including one and three.

1:42:46

First, we used to write down our variable x, that it is important to locate the key

1:42:53

values for this problem that is one and three. Now here we see that x is between these two

1:43:01

numbers, meaning we will have one inequality statement on each side of the variable. Here

1:43:07

it says not including one and three, which means that we do not have an or equal to sign

1:43:12

beneath each inequality. Here we will put one the lowest key value, and here we will

1:43:18

put three the highest key value. Next, we're going to graph this inequality on a number

1:43:24

line.

1:43:27

Here we write our key values one, and three. Because it is not including one, and three,

1:43:34

we have an empty circle around each number. Because it isn't the numbers between we have

1:43:40

a line connecting them.

1:43:41

The last step of this problem is writing this inequality in interval notation.

1:43:50

writing things in interval notation is kind of like writing an ordered pair. Here we will

1:43:54

put one and then here we will put three.

1:43:57

Next we need to put brackets around these numbers.

1:44:03

For this problem, it is not including one in three, which means we will use soft brackets.

1:44:09

However, if it were including one and three, we will use hard brackets.

1:44:15

It is important to note for interval notation, that the smallest value always goes on the

1:44:22

left and the biggest value always goes on the right.

1:44:29

You also include a comma between your two key values.

1:44:32

Now let's work on problem B. Once again we write our variable x. Again it is between

1:44:41

negative four and two but this time it is including a negative four into this requires

1:44:47

us to have the or equal to assign below the inequality. Then we put our lowest key value

1:44:53

here and their highest here. The number line graph for this problem is slightly different.

1:44:59

We still

1:45:00

have our key values negative four and two. But instead of an open circle, like we have

1:45:07

right here, we instead use a closed circle, representing that it is including negative

1:45:15

four and two, you complete this with a line in between. Last we write this in interval

1:45:21

notation. In the last problem, I said that for numbers, including the outside values,

1:45:27

we would use these hard brackets. Now we are actually using this. So we have or hard bracket

1:45:35

on each side because isn't including foreign zoo, then we put our smallest value on the

1:45:40

left, and our highest value on the right, with the comma in between.

1:45:46

You now know how to correctly write these two types of intervals.

1:45:51

Now we are going to practice transforming these from inequality notation to interval

1:45:58

notation, and vice versa. This is slightly more difficult than our previous examples

1:46:05

of having two soft brackets or two hard brackets, we'll have two different types. Now on the

1:46:11

left side will have a hard bracket because it is including the three. However, on the

1:46:18

right side, it is a soft bracket because it is it it is not including the one that we

1:46:24

put a comma in the middle, a lower key value and our higher key value. For the next problem,

1:46:31

we're taking an equation already written an interval notation and putting it back into

1:46:37

inequality notation.

1:46:38

For the second problem, we can see are key values as being five and negative infinity.

1:46:47

Now, this sounds a little bit weird, but let's just set up or an equality, we have our variable

1:46:54

x, which is less than because of the soft bracket five, and greater than without the

1:47:02

or equal to because it has a soft bracket there to infinity. But because x is always

1:47:09

greater than negative infinity, we can take out this part, leaving us with x is less than

1:47:16

five.

1:47:17

It is important to note that a soft bracket always accompanies infinity. This is because

1:47:25

infinity is not a real number, so we cannot include it just go as far up to it as we can.

1:47:32

The next problem is a little tricky, because we don't see another key value over here.

1:47:38

But let's just start off the equation as a soft bracket on this side due to the absence

1:47:43

of an order equal to sign. And negative 15 is the lower key value.

1:47:51

On the other side, we put the highest possible number infinity, and then close it off with

1:47:57

a soft bracket. We can see the relation of this to the previous problem. And how it goes

1:48:05

into this problem when x is greater than a key value instead of less than a key value.

1:48:12

Once again, we have our sauce bracket for infinity, which is always true.

1:48:19

Part D brings up an important point about which number goes on the left. For all of

1:48:26

our other problems, we have had the inequalities pointing left

1:48:30

instead of rights. When seeing Part D, you might think oh, four is on the left, so it

1:48:41

would go here, and zero is on the right who would go here. But that is not true.

1:48:47

When writing inequalities, you must always have the lower value on the left, which for

1:48:53

this problem is zero. To fix this, we can simply flip this inequality. So it reads like

1:49:01

this, it is still identical just written in an easier form to transform it into interval

1:49:06

notation, then we can see that zero has a soft bracket, because it is not including

1:49:13

zero, but r comma four or other key value and a hard backup because it is or equal to

1:49:19

four.

1:49:21

For interval notation, you must always have the smaller number on the left side.

1:49:29

This was our video on interval notation, an alternative way to write inequalities.

1:49:36

This video is about solving inequalities that have absolute value signs in them.

1:49:41

Let's look at the inequality absolute value of x is less than five on the number line.

1:49:45

Thinking of absolute value as distance. This means that the distance between x and zero

1:49:52

is less than five units. So x has to live somewhere in between negative five and five

1:50:01

We can express this as an inequality without absolute value signs by saying negative five

1:50:06

is less than x, which is less than five. Or we can use interval notation, soft bracket

1:50:12

negative five, five, soft bracket.

1:50:16

Both of these formulations are equivalent to the original one, but don't involve the

1:50:20

absolute value signs.

1:50:23

In the second example, we're looking for the values of x for which the absolute value of

1:50:28

x is greater than or equal to five.

1:50:32

on the number line, this means that the distance of x from the zero, it's got to be bigger

1:50:39

than or equal to five units.

1:50:42

A distance bigger than five units means that x has to live somewhere over here, or somewhere

1:50:48

over here, where it's farther than five units away from zero. Course x could also have a

1:50:54

distance equal to five units. So I'll fill in that dot and shade in the other parts of

1:51:00

the number line that satisfy my inequality.

1:51:04

Now I can rewrite the inequality without the absolute value symbols by saying that x is

1:51:10

less than or equal to negative five, or x is greater than or equal to five. I could

1:51:17

also write this in interval notation,

1:51:20

soft bracket, negative infinity, negative five, hard bracket. And the second part is

1:51:26

hard bracket five infinity soft bracket, I combine these with a u for union. Because

1:51:33

I'm trying to describe all these points on the number line together with all these other

1:51:38

points.

1:51:39

Let's take this analysis a step further with a slightly more complicated problem. Now I

1:51:44

want the absolute value of three minus two t to be less than four.

1:51:49

And an absolute value less than four means a distance less than four on the number line.

1:51:56

But it's not the variable t that lives in here at a distance of less than four from

1:52:00

zero, it's the whole expression, three minus two t.

1:52:04

So three minus two t, live somewhere in here. And I can rewrite this as an inequality without

1:52:13

absolute value signs by saying negative four is less than three minus two t is less than

1:52:21

four. Now I have a compound inequality that I can solve the usual way. First, I subtract

1:52:27

three from all three sides to get negative seven is less than negative two t is less

1:52:33

than one. And now I'll divide all three sides by negative two. Since negative two is a negative

1:52:40

number, this reverses the directions of the inequalities.

1:52:46

Simplifying, I get seven halves is greater than t is greater than negative one half.

1:52:52

So my final answer on the number line looks like all the stuff between negative a half

1:52:58

and seven halves.

1:53:01

But not including the endpoints, and an interval notation, I can write this soft bracket negative

1:53:07

a half, seven, half soft bracket,

1:53:09

please pause the video and try the next problem on your own.

1:53:14

thinking in terms of distance, this inequality says that the distance between the expression

1:53:19

three minus two t and zero is always bigger than four. Let me draw this on the number

1:53:25

line.

1:53:27

If three minus two t has a distance bigger than four from zero, then it can't be in this

1:53:33

region that's near zero, it has to be on the outside in one of these two regions.

1:53:39

That is three minus two t is either less than negative four, or three minus two t is bigger

1:53:48

than four. I solve these two inequalities separately, the first one, subtracting three

1:53:54

from both sides, I get negative two t is less than negative seven divided by negative two,

1:54:01

I get T is bigger than seven halves. And then on the other side, I get negative two t is

1:54:08

greater than one. So t is less than negative one half, I had to flip inequalities in the

1:54:15

last step due to dividing by a negative number. So let's check this out on the number line

1:54:21

again, the first piece says that t is greater than seven halves. I'll draw that over here.

1:54:28

And the second piece says that t is less than negative one half. I'll draw that down here.

1:54:41

Because these two statements are joined with an or I'm looking for the t values that are

1:54:46

in this one, or in this one. That is I want both of these regions put together. So an

1:54:52

interval notation, this reads negative infinity to negative one half soft bracket union soft

1:55:00

bracket

1:55:01

Seven halves to infinity.

1:55:03

This last example looks more complicated. But if I simplify first and isolate the absolute

1:55:09

value part, it looks pretty much like the previous ones. So I'll start by subtracting

1:55:14

seven from both sides. And then I'll divide both sides by two. Now I'll draw my number

1:55:20

line.

1:55:21

And I'm looking for this expression for x plus five, to always have distance greater

1:55:28

than or equal to negative three from zero, wait a second, distance greater than equal

1:55:33

to negative three, well, distance is always greater than or equal to negative three is

1:55:36

always greater than equal to zero. So this, in fact, is always true.

1:55:43

And so the answer to my inequality is all numbers between negative infinity and infinity.

1:55:50

In other words, all real numbers.

1:55:55

Once solving absolute value inequalities, it's good to think about distance. And absolute

1:56:01

value of something that's less than a number means that whatever's inside the absolute

1:56:08

value signs is close to zero.

1:56:12

On the other hand, an absolute value is something and being greater than a number means that

1:56:18

whatever's inside the absolute value sign is far away from zero, because its distance

1:56:25

from zero is bigger than that certain number.

1:56:29

drawing these pictures on the number line is a helpful way to rewrite the absolute value

1:56:34

and equality as an inequality that doesn't contain an absolute value sign. In this case,

1:56:40

it would be negative three is less than x plus two is less than three. And in the other

1:56:47

case, it would be either x plus two is less than negative three, or x plus two is greater

1:56:53

than three.

1:56:55

This video is about solving linear inequalities. Those are inequalities like this one that

1:57:01

involve a variable here x, but don't involve any x squared or other higher power terms.

1:57:09

The good news is, we can solve linear inequalities, just like we solve linear equations by distributing,

1:57:15

adding and subtracting terms to both sides and multiplying and dividing by numbers on

1:57:19

both sides. The only thing that's different is that if you multiply or divide by a negative

1:57:26

number, then you need to reverse the direction of the inequality. For example, if we had

1:57:32

the inequality, negative x is less than negative five, and we wanted to multiply both sides

1:57:39

by negative one to get rid of the negative signs, we'd have to also switch or reverse

1:57:45

the inequality. With this caution in mind, let's look at our first example.

1:57:50

Since our variable x is trapped in parentheses, I'll distribute the negative five to free

1:57:56

it from the parentheses.

1:57:58

That gives me negative 5x minus 10 plus three is greater than eight.

1:58:04

Negative 10 plus three is negative seven, so I'll rewrite this as negative 5x minus

1:58:10

seven is greater than eight. Now I'll add seven to both sides to get negative 5x is

1:58:16

greater than 15. Now I'd like to divide both sides by negative five. Since negative five

1:58:23

is a negative number that reverses the inequality. So I get x is less than 15 divided by negative

1:58:31

five. In other words, x is less than negative three.

1:58:36

If I wanted to graph this on a number line, I just need to put down a negative three,

1:58:40

an open circle around it, and shade in to the left,

1:58:44

I use an open circle, because x is strictly less than negative three and can't equal negative

1:58:51

three. If I wanted to write this in interval notation, I've added a soft bracket negative

1:58:56

infinity, negative three soft bracket. Again, the soft bracket is because the negative three

1:59:01

is not included. This next example is an example of a compound inequality. It has two parts.

1:59:10

Either this statement is true, or this statement is true. I want to find the values of x that

1:59:15

satisfy either one. I'll solve this by working out each part separately, and then combining

1:59:21

them at the end. For the inequality on the left, I'll copy it over here. I'm going to

1:59:28

add four to both sides.

1:59:32

then subtract x from both sides

1:59:36

and then divide both sides by two.

1:59:40

I didn't have to reverse the inequality sign because I divided by a positive number

1:59:45

on the right side, I'll copy the equation over, subtract one from both sides, and divide

1:59:51

both sides by 696 is the same as three halves. Now I'm looking for the x values that make

1:59:59

this statement.

2:00:00

row four, make this statement true. Let me graph this on a number line.

2:00:07

x is less than or equal to negative two, means I put a filled in circle there and graph everything

2:00:14

to the left. X is greater than three halves means I put a empty circle here and shaded

2:00:22

and everything to the right.

2:00:25

My final answer includes both of these pieces, which I'll reshade in green, because x is

2:00:33

allowed to be an either one or the other. Finally, I can write this in interval notation.

2:00:38

The first piece on the number line can be described as soft bracket negative infinity,

2:00:44

negative two hard bracket. And the second piece can be described as soft bracket three

2:00:50

halves, infinity soft bracket to indicate that x can be in either one of these pieces,

2:00:57

I use the union side, which is a U.

2:01:01

That means that my answer includes all x values in here together with all x values in here.

2:01:09

This next example is also a compound inequality. This time I'm joined by an ad, the and means

2:01:15

I'm looking for all y values that satisfy both this and this at the same time. Again,

2:01:22

I can solve each piece separately

2:01:25

on the left, to isolate the Y, I need to multiply by negative three halves on both sides. So

2:01:34

that gives me Why is less than negative 12 times negative three halves, the greater than

2:01:40

flip to a less than because I was multiplying by negative three halves, which is a negative

2:01:46

number.

2:01:47

By clean up the right side, I get why is less than 18.

2:01:54

On the right side, I'll start by subtracting two from both sides.

2:01:59

And now I'll divide by negative four, again, a negative number, so that flips the inequality.

2:02:07

So that's why is less than three over negative four. In other words, y is less than negative

2:02:13

three fourths. Again, I'm looking for the y values that make both of these statements

2:02:19

true at the same time.

2:02:22

Let me graph this on the number line,

2:02:25

the Y is less than 18. I can graph that by drawing the number 18.

2:02:34

I don't want to include it. So I use an empty circle and I shaded everything to the left

2:02:41

the statement y is less than negative three fourths, I need to draw a negative three fourths.

2:02:47

And again, I don't include it, but I do include everything on the left. Since I want the y

2:02:54

values for which both of these statements are true, I need the y values that are in

2:02:59

both colored blue and colored red. And so that would be this part right here. I'll just

2:03:05

draw it above so you can see it easily. So that would be all the numbers from negative

2:03:10

three fourths and lower, not including negative three, four, since those are the parts of

2:03:15

the number line that have both red and blue colors on them. In interval notation, my final

2:03:21

answer will be soft bracket negative infinity to negative three fourths soft bracket.

2:03:26

As my final example, I have an inequality that has two inequality signs in it negative

2:03:34

three is less than or equal to 6x minus two is less than 10. I can think of this as being

2:03:40

a compound inequality with two parts, negative 3x is less than or equal to 6x minus two.

2:03:47

And at the same time 6x minus two is less than 10. I could solve this into pieces as

2:03:54

before. But instead, it's a little more efficient to just solve it all at once, by doing the

2:04:00

same thing to all three sides. So as a first step, I'll add two to all three sides. That

2:04:08

gives me negative one is less than or equal to 6x is less than 12. And now I'm going to

2:04:14

divide all three sides by six to isolate the x. So that gives me negative one, six is less

2:04:20

than or equal to x is less than two.

2:04:24

If we solved it, instead, in two pieces above, we'd end up with the same thing because we

2:04:29

get negative one six is less than or equal to x from this piece, and we'd get x is less

2:04:35

than two on this piece. And because of the and statement, that's the same thing as saying

2:04:40

negative one, six is less than or equal to x, which is less than two.

2:04:45

Either way we do it. Let's see if what it looks like on the number line. So on the number

2:04:50

line, we're looking for things that are between two and negative one six, including the negative

2:04:58

one, six

2:04:59

But not including the to interval notation, we can write this as hard bracket, negative

2:05:05

162 soft bracket.

2:05:08

In this video, we solve linear inequalities, including some compound inequalities, joined

2:05:13

by the conjunctions and, and or,

2:05:18

remember when we're working with and we're looking for places on the number line where

2:05:22

both statements are true.

2:05:24

That is, we're looking for the overlap on the number line.

2:05:30

In this case, the points on the number line that are colored both red and blue at the

2:05:35

same time.

2:05:36

When we're working with oral statements, we're looking for places where either one or the

2:05:43

other statement is true or both

2:05:46

on the number line,

2:05:49

this corresponds to points that are colored either red or blue, or both. And in this picture,

2:05:54

that will actually correspond to the entire number line.

2:05:57

In this video, we'll solve inequalities involving polynomials like this one, and inequalities

2:06:04

involving rational expressions like this one.

2:06:09

Let's start with a simple example. Maybe a deceptively simple example. If you see the

2:06:14

inequality, x squared is less than four, you might be very tempted to take the square root

2:06:19

of both sides and get something like x is less than two as your answer. But in fact,

2:06:26

that doesn't work.

2:06:27

To see why it's not correct, consider the x value of negative 10.

2:06:35

Negative 10 satisfies the inequality, x is less than two since negative 10 is less than

2:06:43

two. But it doesn't satisfy the inequality x squared is less than four, since negative

2:06:51

10 squared is 100, which is not less than four. So these two inequalities are not the

2:06:58

same. And it doesn't work to solve a quadratic inequality just to take the square root of

2:07:04

both sides, you might be thinking part of why this reasoning is wrong, as we've ignored

2:07:09

the negative two option, right? If we had the equation, x squared equals four, then

2:07:15

x equals two would just be one option, x equals negative two would be another solution. So

2:07:21

somehow, our solution to this inequality should take this into account. In fact, a good way

2:07:27

to solve an inequality involving x squares or higher power terms, is to solve the associated

2:07:33

equation first. But before we even do that, I like to pull everything over to one side,

2:07:40

so that my inequality has zero on the other side.

2:07:44

So for our equation, I'll subtract four from both sides to get x squared minus four is

2:07:49

less than zero.

2:07:50

Now, I'm going to actually solve the associated equation, x squared minus four is equal to

2:07:58

zero, I can do this by factoring to x minus two times x plus two is equal to zero. And

2:08:06

I'll set my factors equal to zero, and I get x equals two and x equals minus two.

2:08:12

Now, I'm going to plot the solutions to my equation on the number line. So I write down

2:08:17

negative two and two, those are the places where my expression x squared minus four is

2:08:23

equal to zero.

2:08:26

Since I want to find where x squared minus four is less than zero, I want to know where

2:08:30

this expression x squared minus four is positive or negative, a good way to find that out is

2:08:36

to plug in test values. So first, a plug in a test value in this area of the number line,

2:08:43

something less than negative to say x equals negative three.

2:08:47

If I plug in negative three into x squared minus four, I get negative three squared minus

2:08:53

four, which is nine minus four, which is five, that's a positive number. So at negative three,

2:09:02

the expression x squared minus four is positive. And in fact, everywhere on this region of

2:09:08

the number line, my expression is going to be positive, because it can jump from positive

2:09:12

to negative without going through a place where it's zero, I can figure out whether

2:09:17

x squared minus four is positive or negative on this region, and on this region of the

2:09:21

number line by plugging in test value similar way,

2:09:25

evaluate the plug in between negative two and two, a nice value is x equals 00 squared

2:09:31

minus four, that's negative four and negative number. So I know that my expression x squared

2:09:36

minus four is negative on this whole interval. Finally, I can plug in something like x equals

2:09:43

10, something bigger than two, and I get 10 squared minus four. Without even computing

2:09:49

that I can tell that that's going to be a positive number. And that's all that's important.

2:09:53

Again, since I want x squared minus four to be less than zero, I'm looking for the places

2:09:57

on this number line where I'm getting

2:09:59

negatives. So I will share that in on my number line. It's in here, not including the endpoints.

2:10:06

Because the endpoints are where my expression x squared minus four is equal to zero, and

2:10:10

I want it strictly less than zero,

2:10:12

I can write my answer as an inequality, negative two is less than x is less than two, or an

2:10:19

interval notation as soft bracket negative two to soft bracket.

2:10:25

Our next example, we can solve similarly, first, we'll move everything to one side,

2:10:32

so that our inequality is x cubed minus 5x squared minus 6x is greater than or equal

2:10:39

to zero. Next, we'll solve the associated equation by factoring. So first, I'll write

2:10:47

down the equation. Now I'll factor out an x. And now I'll factor the quadratic. So the

2:10:55

solutions to my equation are x equals 0x equals six and x equals negative one,

2:11:01

I'll write the solutions to the equation on the number line.

2:11:05

So that's negative one, zero, and six. That's where my expression x times x minus six times

2:11:17

x plus one is equal to zero.

2:11:21

But I want to find where it's greater than or equal to zero. So again, I can use test

2:11:28

values, I can plug in, for example, x equals negative two, either to this version of expression,

2:11:35

or to this factored version. Since I only care whether my answer is positive or negative,

2:11:40

it's sometimes easier to use the factored version. For example, when x is negative two,

2:11:45

this factor is negative.

2:11:48

But this factor, x minus six is also negative when I plug in negative two for x.

2:11:55

Finally, x plus one, when I plug in negative two for x, that's negative one, that's also

2:12:03

negative. And a negative times a negative times a negative gives me a negative number.

2:12:08

If I plug in something, between negative one and zero, say x equals negative one half,

2:12:16

then I'm going to get a negative for this factor, a negative for this factor, but a

2:12:21

positive for this third factor.

2:12:25

Negative times negative times positive gives me a positive

2:12:28

for a test value between zero and six, let's try x equals one.

2:12:34

Now I'll get a positive for this factor a negative for this factor, and a positive for

2:12:41

this factor.

2:12:43

positive times a negative times a positive gives me a negative. Finally, for a test value

2:12:50

bigger than six, we could use say x equals 100. That's going to give me positive, positive

2:12:56

positive.

2:12:59

So my product will be positive.

2:13:03

Since I want values where my expression is greater than or equal to zero, I want the

2:13:08

places where it equals zero.

2:13:12

And the places where it's positive.

2:13:16

So my final answer will be close bracket negative one to zero, close bracket union, close bracket

2:13:24

six to infinity.

2:13:25

As our final example, let's consider the rational inequality, x squared plus 6x plus nine divided

2:13:34

by x minus one is less than or equal to zero.

2:13:38

Although it might be tempting to clear the denominator and multiply both sides by x minus

2:13:43

one, it's dangerous to do that, because x minus one could be a positive number. But

2:13:49

it could also be a negative number. And when you multiply both sides by a negative number,

2:13:53

you have to reverse the inequality. Although it's possible to solve the inequality this

2:13:59

way, by thinking of cases where x minus one is less than zero or bigger than zero, I think

2:14:04

it's much easier just to solve the same way as we did before. So we'll start by rewriting

2:14:10

so that we move all terms to the left and have zero on the right. Well, that's already

2:14:15

true. So the next step would be to solve the associated equation.

2:14:20

That is x squared plus 6x plus nine over x minus one is equal to zero.

2:14:29

That would be where the numerator is 0x squared plus 6x plus nine is equal to zero. So we're

2:14:35

x plus three squared is zero, or x equals negative three. There's one extra step we

2:14:42

have to do for rational expressions. And that's we need to find where the expression does

2:14:48

not exist. That is, let's find where the denominator is zero. And that said, x equals one.

2:14:55

I'll put all those numbers on the number line. The places where am I

2:14:59

rational expression is equal to zero, and the place where my rational expression doesn't

2:15:04

exist, then I can start in with test values. For example, x equals minus four, zero and

2:15:13

to work. If I plug those values into this expression here, I get a negative answer a

2:15:20

negative answer and a positive answer. The reason I need to conclude the values on my

2:15:26

number line where my denominators zero is because I can my expression can switch from

2:15:32

negative to positive by passing through a place where my rational expression doesn't

2:15:36

exist, as well as passing by passing flew through a place where my rational expression

2:15:41

is equal to zero.

2:15:43

Now I'm looking for where my original expression was less than or equal to zero. So that means

2:15:49

I want the places on the number line where my expression is equal to zero, and also the

2:15:56

places where it's negative.

2:15:59

So my final answer is x is less than one, or an interval notation, negative infinity

2:16:06

to one.

2:16:08

In this video, we solved polynomial and rational inequalities by making a number line and you

2:16:14

think test values to make a sine chart.

2:16:19

The distance formula can be used to find the distance between two points, if we know their

2:16:26

coordinates,

2:16:27

if this first point has coordinates, x one, y one, and the second point has coordinates

2:16:34

x two, y two, then the distance between them is given by the formula, the square root of

2:16:42

x two minus x one squared plus y two minus y one squared.

2:16:49

This formula actually comes from the Pythagorean Theorem, let me draw a right triangle with

2:16:55

these two points as two of its vertices,

2:17:01

then the length of this side is the difference between the 2x coordinates. Similarly, the

2:17:09

length of this vertical side is the difference between the two y coordinates.

2:17:14

Now that Pythagoras theorem says that for any right triangle, with sides labeled A and

2:17:20

B, and hypotony is labeled C, we have that a squared plus b squared equals c squared.

2:17:27

Well, if we apply that to this triangle here,

2:17:31

this hypotony News is the distance between the two points that we're looking for.

2:17:38

So but Tyrion theorem says, the square of this side length, that is x two minus x one

2:17:45

squared, plus the square of this side length, which is y two minus y one squared has to

2:17:55

equal the square of the hypothesis, that is d squared.

2:18:01

taking the square root of both sides of this equation, we get the square root of x two

2:18:07

minus x one squared plus y two minus y one squared is equal to d,

2:18:13

we don't have to worry about using plus or minus signs when we take the square root because

2:18:18

distance is always positive.

2:18:21

So we've derived our distance formula. Now let's use it as an example.

2:18:26

Let's find the distance between the two points, negative one five, for negative one, five,

2:18:34

maybe over here, and for two,

2:18:39

this notation just means that P is the point with coordinates negative one five, and q

2:18:48

is the point with coordinates for two.

2:18:52

We have the distance formula,

2:18:54

let's think of P being the point with coordinates x one y one, and Q being the point with coordinates

2:19:02

x two y two.

2:19:04

But as we'll see, it really doesn't matter which one is which, plugging into our formula,

2:19:10

we get d is the square root of n x two minus x one. So that's four minus negative one squared,

2:19:23

and then we add

2:19:26

y two minus y one, so that's two minus five squared.

2:19:37

Working out the arithmetic a little bit for minus negative one, that's four plus one or

2:19:41

five squared, plus negative three squared. So that's the square root of 25 plus nine

2:19:49

are the square root of 34. Let's see what would have happened if we'd call this first

2:19:55

point, x two y two instead, and a second.

2:20:00

Why'd x one y one, then we would have gotten the same distance formula, but we would have

2:20:06

taken negative one minus four

2:20:12

and added the difference of y's squared. So that's why two five minus two squared.

2:20:22

That gives us the square root of negative five squared plus three squared, which works

2:20:27

out to the same thing.

2:20:31

In this video, we use the distance formula to find the distance between two points. When

2:20:36

writing down the distance formula, sometimes students forget whether this is a plus or

2:20:40

a minus and whether these are pluses or minuses. One way to remember is to think that the distance

2:20:45

formula comes from the Pythagorean Theorem. That's why there's a plus in here.

2:20:51

And then a minus in here, because we're finding the difference between the coordinates to

2:20:56

find the lengths of the sides.

2:20:59

The midpoint formula helps us find the coordinates of the midpoint of a line segment, as long

2:21:05

as we know the coordinates of the endpoints.

2:21:09

So let's call the coordinates of this endpoint x one, y one, and the coordinates of this

2:21:15

other endpoint x two, y two,

2:21:17

the x coordinate of the midpoint is going to be exactly halfway in between the x coordinates

2:21:25

of the endpoints. To get a number halfway in between two other numbers, we just take

2:21:31

the average.

2:21:34

Similarly, the y coordinate of this midpoint is going to be exactly halfway in between

2:21:42

the y coordinates of the endpoints. So the y coordinate of the midpoint is going to be

2:21:47

the average of those y coordinates.

2:21:53

So we see that the coordinates of the midpoint are x one plus x two over two, y one plus

2:22:00

y two over two.

2:22:04

Let's use this midpoint formula in an example,

2:22:07

we want to find the midpoint of the segment between the points, negative one, five,

2:22:17

and four

2:22:20

to

2:22:24

me draw the line segment between them. So visually, the midpoint is going to be somewhere

2:22:29

around here. But to find its exact coordinates, we're going to use x one plus x two over two,

2:22:36

and y one plus y two over two, where this first point is coordinates, x one, y one,

2:22:44

and the second point has coordinates x two y two, it doesn't actually matter which point

2:22:49

you decide is x one y one, which one is x two, y two, the formula will still give you

2:22:54

the same answer for the midpoint.

2:22:56

So let's see, I take my average of my x coordinates. So that's negative one, plus four over two,

2:23:06

and the average of my Y coordinates, so that's five plus two over two, and I get three halves,

2:23:15

seven halves, as the coordinates of my midpoint.

2:23:19

In this video, we use the midpoint formula to find the midpoint of a line segment, just

2:23:25

by taking the average of the x coordinates and the average of the y coordinates. This

2:23:30

video is about graphs and equations of circles. Suppose we want to find the equation of a

2:23:36

circle of radius five, centered at the point three, two

2:23:42

will look something like this.

2:23:47

For any point, x, y on the circle, we know that

2:23:53

the distance of that point xy from the center is equal to the radius that is five from the

2:24:00

distance formula, that distance of five is equal to the square root of the difference

2:24:07

of the x coordinates. That's x minus three

2:24:11

squared plus the difference of the y coordinates, that's y minus two squared.

2:24:20

And if I square both sides of that equation, I get five squared is this square root squared.

2:24:31

In other words, 25 is equal to x minus three squared plus y minus two squared. Since the

2:24:40

square root and the squared undo each other.

2:24:43

A lot of times people will write the X minus three squared plus y minus two squared on

2:24:49

the left side of the equation and the 25 on the right side of the equation. This is the

2:24:53

standard form for the equation of the circle.

2:24:58

The same reasoning can be generalized

2:24:59

Find the general equation for a circle with radius R centered at a point HK.

2:25:07

For any point with coordinates, x, y on the circle, the distance between the point with

2:25:14

coordinates x, y and the point with coordinates HK is equal to the radius r. So we have that

2:25:22

the distance is equal to r, but by the distance formula, that's the square root of the difference

2:25:28

between the x coordinates x minus h squared plus the difference in the y coordinates y

2:25:36

minus k squared. squaring both sides as before, we get r squared is the square root squared,

2:25:48

canceling the square root and the squared, and rearranging the equation, that gives us

2:25:53

x minus h squared plus y minus k squared equals r squared. That's the general formula for

2:26:01

a circle with radius r, and center HK.

2:26:06

Notice that the coordinates h and k are subtracted here, the two squares are added because they

2:26:13

are in the distance formula, and the radius is squared on the other side, if you remember

2:26:18

this general formula, that makes it easy to write down the equation for a circle, for

2:26:23

example, if we want the equation for a circle,

2:26:29

of radius six, and center at zero, negative three,

2:26:35

then we have our equal six, and h k is our zero negative three. So plugging into the

2:26:45

formula,

2:26:47

we get x minus zero squared, plus y minus negative three squared equals six squared,

2:26:57

or simplified this is x squared plus y plus three squared equals 36.

2:27:04

Suppose we're given an equation like this one, and we want to decide if it's the equation

2:27:09

of a circle, and if so what's the center was the radius.

2:27:13

Well, this equation matches the form for a circle, x minus h squared plus y minus k squared

2:27:21

equals r squared. If we let h, b five, Why be negative sex since that way, subtracting

2:27:30

a negative six is like adding a six,

2:27:34

five must be our r squared. So that means that R is the square root of five. So this

2:27:41

is our radius. And our center is the point five, negative six. And this is indeed the

2:27:49

equation for a circle,

2:27:52

which we could then graph by putting down the center and estimating the radius, which

2:27:57

is a little bit more than two.

2:28:03

This equation might not look like the equation of a circle, but it actually can be transformed.

2:28:08

To look like the equation of a circle, we're trying to make it look like something of the

2:28:13

form x minus h squared plus y minus k squared equals r squared.

2:28:19

First, I'd like to get rid of the coefficients in front of the x squared and the y squared,

2:28:24

so I'm going to divide everything by nine. This gives me x squared plus y squared plus

2:28:30

8x minus two y plus four equals zero. Next, I'm going to group the x terms together the

2:28:38

x squared and the 8x. So write x squared plus 8x. I'll group the y terms together the Y

2:28:45

squared minus two y.

2:28:49

And I'll subtract the four over to the other side.

2:28:54

This still doesn't look much like the equation of a circle. But as the next step, I'm going

2:28:59

to do something called completing the square, I'm going to take this coefficient of x eight

2:29:05

divided by two and then square it. In other words, eight over two squared, that's four

2:29:12

squared, which is 16. I'm going to add 16 to both sides of my equation.

2:29:18

So the 16 here,

2:29:22

and on the other side,

2:29:24

now I'm going to do the same thing to the coefficient of y, negative two divided by

2:29:31

two is negative one, square that and I get one. So now I'm going to add a one to both

2:29:38

sides, but I'll put it near the y terms. So add a one there. And then to keep things balanced,

2:29:44

I have to add it to the other side. On the right side, I have the number 13. On the left

2:29:50

side, I can wrap up this expression x squared plus 8x plus 16 into x plus four squared.

2:29:59

To convince you, that's correct. If we multiply out x plus four times x plus four, we get

2:30:07

x squared plus 4x plus 4x plus 16, or x squared plus 8x plus 16, the same thing we have right

2:30:16

here. Similarly, we can wrap up y squared minus two y plus one as

2:30:25

y minus one squared. Again, I'll work out the distribution down here, that's y squared

2:30:32

minus y minus y plus one, or y squared minus two y plus one, just like we have up here.

2:30:40

If you're wondering how I knew to use four and minus one, the four came from taking half

2:30:46

of eight, and the minus one came from taking half the negative two. Now we have an equation

2:30:53

for a circle and standard form. And we can read off the center, which is negative four,

2:31:01

negative one, and the radius, which is the square root of 13.

2:31:10

It might seem like magic that this trick of taking half of this coefficient and squaring

2:31:14

it and adding it to both sides lets us wrap up this expression into this expression a

2:31:20

perfect square. But to see why that works, let's take a look at what happens if we expand

2:31:26

out x minus h squared, the thing that we're trying to get to. So if we expand that out,

2:31:33

x minus h squared, which is x minus h times x minus h is, which multiplies out to x squared

2:31:43

minus HX minus h x plus h squared, or x squared minus two h x plus h squared. Now if we start

2:31:53

out with this part, with some x squared term and some coefficient of x, we're trying to

2:32:01

decide what to add on in order to wrap it up into x minus h squared. The thing to add

2:32:06

on is h squared here, which comes from half of this coefficient squared, right, because

2:32:13

half of negative two H is a negative H, square that you get the H squared. And then when

2:32:18

we do wrap it up, it's half of this coefficient of x, that appears right here.

2:32:26

This trick of completing the square is really handy for turning an equation for a circle

2:32:30

in disguise, into the standard equation for the circle. In this video, we found the standard

2:32:37

equation for a circle x minus h squared plus y minus k squared equals r squared, where

2:32:44

r is the radius, and h k is the center.

2:32:52

We also showed a method of completing the square.

2:32:56

When you have an equation for a circle in disguise, completing the square will help

2:33:01

you rewrite it into the standard form.

2:33:06

This video is about graphs and equations of lines.

2:33:10

Here we're given the graph of a line, we want to find the equation, one standard format

2:33:16

for the equation of a line is y equals mx plus b. here am represents the slope, and

2:33:25

B represents the y intercept, the y value where the line crosses the y axis, the slope

2:33:34

is equal to the rise over the run. Or sometimes this is written as the change in y values

2:33:41

over the change in x values. Or in other words, y two minus y one over x two minus x one,

2:33:49

where x one y one and x two y two are points on the line.

2:33:56

While we could use any two points on the line, to find the slope, it's convenient to use

2:34:01

points where the x and y coordinates are integers, that is points where the line passes through

2:34:08

grid points. So here would be one convenient point to use. And here's another convenient

2:34:13

point to use.

2:34:15

The coordinates of the first point are one, two, and the next point, this is let's say,

2:34:22

five, negative one. Now I can find the slope by looking at the rise over the run. So as

2:34:29

I go through a run of this distance, I go through a rise of that distance especially

2:34:35

gonna be a negative rise or a fall because my line is pointing down. So let's see counting

2:34:41

off squares. This is a run of 1234 squares and a rise of 123. So negative three, so my

2:34:48

slope is going to be

2:34:52

negative three over four.

2:34:56

I got that answer by counting squares, but I could have also gotten it by

2:35:00

Looking at the difference in my y values over the difference in my x values, that is, I

2:35:05

could have done

2:35:08

negative one minus two, that's from my difference in Y values, and divide that by my difference

2:35:15

in x values,

2:35:18

which is five minus one, that gives me negative three over four, as before.

2:35:26

So my M is negative three fourths.

2:35:30

Now I need to figure out the value of b, my y intercept, well, I could just read it off

2:35:36

the graph, it looks like approximately 2.75. But if I want to be more accurate, I can again

2:35:41

use a point that has integer coordinates that I know it's exact coordinates. So either this

2:35:47

point or that point, let's try this point. And I can start off with my equation y equals

2:35:53

mx plus b, that is y equals negative three fourths x plus b. And I can plug in the point

2:36:02

one, two, for my x and y. So that gives me two equals negative three fourths times one

2:36:12

plus b, solving for B. Let's see that's two equals negative three fourths plus b. So add

2:36:20

three fourths to both sides, that's two plus three fourths equals b. So b is eight fourths

2:36:28

plus three fourths, which is 11. Four switches actually, just what I eyeballed it today.

2:36:33

So now I can write out my final equation for my line y equals negative three fourths x

2:36:40

plus 11 fourths by plugging in for m and b. Next, let's find the equation for this horizontal

2:36:48

line.

2:36:49

a horizontal line has slope zero. So if we think of it as y equals mx plus b, m is going

2:36:57

to be zero. In other words, it's just y equals b, y is some constant. So if we can figure

2:37:03

out what that that constant y value is, it looks like it's

2:37:08

two, let's see this three, three and a half, we can just write down the equation directly,

2:37:14

y equals 3.5.

2:37:18

For a vertical line, like this one, it doesn't really have a slope. I mean, if you tried

2:37:24

to do the rise over the run,

2:37:26

there's no run. So you'd I guess you'd be dividing by zero and get an infinite slope.

2:37:32

But But instead, we just think of it as an equation of the form x equals something. And

2:37:39

in this case, x equals negative two, notice that all of the points on our line have the

2:37:45

same x coordinate of negative two and the y coordinate can be anything.

2:37:50

So this is how we write the equation for a vertical line.

2:37:54

In this example, we're not shown a graph of the line, we're just get told that it goes

2:37:59

through two points. But knowing that I go through two points is enough to find the equation

2:38:04

for the line. First, we can find the slope by taking the difference in Y values over

2:38:12

the difference in x values. So that's negative three minus two over four minus one, which

2:38:20

is negative five thirds.

2:38:24

So we can use the

2:38:26

standard equation for the line, this is called the slope intercept form.

2:38:34

And we can plug in negative five thirds. And we can use one point, either one will do will

2:38:42

still get the same final answer. So let's use one two and plug that in to get two equals

2:38:48

negative five thirds times one plus b. And so B is two plus five thirds, which is six

2:38:57

thirds plus five thirds, which is 11 thirds. So our equation is y equals negative five

2:39:03

thirds x plus 11 thirds.

2:39:07

This is method one.

2:39:11

method two uses a slightly different form of the equation, it's called the point slope

2:39:17

form and it goes y minus y naught is equal to m times x minus x naught, where x naught

2:39:25

y naught is a point on the line and again is the slope. So we calculate the slope the

2:39:33

same way by taking a difference in Y values over a difference in x values. But then, we

2:39:40

can simply plug in any point.

2:39:43

For example, the point one two will work we can plug one in for x naught and two in for

2:39:53

Y not in this point slope form. That gives us y

2:39:59

minus

2:40:00

Two is equal to minus five thirds x minus one.

2:40:06

Notice that these two equations, while they may look different, are actually equivalent,

2:40:11

because if I distribute the negative five thirds, and then add the two to both sides,

2:40:25

I get the same equation as above.

2:40:28

So we've seen two ways of finding the equation for the line using the slope intercept form.

2:40:37

And using the point slope form.

2:40:40

In this video, we saw that you can find the equation for a line, if you know the slope.

2:40:48

And you know one point,

2:40:52

you can also find the equation for the line if you know two points, because you can use

2:40:57

the two points to get the slope and then plug in one of those points. To figure out the

2:41:03

rest of the equation.

2:41:06

We saw two standard forms for the equation of a line

2:41:08

the slope intercept form y equals mx plus b, where m is the slope, and B is the y intercept.

2:41:19

And the point slope form y minus y naught equals m times x minus x naught, where m again

2:41:27

is the slope and x naught y naught is a point on the line.

2:41:35

This video is about finding parallel and perpendicular lines. Suppose we have a line of slope three

2:41:43

fourths, in other words, the rise

2:41:47

over the run

2:41:50

is three over four,

2:41:52

than any other line that's parallel to this line will have the same slope, the same fries

2:42:01

over run.

2:42:03

So that's our first fact to keep in mind, parallel lines have the same slopes. Suppose

2:42:09

on the other hand, we want to find a line perpendicular to our original line with its

2:42:15

original slope of three fourths.

2:42:18

A perpendicular line, in other words, align at a 90 degree angle to our original line

2:42:25

will have a slope, that's the negative reciprocal of the opposite reciprocal of our original

2:42:32

slope. So we take the reciprocal of three for us, that's four thirds, and we make it

2:42:37

its opposite by changing it from positive to negative.

2:42:41

So I'll write this as a principle that perpendicular lines have opposite reciprocal slopes, to

2:42:49

get the hang of what it means to be an opposite reciprocal, let's look at a few examples.

2:42:54

So here's the original slope, and this will be the opposite reciprocal, which would represent

2:43:01

the slope of our perpendicular line. So for example, if one was to the opposite reciprocal,

2:43:07

reciprocal of two is one half opposite means I change the sign, if the first slope turned

2:43:14

out to be, say, negative 1/3, the reciprocal of 1/3 is three over one or just three. And

2:43:23

the opposite means I change it from a negative to a positive, so my perpendicular slope would

2:43:29

be would be three.

2:43:32

One more example, if I started off with a slope of, say, seven halves, then the reciprocal

2:43:38

of that would be two sevenths. And I change it to an opposite reciprocal, by changing

2:43:43

the positive to a negative.

2:43:47

Let's use these two principles in some examples.

2:43:52

In our first example, we need to find the equation of a line that's parallel to this

2:43:56

slump, and go through the point negative three two. Well, in order to get started, I need

2:44:02

to figure out the slope of this line. So let me put this line into a more standard form

2:44:07

the the slope intercept form. So I'll start with three y minus 4x plus six equals zero,

2:44:14

I'm going to solve for y to put it in this this more standard form. So I'll say three

2:44:19

y equals 4x minus six and then divide by three to get y equals four thirds x minus six thirds

2:44:29

is divided all my turns by three, I can simplify that a little bit y equals four thirds x minus

2:44:34

two. Now I can read off the slope of my original line, and one is four thirds. That means my

2:44:43

slope of my new line, my parallel line will also be four thirds. So my new line will have

2:44:51

the equation y equals four thirds ax plus b for some B, of course, B will probably not

2:44:58

be negative two like it was for the

2:45:00

first line, it'll be something else determined by the fact that it goes through this point

2:45:04

negative three to, to figure out what b is, I plug in the point negative three to four,

2:45:10

negative three for x, and two for y. So that gives me two equals four thirds times negative

2:45:16

three plus b, and I'll solve for b. So let's see. First, let me just simplify. So two equals,

2:45:25

this is negative 12 thirds plus b, or in other words, two is negative four plus b. So that

2:45:32

means that B is going to be six, and by my parallel line will have the equation y equals

2:45:38

four thirds x plus six.

2:45:42

Next, let's find the equation of a line that's perpendicular to a given line through and

2:45:49

go through a given point. Again, in order to get started, I need to find what the slope

2:45:54

of this given line is. So I'll rewrite it, well, first, I'll just copy it over 6x plus

2:45:59

three, y equals four. And then I can put it in the standard slope intercept form by solving

2:46:06

for y. So three, y is negative 6x plus four, divide everything by three, I get y equals

2:46:14

negative six thirds x plus four thirds. So y is negative 2x, plus four thirds. So my

2:46:22

original slope of my original line is negative two, which means my perpendicular slope is

2:46:28

the opposite reciprocal, so I take the reciprocal of of negative two, that's negative one half

2:46:34

and I change the sign so that gives me one half as the slope of my perpendicular line.

2:46:39

Now, my new line, I know is going to be y equals one half x plus b for some B, and I

2:46:47

can plug in my point on my new line, so one for y and four for x, and solve for b, so

2:46:57

I get one equals two plus b, So b is negative one. And so my new line has equation one half

2:47:04

x minus one.

2:47:07

These next two examples are a little bit different, because now we're looking at a line parallel

2:47:14

to a completely horizontal line, let me draw this line y equals three, so the y coordinate

2:47:23

is always three, which means that my line will be a horizontal line through that height

2:47:29

at height y equals three, if I want something parallel to this line, it will also be a horizontal

2:47:36

line. Since it goes through the point negative two, one, C, negative two, one has to go through

2:47:42

that point there,

2:47:45

it's gonna have always have a y coordinate of one the same as the y coordinate of the

2:47:51

point it goes through, so my answer will just be y equals one.

2:47:55

In the next example, we want a line that's perpendicular to the line y equals four another

2:48:06

horizontal line y equals four, but perpendicular means I'm going to need a vertical line. So

2:48:12

I need a vertical line that goes through the point three, four.

2:48:16

Okay, and so I'm going to draw a vertical line there. Now vertical lines have the form

2:48:27

x equals something for the equation. And to find what x equals, I just need to look at

2:48:32

the x coordinate of the point I'm going through this point three four, so that x coordinate

2:48:38

is three and all the points on this, this perpendicular and vertical line have X coordinate

2:48:45

three so my answer is x equals three.

2:48:53

In this video will use the fact that parallel lines have the same slope and perpendicular

2:48:58

lines have opposite reciprocal slopes, to find the equations of parallel and perpendicular

2:49:03

lines.

2:49:05

This video introduces functions and their domains and ranges.

2:49:09

A function is a correspondence between input numbers usually the x values and output numbers,

2:49:17

usually the y values, that sends each input number to exactly one output number. Sometimes

2:49:25

a function is thought of as a rule or machine

2:49:28

in which you can feed in x values as input and get out y values as output. So, non mathematical

2:49:38

example of a function might be the biological mother function, which takes as input any

2:49:44

person and it gives us output their biological mother.

2:49:49

This function satisfies the condition that each input number object person in this case,

2:49:57

get sent to exactly one output per

2:50:01

Because if you take any person, they just have one biological mother. So that rule does

2:50:07

give you a function. But if I change things around and just use the the mother function,

2:50:14

which sends to each person, their mother, that's no longer going to be a function because

2:50:19

there are some people who have more than one mother, right, they could have a biological

2:50:23

mother and adopted mother, or a mother and a stepmother, or any number of situations.

2:50:28

So since there's, there's at least some people who you would put it as input, and then you'd

2:50:33

get like more than one possible output that violates this, this rule of functions, that

2:50:39

would not be a function. Now, most of the time, we'll work with functions that are described

2:50:43

with equations, not in terms of mothers. So for example, we can have the function y equals

2:50:48

x squared plus one.

2:50:51

This can also be written as f of x equals x squared plus one. Here, f of x is function

2:50:57

notation that stands for the output value of y.

2:51:02

Notice that this notation is not representing multiplication, we're not multiplying f by

2:51:07

x, instead, we're going to be putting in a value for x as input and getting out a value

2:51:14

of f of x or y. For example, if we want to evaluate f of two, we're plugging in two as

2:51:21

input for x either in this equation, or in that equation. Since F of two means two squared

2:51:30

plus one, f of two is going to equal five. Similarly, f of five means I plug in five

2:51:38

for x, so that's gonna be five squared plus one, or 26. Sometimes it's useful to evaluate

2:51:46

a function on a more complicated expression involving other variables. In this case, remember,

2:51:52

the functions value on any expression, it's just what you what you get when you plug in

2:51:57

that whole expression for x.

2:52:00

So f of a plus three is going to be the quantity a plus three squared plus one,

2:52:09

we could rewrite that as a squared plus six a plus nine plus one, or a squared plus six

2:52:17

a plus 10.

2:52:19

When evaluating a function on a complex expression, it's important to keep the parentheses when

2:52:25

you plug in for x. That way, you evaluate the function on the whole expression. For

2:52:32

example, it would be wrong

2:52:35

to write f of a plus three equals a plus three squared plus one without the parentheses,

2:52:42

because that would imply that we were just squaring the three and not the whole expression,

2:52:47

a plus three.

2:52:52

Sometimes a function is described with a graph instead of an equation. In this example, this

2:52:58

graph is supposed to represent the function g of x. Not all graphs actually represent

2:53:04

functions. For example, the graph of a circle doesn't represent a function. That's because

2:53:11

the graph of a circle violates the vertical line test, you can draw a vertical line and

2:53:15

intersect the graph in more than one point.

2:53:18

But our graph, it left satisfies the vertical line test, any vertical line intersects the

2:53:24

graph, and at most one point, that means is a function, because every x value will have

2:53:29

at most one y value that corresponds to it. Let's evaluate g have to note that two is

2:53:37

an x value. And we'll use the graph to find the corresponding y value. So I look for two

2:53:44

on the x axis, and find the point on the graph that has that x value that's right here. Now

2:53:51

I can look at the y value of that point looks like three, and therefore gf two is equal

2:53:58

to three. If I try to do the same thing to evaluate g of five, I run into trouble. Five

2:54:05

is an x value, I look for it on the x axis, but there's no point on the graph that has

2:54:11

that x value.

2:54:13

Therefore, g of five is undefined, or we can say it does not exist. The question of what

2:54:21

x values and y values make sense for a function leads us to a discussion of domain and range.

2:54:27

The domain of a function is all possible x values that makes sense for that function.

2:54:32

The range is the y values that make sense for the function.

2:54:36

In this example, we saw that the x value of five didn't have a corresponding y value for

2:54:42

this function. So the x value of five is not in the domain of our function J. To find the

2:54:49

x values in the domain, we have to look at the x values that correspond to points on

2:54:54

the graph. One way to do that is to take the shadow or projection of the graph

2:55:00

onto the x axis and see what x values are hit. It looks like we're hitting all x values

2:55:07

starting at negative eight, and continuing up to four. So our domain

2:55:15

is the x's between negative eight, and four, including those endpoints. Or we can write

2:55:21

this in interval notation as negative eight comma four with square brackets.

2:55:27

To find the range of the function, we look for the y values corresponding to points on

2:55:31

this graph, we can do that.

2:55:35

By taking the shadow or projection of the graph onto the y axis,

2:55:42

we seem to be hitting our Y values from negative five, up through three.

2:55:51

So our range is wise between negative five and three are an interval notation negative

2:55:59

five, three with square brackets. If we meet a function that's described as an equation

2:56:05

instead of a graph, one way to find the domain and range are to graph the function. But it's

2:56:11

often possible to find the domain at least more quickly, by using algebraic considerations.

2:56:17

We think about what x values that makes sense to plug into this expression, and what x values

2:56:22

need to be excluded, because they make the algebraic expression impossible to evaluate.

2:56:27

Specifically, to find the domain of a function, we need to exclude x values that make the

2:56:35

denominator zero. Since we can't divide by zero,

2:56:38

we also need to exclude x values that make an expression inside a square root sign negative.

2:56:47

Since we can't take the square root of a negative number. In fact, we need to exclude values

2:56:52

that make the expression inside any even root negative because we can't take an even root

2:56:58

of a negative number, even though we can take an odd root like a cube root of a negative

2:57:02

number. Later, when we look at logarithmic functions, we'll have some additional exclusions

2:57:07

that we have to make. But for now, these two principles should handle all functions We'll

2:57:12

see. So let's apply them to a couple examples.

2:57:16

For the function in part A, we don't have any square root signs, but we do have a denominator.

2:57:23

So we need to exclude x values that make the denominator zero. In other words, we need

2:57:28

x squared minus 4x plus three to not be equal to zero.

2:57:34

If we solve x squared minus 4x plus three equal to zero, we can do that by factoring.

2:57:42

And that gives us x equals three or x equals one, so we need to exclude these values.

2:57:48

All other x values should be fine. So if I draw the number line, I can put on one and

2:57:54

three and just dig out a hole at both of those. And my domain includes everything else on

2:58:00

the number line. In interval notation, this means my domain is everything from negative

2:58:06

infinity to one, together with everything from one to three, together with everything

2:58:12

from three to infinity.

2:58:15

In the second example, we don't have any denominator to worry about, but we do have a square root

2:58:21

sign. So we need to exclude any x values that make three minus 2x less than zero. In other

2:58:29

words, we can include all x values for which three minus 2x is greater than or equal to

2:58:35

zero. Solving that inequality gives us three is greater than equal to 2x. In other words,

2:58:43

x

2:58:44

is less than or equal to three halves,

2:58:48

I can draw this on the number line,

2:58:51

or write it in interval notation.

2:58:54

Notice that three halves is included, and that's because three minus 2x is allowed to

2:59:00

be zero, I can take the square root of zero, that's just zero. And that's not a problem.

2:59:05

Finally, let's look at a more complicated function that involves both the square root

2:59:10

and denominator. Now there are two things I need to worry about.

2:59:14

I need the denominator

2:59:18

to not be equal to zero, and I need the stuff inside the square root side to be greater

2:59:24

than or equal to zero. from our earlier work, we know the first condition means that x is

2:59:30

not equal to three, and x is not equal to one. And the second condition means that x

2:59:36

is less than or equal to three halves. Let's try both of those conditions on the number

2:59:41

line.

2:59:44

x is not equal to three and x is not equal to one means we've got everything except those

2:59:51

two dug out points. And the other condition x is less than or equal to three halves means

2:59:56

we can have three halves and everything

3:00:00

To the left of it. Now to be in our domain and to be legit for our function, we need

3:00:05

both of these conditions to be true. So I'm going to connect these conditions with N and

3:00:09

N, that means we're looking for a numbers on the number line that are colored both red

3:00:13

and blue. So I'll draw that above in purple. So that's everything from three halves to

3:00:19

one, I have to dig out one because one was a problem for the denominator. And then I

3:00:24

can continue for all the things that are colored both both colors red and blue. So my final

3:00:30

domain is going to be, let's see negative infinity, up to but not including one together

3:00:37

with one, but not including it to three halves, and I include three half since that was colored,

3:00:44

both red and blue. Also,

3:00:46

in this video, we talked about functions, how to evaluate them, and how to find domains

3:00:51

and ranges. This video gives the graphs of some commonly used functions that I call the

3:00:57

toolkit functions. The first function is the function y equals x, let's plot a few points

3:01:03

on the graph of this function. If x is zero, y zero, if x is one, y is one, and so on y

3:01:13

is always equal to x doesn't have to just be an integer, it can be any real number,

3:01:19

and we'll connecting the dots, we get a straight line through the origin.

3:01:24

Let's look at the graph of y equals x squared. If x is zero, y is zero. So we go through

3:01:30

the origin again, if x is one, y is one, and x is negative one, y is also one,

3:01:40

the x value of two gives a y value of four and the x value of negative two gives a y

3:01:45

value of four also connecting the dots, we get a parabola.

3:01:51

That is this, this function is an even function.

3:01:56

That means it has mirror symmetry across the y axis, the left side it looks like exactly

3:02:02

like the mirror image of the right side. That happens because when you square a positive

3:02:08

number, like two, you get the exact same y value as when you square it's a mirror image

3:02:14

x value of negative two.

3:02:17

The next function y equals x cubed. I'll call that a cubic. Let's plot a few points when

3:02:24

x is zero, y is zero. When x is one, y is one, when x is negative one, y is negative

3:02:31

one, two goes with the point eight way up here and an x value of negative two is going

3:02:37

to give us negative eight. If I connect the dots, I get something that looks kind of like

3:02:42

this.

3:02:45

This function is what's called an odd function, because it has 180 degree rotational symmetry

3:02:53

occur around the origin. If I rotate this whole graph by 180 degrees, or in other words,

3:02:58

turn the paper upside down, I'll get exactly the same shape. The reason this function has

3:03:04

this odd symmetry is because when I cube a positive number, and get its y value, I get

3:03:12

n cube the corresponding negative number to get its y value, the negative number cubed

3:03:17

gives us exactly the negative of the the y value we get with cubing the positive number.

3:03:25

Let's look at the next example. Y equals the square root of x.

3:03:29

Notice that the domain of this function is just x values bigger than or equal to zero

3:03:37

because we can't take the square root of a negative number. Let's plug in a few x values.

3:03:42

X is zero gives y is 0x is one square root of one is one, square root of four is two,

3:03:49

and connecting the dots, I get a function that looks like this.

3:03:55

The absolute value function is next. Again, if we plug in a few points, x is zero goes

3:04:01

with y equals 0x is one gives us one, the absolute value of negative one is 122 is on

3:04:11

the graph, and the absolute value of negative two is to I'm ending up getting a V shaped

3:04:17

graph. It also has even or a mirror symmetry.

3:04:24

Y equals two to the x is what's known as an exponential function. That's because the variable

3:04:30

x is in the exponent. If I plot a few points,

3:04:35

two to the zero is one,

3:04:41

two to the one is two, two squared is four, two to the minus one is one half. I'll plot

3:04:49

these on my graph, you know, let me fill in a few more points. So let's see. two cubed

3:04:55

is eight. That's way up here and negative two gives

3:04:59

Maybe 1/4 1/8 connecting the dots, I get something that's shaped like this.

3:05:07

You might have heard the expression exponential growth, when talking about, for example, population

3:05:11

growth, this is function is represents exponential growth, it grows very, very steeply. Every

3:05:19

time we increase the x coordinate by one, we double the y coordinate.

3:05:26

We could also look at a function like y equals three dx, or sometimes y equals e to the x,

3:05:32

where E is just a number about 2.7. These functions look very similar. It's just the

3:05:40

bigger bass makes us rise a little more steeply.

3:05:45

Now let's look at the function y equals one over x. It's not defined when x is zero, but

3:05:52

I can plug in x equals one half, one over one half is to

3:05:58

whenever one is one, and one or two is one half. By connect the dots, I get this shape

3:06:06

in the first quadrant, but I haven't looked at negative values of x, when x is negative,

3:06:11

whenever negative one is negative one, whenever negative half is negative two, and I get a

3:06:18

similar looking piece in the third quadrant.

3:06:22

This is an example of a hyperbola.

3:06:27

And it's also an odd function, because it has that 180 degree rotational symmetry, if

3:06:34

I turn the page upside down, it'll look exactly the same. Finally, let's look at y equals

3:06:40

one over x squared.

3:06:42

Again, it's not defined when x is zero, but I can plug in points like one half or X, let's

3:06:48

see one over one half squared is one over 1/4, which is four.

3:06:55

And one over one squared, one over two squared is a fourth, it looks pretty much like the

3:07:03

previous function is just a little bit more extreme rises a little more steeply, falls

3:07:08

a little more dramatically. But for negative values of x, something a little bit different

3:07:12

goes on. For example, one over negative two squared is just one over four, which is a

3:07:19

fourth. So I can plot that point there, and one over a negative one squared, that's just

3:07:27

one. So my curve for negative values of x is gonna lie in the second quadrant, not the

3:07:32

third quadrant. This is an example of an even function

3:07:38

because it has perfect mirror symmetry across the y axis.

3:07:42

These are the toolkit functions, and I recommend you memorize the shape of each of them.

3:07:49

That way, you can draw at least a rough sketch very quickly without having to plug in points.

3:07:56

That's all for the graphs of the toolkit functions.

3:07:58

If we change the equation of a function, then the graph of the function changes or transforms

3:08:07

in predictable ways.

3:08:09

This video gives some rules and examples for transformations of functions. To get the most

3:08:14

out of this video, it's helpful if you're already familiar with the graph of some typical

3:08:19

functions, I call them toolkit functions like y equals square root of x, y equals x squared,

3:08:25

y equals the absolute value of x and so on. If you're not familiar with those graphs,

3:08:30

I encourage you to watch my video called toolkit functions first, before watching this one.

3:08:36

I want to start by reviewing function notation.

3:08:39

If g of x represents the function, the square root of x, then we can rewrite these expressions

3:08:46

in terms of square roots. For example, g of x minus two is the same thing as the square

3:08:53

root of x minus two.

3:08:55

g of quantity x minus two means we plug in x minus two, everywhere we see an x, so that

3:09:05

would be the same thing as the square root of quantity x minus two.

3:09:11

In this second example, I say that we're subtracting two on the inside of the function, because

3:09:16

we're subtracting two before we apply the square root function. Whereas on the first

3:09:21

example, I say that the minus two is on the outside of the function, we're doing the square

3:09:26

root function first and then subtracting two. In this third example, g of 3x, we're multiplying

3:09:33

by three on the inside of the function. To evaluate this in terms of square root, we

3:09:38

plug in the entire 3x for x and the square root function. That gives us the square root

3:09:44

of 3x.

3:09:45

In the next example, we're multiplying by three on the outside of the function. This

3:09:51

is just three times the square root of x. Finally, g of minus x means the square root

3:09:58

of minus x.

3:10:00

Now, this might look a little odd because we're not used to taking the square root of

3:10:03

a negative number. But remember that if x itself is negative, like negative to the negative

3:10:09

x will be negative negative two or positive two. So we're really be taking the square

3:10:13

root of a positive number in that case, let me record which of these are inside and which

3:10:18

of these are outside of my function.

3:10:21

In this next set of examples, we're using the same function g of x squared of x. But

3:10:26

this time, we're starting with an expression with square roots in it, and trying to write

3:10:29

it in terms of g of x. So the first example, the plus 17 is on the outside of my function,

3:10:37

because I'm taking the square root of x first, and then adding 17. So I can write this as

3:10:43

g of x plus 17.

3:10:46

In the second example, I'm taking x and adding 12 first, then I take the square root of the

3:10:53

whole thing. Since I'm adding the 12 to x first that's on the inside of my function.

3:10:58

So I write that as g of the quantity x plus 12.

3:11:04

Remember that this notation means I plug in the entire x plus 12, into the square root

3:11:09

sign, which gives me exactly square root of x plus 12. And this next example, undoing

3:11:14

the square root first and then multiplying by negative 36. So i minus 36, multiplications,

3:11:20

outside my function, I can rewrite this as minus 36 times g of x. Finally, in this last

3:11:27

example, I take x multiplied by a fourth and then apply the square root of x. So that's

3:11:33

the same thing as g of 1/4 X, my 1/4 X is on the inside of my function. In other words,

3:11:40

it's inside the parentheses when I use function notation.

3:11:44

Let's graph the square root of x and two transformations have this function,

3:11:50

y equals the square root of x goes to the point 0011. And for two, since the square

3:11:56

root of four is two, it looks something like this.

3:12:01

In order to graph y equals the square root of x minus two, notice that the minus two

3:12:06

is on the outside of the function, that means we're going to take the square root of x first

3:12:10

and then subtract two. So for example, if we start with the x value of zero, and compute

3:12:17

the square root of zero, that's zero, then we subtract two to give us a y value of negative

3:12:21

two,

3:12:23

an x value of one, which under the square root function had a y value of one now has

3:12:29

a y value that's decreased by two, one minus two is negative one. And finally, an x value

3:12:36

of four, which under the square root function had a y value of two now has a y value of

3:12:42

two minus two or zero, its y value is also decreased by two. If I plot these new points,

3:12:49

zero goes with negative two, one goes with negative one, and four goes with zero, I have

3:12:56

my transformed graph.

3:13:00

Because I subtracted two on the outside of my function, my y values were decreased by

3:13:04

two, which brought my graph down by two units. Next, let's look at y equals the square root

3:13:12

of quantity x minus two. Now we're subtracting two on the inside of our function, which means

3:13:17

we subtract two from x first and then take the square root. In order to get the same

3:13:21

y value of zero as we had in our blue graph, we need our x minus two to be zero, so we

3:13:28

need our original x to be two.

3:13:30

In order to get the y value of one that we had in our blue graph, we need to be taking

3:13:36

the square root of one, so we need x minus two to be one, which means that we need to

3:13:41

start with an x value of three.

3:13:43

And in order to reproduce our y value of two from our original graph, we need the square

3:13:50

root of x minus two to be two, which means we need to start out by taking the square

3:13:55

root of four, which means our x minus two is four, so our x should be up there at six.

3:14:04

If I plot my x values, with my corresponding y values of square root of x minus two, I

3:14:10

get the following graph. Notice that the graph has moved horizontally to the right by two

3:14:18

units.

3:14:19

To me, moving down by two units, makes sense because we're subtracting two, so we're decreasing

3:14:25

y's by two units, but the minus two on the inside that kind of does the opposite of what

3:14:30

I expect, I might expected to to move the graph left, I might expect the x values to

3:14:35

be going down by two units, but instead, it moves the graph to the right,

3:14:41

because the x units have to go up by two units in order to get the right square root when

3:14:46

I then subtract two units again, the observations we made for these transformations of functions

3:14:52

hold in general, according to the following rules. First of all, numbers on the outside

3:14:58

of the function like

3:14:59

In our example, y equals a squared of x minus two, those numbers affect the y values. And

3:15:06

a result in vertical motions, like we saw, these motions are in the direction you expect.

3:15:12

So subtracting two was just down by two. If we were adding two instead, that would move

3:15:18

us up by two

3:15:19

numbers on the inside of the function. That's like our example, y equals the square root

3:15:24

of quantity x minus two, those affect the x values and results in a horizontal motion,

3:15:31

these motions go in the opposite direction from what you expect. Remember, the minus

3:15:35

two on the inside actually shifted our graph to the right, if it had had a plus two on

3:15:40

the inside, that would actually shift our graph to the left.

3:15:45

Adding results in a shift those are called translations, but multiplying like something

3:15:51

like y equals three times the square root of x, that would result in a stretch or shrink.

3:15:59

In other words, if I start with the square root of x,

3:16:03

and then when I graph y equals three times the square root of x, that stretches my graph

3:16:08

vertically by a factor of three.

3:16:12

Like this,

3:16:13

if I want to graph y equals 1/3, times the square root of x, that shrinks my graph vertically

3:16:20

by a factor of 1/3.

3:16:22

Finally, a negative sign results in reflection. For example, if I start with the graph of

3:16:29

y equals the square root of x, and then when I graph y equals the square root of negative

3:16:34

x, that's going to do a reflection in the horizontal direction because the negative

3:16:39

is on the inside of the square root sign.

3:16:42

A reflection in the horizontal direction means a reflection across the y axis.

3:16:49

If instead, I want to graph y equals negative A squared of x, that negative sign on the

3:16:53

outside means a vertical reflection, a reflection across the x axis.

3:17:00

Pause the video for a moment and see if you could describe what happens in these four

3:17:06

transformations.

3:17:07

In the first example, we're subtracting four on the outside of the function. adding or

3:17:12

subtracting means a translation or shift. And since we're on the outside of the function

3:17:18

affects the y value, so that's moving us vertically. So this transformation should take the square

3:17:24

root of graph and move it down by four units, that would look something like this.

3:17:31

In the next example, we're adding 12 on the inside, that's still a translation. But now

3:17:39

we're moving horizontally. And so since we go the opposite direction, we expect we are

3:17:44

going to go to the left by 12 units, that's going to look something like this.

3:17:51

And the next example, we're multiplying by three and introducing a negative sign both

3:17:55

on the outside of our function outside our function means we're affecting the y values.

3:18:00

So in multiplication means we're stretching by a factor of three, the negative sign means

3:18:06

we reflect in the vertical direction,

3:18:10

here's stretching by a factor of three vertically, before I apply the minus sign, and now the

3:18:15

minus sign reflects in the vertical direction.

3:18:19

Finally, in this last example, we're multiplying by one quarter on the inside of our function,

3:18:26

we know that multiplication means stretch or shrink. And since we're on the inside,

3:18:31

it's a horizontal motion, and it does the opposite of what we expect. So instead of

3:18:36

shrinking by a factor of 1/4, horizontally, it's actually going to stretch by the reciprocal,

3:18:41

a factor of four horizontally.

3:18:44

that'll look something like this. Notice that stretching horizontally by a factor of four

3:18:50

looks kind of like shrinking vertically by a factor of one half.

3:18:55

And that's actually borne out by the algebra, because the square root of 1/4 x is the same

3:19:01

thing as the square root of 1/4 times the square root of x, which is the same thing

3:19:05

as one half times the square root of x. And so now we can see algebraically that vertical

3:19:12

shrink by a factor of one half is the same as a horizontal stretch by a factor of four,

3:19:18

at least for this function, the square root function.

3:19:22

This video gives some rules for transformations of functions, which I'll repeat below. numbers

3:19:27

on the outside correspond to changes in the y values, or vertical motions.

3:19:37

numbers on the inside of the function, affect the x values, and result in horizontal motions.

3:19:44

Adding and subtracting

3:19:47

corresponds to translations or shifts.

3:19:52

multiplying and dividing by numbers corresponds to stretches and shrinks

3:19:57

and putting in a negative sign.

3:20:01

corresponds to a reflection,

3:20:04

horizontal reflection, if the negative sign is on the inside, and a vertical reflection,

3:20:10

if the negative sign is on the outside,

3:20:13

knowing these basic rules about transformations empowers you to be able to sketch graphs of

3:20:17

much more complicated functions, like y equals three times the square root of x plus two,

3:20:24

by simply considering the transformations, one at a time.

3:20:29

a quadratic function is a function that can be written in the form f of x equals A x squared

3:20:35

plus bx plus c, where a, b and c are real numbers. And a is not equal to zero. The reason

3:20:46

we require that A is not equal to zero is because if a were equal to zero, we'd have

3:20:53

f of x is equal to b x plus c, which is called a linear function. So by making sure that

3:21:01

a is not zero, we make sure there's really an x squared term which is the hallmark of

3:21:06

a quadratic function.

3:21:08

Please pause the video for a moment and decide which of these equations represent quadratic

3:21:13

functions.

3:21:15

The first function can definitely be written in the form g of x equals A x squared plus

3:21:23

bx plus c. fact it's already written in that form, where a is negative five, B is 10, and

3:21:30

C is three.

3:21:32

The second equation is also a quadratic function, because we can think of it as f of x equals

3:21:39

one times x squared plus zero times x plus zero.

3:21:46

So it is in the right form, where A is one, B is zero, and C is zero.

3:21:53

It's perfectly fine for the coefficient of x and the constant term to be zero for a quadratic

3:21:59

function, we just need the coefficient of x squared to be nonzero, so the x squared

3:22:03

term is preserved.

3:22:06

The third equation is not a quadratic function. It's a linear function, because there's no

3:22:13

x squared term.

3:22:17

The fourth function might not look like a quadratic function. But if we rewrite it by

3:22:21

expanding out the X minus three squared, let's see what happens. We get y equals two times

3:22:28

x minus three times x minus three plus four. So that's two times x squared minus 3x minus

3:22:36

3x, plus nine plus four, continuing, I get 2x squared,

3:22:45

minus 12 access, plus 18 plus four, in other words, y equals 2x squared minus 12x plus

3:22:53

22. So in fact, our function can be written in the right form, and it is a quadratic function.

3:23:02

A function that is already written in the form y equals A x squared plus bx plus c is

3:23:08

said to be in standard form.

3:23:12

So our first example g of x is in standard form

3:23:17

a function that's written in the format of the last function that is in the form of y

3:23:23

equals a times x minus h squared plus k for some numbers, a, h, and K. That said to be

3:23:33

in vertex form,

3:23:35

I'll talk more about standard form and vertex form in another video.

3:23:40

In this video, we identified some quadratic functions, and talked about the difference

3:23:45

between standard form

3:23:49

and vertex form.

3:23:54

This video is about graphing quadratic functions. a quadratic function, which is typically written

3:23:59

in standard form, like this, or sometimes in vertex form, like this, always has the

3:24:08

graph that looks like a parabola.

3:24:11

This video will show how to tell whether the problem is pointing up or down. How to find

3:24:17

its x intercepts, and how to find its vertex.

3:24:21

The bare bones basic quadratic function is f of x equals x squared, it goes to the origin,

3:24:30

since f of zero is zero squared, which is zero, and it is a parabola pointing upwards

3:24:38

like this.

3:24:40

The vertex of a parabola is its lowest point if it's pointing upwards, and its highest

3:24:46

point if it's pointing downwards. So in this case, the vertex is at 00.

3:24:53

The x intercepts are where the graph crosses the x axis. In other words, where y is zero

3:24:58

In this function, y equals zero means that x squared is zero, which happens only when

3:25:05

x is zero. So the x intercept, there's only one of them is also zero.

3:25:11

The second function, y equals negative 3x squared also goes to the origin. Since the

3:25:17

functions value when x is zero, is y equals zero. But in this case, the parabola is pointing

3:25:25

downwards.

3:25:27

That's because thinking about transformations of functions, a negative sign on the outside

3:25:32

reflects the function vertically over the x axis, making the problem instead of pointing

3:25:38

upwards, reflecting the point downwards.

3:25:43

The number three on the outside stretches the graph vertically by a factor of three.

3:25:49

So it makes it kind of long and skinny like this.

3:25:55

In general, a negative coefficient to the x squared term means the problem will be pointing

3:26:00

down. Whereas a positive coefficient, like here, the coefficients, one means the parabola

3:26:05

is pointing up.

3:26:06

Alright, that roll over here.

3:26:09

So if a is bigger than zero, the parabola opens up.

3:26:16

And if the value of the coefficient a is less than zero, then the parabola opens down.

3:26:22

In this second example, we can see again that the vertex is at 00. And the x intercept is

3:26:27

x equals zero.

3:26:30

Let's look at this third example.

3:26:33

If we multiplied our expression out, we'd see that the coefficient of x squared would

3:26:38

be to a positive number. So that means our parabola is going to be opening up.

3:26:43

But the vertex of this parabola will no longer be at the origin. In fact, we can find the

3:26:50

parabolas vertex by thinking about transformations of functions.

3:26:55

Our function is related to the function y equals to x squared, by moving it to the right

3:27:02

by three and up by four. Since y equals 2x squared has a vertex at 00. If we move that

3:27:13

whole parabola including the vertex, right by three, and up by four, the vertex will

3:27:21

end up at the point three, four.

3:27:26

So a parabola will look something like this.

3:27:30

Notice how easy it was to just read off the vertex when our quadratic function is written

3:27:36

in this form. In fact, any parabola any quadratic function written in the form a times x minus

3:27:44

h squared plus k has a vertex at h k. By the same reasoning, we're moving the parabola

3:27:53

with a vertex at the origin to the right by H, and by K.

3:27:59

That's why this form of a quadratic function is called the vertex form.

3:28:06

Notice that this parabola has no x intercepts because it does not cross the x axis.

3:28:11

For our final function, we have g of x equals 5x squared plus 10x plus three,

3:28:19

we know the graph of this function will be a parabola pointing upwards, because the coefficient

3:28:23

of x squared is positive.

3:28:26

To find the x intercepts, we can set y equals zero, since the x intercepts is where our

3:28:32

graph crosses the x axis, and that's where the y value is zero.

3:28:37

So zero equals 5x squared plus 10x plus three, I'm going to use the quadratic formula to

3:28:43

solve that. So x is negative 10 plus or minus the square root of 10 squared minus four times

3:28:52

five times three, all over two times five. That simplifies to x equals negative 10 plus

3:28:59

or minus the square root of 40 over 10,

3:29:03

which simplifies further to x equals negative 10 over 10, plus or minus the square root

3:29:08

of 40 over 10, which is negative one plus or minus two square root of 10 over 10, or

3:29:16

negative one plus or minus square root of 10 over five.

3:29:21

Since the square root of 10 is just a little bit bigger than three, this works out to approximately

3:29:27

about negative two fifths and negative eight foot or so somewhere around right here. So

3:29:34

our parabola is going to look something like this. Notice that it goes through crosses

3:29:40

the x axis at y equals three, that's because when we plug in x equals zero, and two here

3:29:48

we get y equal three, so the y intercept is at three. Finally, we can find the vertex.

3:29:56

Since this function is written in standard form,

3:30:00

On the Y equals a x plus a squared plus bx plus c form, and not in vertex form, we can't

3:30:10

just read off the vertex like we couldn't before. But there's a trick called the vertex

3:30:15

formula, which says that whenever you have a function in a quadratic function in standard

3:30:20

form, the vertex has an x coordinate

3:30:24

of negative B over two A. So in this case, that's an x coordinate of negative 10 over

3:30:34

two times five or negative one, which is kind of like where I put it on the graph. To find

3:30:39

the y coordinate of that vertex, I can just plug in negative one into my equation for

3:30:47

x, which gives me at y equals five times negative one squared, plus 10, times negative one plus

3:30:54

three, which is negative two.

3:30:59

So I think I better redraw my graph a little bit to put that vertex down at a y coordinate

3:31:04

of negative two where it's supposed to be.

3:31:10

Let's summarize the steps we use to graph these quadratic functions. First of all, the

3:31:15

graph of a quadratic function has the shape of a parabola.

3:31:19

The parabola opens up, if the coefficient of x squared, which we'll call a is greater

3:31:27

than zero and down if a is less than zero. To find the x intercepts, we set y equal to

3:31:34

zero, or in other words, f of x equal to zero and solve for x, the find the vertex, we can

3:31:44

either read it off as h k, if our function is in vertex form,

3:31:53

or we can use the vertex formula

3:31:57

and get the x coordinate

3:32:01

of the vertex to be negative B over to a if our function is in standard form.

3:32:10

To find the y coordinate of the vertex in this case, we just plug in the x coordinate

3:32:19

and figure out what y is.

3:32:21

Finally, we can always find additional points on the graph by plugging in values of x.

3:32:29

In this video, we learned some tricks for graphing quadratic functions. In particular,

3:32:34

we saw that the vertex can be read off as h k, if our function is written in vertex

3:32:43

form, and the x coordinate of the vertex can be calculated

3:32:50

as negative B over two A, if our function is in standard form. For an explanation of

3:32:57

why this vertex formula works, please see the my other video.

3:33:02

a quadratic and standard form looks like y equals A x squared plus bx plus c, where a,

3:33:09

b and c are real numbers, and a is not zero. a quadratic function in vertex form looks

3:33:16

like y equals a times x minus h squared plus k, where a h and k are real numbers, and a

3:33:24

is not zero. When a functions in vertex form, it's easy to read off the vertex is just the

3:33:30

ordered pair, H, okay.

3:33:34

This video explains how to get from vertex form two standard form and vice versa.

3:33:40

Let's start by converting this quadratic function from vertex form to standard form. That's

3:33:46

pretty straightforward, we just have to distribute out.

3:33:50

So if I multiply out the X minus three squared,

3:33:58

I get minus four times x squared minus 6x plus nine plus one, distributing the negative

3:34:07

four, I get negative 4x squared plus 24x minus 36 plus one. So that works out to minus 4x

3:34:16

squared plus 24x minus 35. And I have my quadratic function now in standard form.

3:34:26

Now let's go the other direction and convert a quadratic function that's already in standard

3:34:31

form into vertex form. That is, we want to put it in the form of g of x equals a times

3:34:40

x minus h squared plus k, where the vertex is at h k. To do this, it's handy to use the

3:34:51

vertex formula.

3:34:53

The vertex formula says that the x coordinate of the vertex is given by negative

3:35:00

over to a, where A is the coefficient of x squared, and B is the coefficient of x. So

3:35:08

in this case, we get an x coordinate of negative eight over two times two or negative two.

3:35:16

To find the y coordinate of the vertex, we just plug in the x coordinate into our formula

3:35:24

for a g of x. So that's g of negative two, which is two times negative two squared plus

3:35:30

eight times negative two plus six. And that works out to be negative two by coincidence.

3:35:37

So the vertex for our quadratic function has coordinates negative two, negative two. And

3:35:44

if I want to write g of x in vertex form, it's going to be a times x minus negative

3:35:50

two squared plus minus two. That's because remember, we subtract H and we add K. So that

3:36:00

simplifies to g of x equals a times x plus two squared minus two. And finally, we just

3:36:08

need to figure out what this leading coefficient as. But notice, if we were to multiply, distribute

3:36:14

this out, then the coefficient of x squared would end up being a. So therefore, the coefficient

3:36:23

of x squared here, which is a has to be the same as the coefficient of x squared here,

3:36:27

which we conveniently also called a, in other words, are a down here needs to be two. So

3:36:33

I'm going to write that as g of x equals two times x plus two squared minus two, lots of

3:36:39

twos in this problem. And that's our quadratic function in vertex form. If I want to check

3:36:46

my answer, of course, I could just distribute out again, I'd get two times x squared plus

3:36:53

4x, plus two, minus two, in other words, 2x squared plus 8x plus six, which checks out

3:37:03

to exactly what I started with. This video showed how to get from vertex form to standard

3:37:09

form by distributing out and how to get from standard form to vertex form by finding the

3:37:14

vertex using the vertex formula.

3:37:17

Suppose you have a quadratic function in the form y equals x squared plus bx plus c, and

3:37:26

you want to find where the vertex is, when you graph it.

3:37:31

The vertex formula says that the x coordinate of this vertex is that negative B over two

3:37:40

A,

3:37:42

this video gives a justification for where that formula comes from.

3:37:47

Let's start with a specific example. Suppose I wanted to find the x intercepts and the

3:37:53

vertex for this quadratic function. To find the x intercepts, I would set y equal to zero

3:38:00

and solve for x. So that's zero equals 3x squared plus 7x minus five. And to solve for

3:38:07

x, I use the quadratic formula. So x is going to be negative seven plus or minus the square

3:38:13

root of seven squared minus four times three times negative five, all over two times three.

3:38:22

That simplifies to negative seven plus or minus a squared of 109 over six, we could,

3:38:29

I could also write this as negative seven, six, plus the square root of 109 over six,

3:38:35

or x equals negative seven, six minus the square root of 109 over six, since the square

3:38:41

root of 109 is just a little bit bigger than 10, this can be approximated by negative seven

3:38:47

six plus 10, six, and negative seven, six minus 10, six.

3:38:54

So pretty close to, I guess about what half over here and pretty close to about negative

3:39:00

three over here, I'm just going to estimate so that I can draw a picture of the function.

3:39:06

Since the leading coefficient three is positive, I know my parabola is going to be opening

3:39:11

up and the intercepts are somewhere around here and here. So roughly speaking, it's gonna

3:39:20

look something like this.

3:39:25

Now the vertex is going to be somewhere in between the 2x intercepts, in fact, it's going

3:39:30

to be by symmetry, it'll be exactly halfway in between the 2x intercepts. Since the x

3:39:37

intercepts are negative seven, six plus and minus 100 squared of 109 over six, the number

3:39:43

halfway in between those is going to be exactly negative seven, six,

3:39:47

right, because on the one hand, I have negative seven six plus something. And on the other

3:39:56

hand, I have negative seven six minus that same thing.

3:39:59

So negative seven, six will be exactly in the middle. So my x coordinate of my vertex

3:40:08

will be at negative seven sex. Notice that I got that number from the quadratic formula.

3:40:16

More generally, if I want to find the x intercepts for any quadratic function, I set y equal

3:40:24

to zero

3:40:27

and solve for x using the quadratic formula, negative b plus or minus the square root of

3:40:33

b squared minus four AC Oliver to a, b, x intercepts will be at these two values, but

3:40:40

the x coordinate of the vertex, which is exactly halfway in between the 2x intercepts will

3:40:47

be at negative B over two A. That's where the vertex formula comes from.

3:40:53

And it turns out that this formula works even when there are no x intercepts, even when

3:40:58

the quadratic formula gives us no solutions. That vertex still has the x coordinate, negative

3:41:04

B over two A.

3:41:06

And that's the justification of the vertex formula.

3:41:11

This video is about polynomials and their graphs.

3:41:14

We call that a polynomial is a function like this one, for example.

3:41:20

His terms are numbers times powers of x. I'll start with some definitions. The degree of

3:41:28

a polynomial is the largest exponent. For example, for this polynomial, the degree is

3:41:35

for

3:41:37

the leading term is the term with the largest exponent. In the same example, the leading

3:41:44

term is 5x to the fourth, it's conventional to write the polynomial in descending order

3:41:51

of powers of x. So the leading term is first. But the leading term doesn't have to be the

3:41:56

first term. If I wrote the same polynomial as y equals, say, two minus 17x minus 21x

3:42:04

cubed plus 5x. Fourth, the leading term would still be the 5x to the fourth, even though

3:42:11

it was last.

3:42:13

The leading coefficient is the number in the leading term. In this example, the leading

3:42:20

coefficient is five.

3:42:21

Finally, the constant term is the term with no x's in it. In our same example, the constant

3:42:28

term is to

3:42:29

please pause the video for a moment and take a look at this next example of polynomial

3:42:34

figure out what's the degree the leading term, the leading coefficient and the constant term.

3:42:40

The degree is again, four, since that's the highest power we have, and the leading term

3:42:48

is negative 7x. to the fourth, the leading coefficient is negative seven, and the constant

3:42:56

term is 18.

3:42:58

In the graph of the polynomial Shown here are the three marks points are called turning

3:43:04

points, because the polynomial turns around and changes direction at those three points,

3:43:12

those same points can also be called local extreme points, or they can be called local

3:43:18

maximum and minimum points. For this polynomial, the degree is four, and the number of turning

3:43:25

points is three.

3:43:28

Let's compare the degree and the number of turning points for these next three examples.

3:43:33

For the first one, the degree is to and there's one turning point.

3:43:39

For the second example, that agree, is three, and there's two turning points.

3:43:50

And for this last example, the degree is four, and there's one turning point.

3:43:56

For this first example, and the next two, the number of turning points is exactly one

3:44:01

less than the degree. So you might conjecture that this is always true. But in fact, this

3:44:07

is not always true. In this last example, the degree is four, but the number of turning

3:44:13

points is one, not three.

3:44:16

In fact, it turns out that while the number of turning points is not always equal to a

3:44:21

minus one, it is always less than or equal to the degree minus one. That's a useful factor.

3:44:28

Remember, when you're sketching graphs are recognizing graphs of polynomials.

3:44:33

The end behavior of a function is how the ends of the function look, as x gets bigger

3:44:39

and bigger heads towards infinity, or x gets goes through larger and larger negative numbers

3:44:45

towards negative infinity.

3:44:47

In this first example, the graph of the function goes down as x gets towards infinity as x

3:44:54

goes towards negative infinity. I can draw this with two little arrows pointing

3:44:59

Down on either side, or I can say in words, that the function is falling as we had left

3:45:05

and falling also as we had right.

3:45:09

In the second example, the graph rises to the left and rises to the right. In the third

3:45:16

example, the graph falls to the left, but rises to the right. And in the fourth example,

3:45:23

it rises to the left and falls to the right. If you study these examples, and others, you

3:45:29

might notice there's a relationship between the degree of the polynomial the leading coefficients

3:45:36

of the polynomials and the end behavior. Specifically, these four types of end behavior are determined

3:45:44

by whether the degree is even or odd. And by whether the leading coefficient is positive

3:45:51

or negative.

3:45:54

When the degree is even, and the leading coefficient is positive, like in this example, where the

3:46:00

leading coefficient is one, we have this sorts of n behavior rising on both sides.

3:46:08

When the degree is even at the leading coefficient is negative, like in this example, we have

3:46:14

the end behavior that's falling on both sides

3:46:18

when the degree is odd, and the leading coefficient is positive, that's like this example, with

3:46:23

the degree three and the leading coefficient three, then we have this sort of end behavior.

3:46:30

And finally, when the degree is odd, and the leading coefficient is negative, like in this

3:46:35

example, we have this sort of NBA havior.

3:46:40

I like to remember this chart just by thinking of the most simple examples, y equals x squared,

3:46:48

y equals negative x squared, y equals x cubed, and y equals negative x cubed. Because I know

3:46:55

by heart what those four examples look like,

3:47:00

then I just have to remember that any polynomial with a even degree and positive leading coefficient

3:47:08

has the same end behavior as x squared.

3:47:11

And similarly, any polynomial with even degree and negative leading coefficient has the same

3:47:18

end behavior as negative x squared. And similar statements for x cubed and negative x cubed.

3:47:25

We can use facts about turning points and behavior, to say something about the equation

3:47:30

of a polynomial just by looking at this graph.

3:47:34

In this example, because of the end behavior, we know that the degree is odd,

3:47:42

we know that the leading coefficient

3:47:46

must be positive.

3:47:49

And finally, since there are 1234, turning points, we know that the degree is greater

3:47:58

than or equal to five.

3:48:01

That's because the number of turning points is less than or equal to the degree minus

3:48:06

one. And in this case, the number of turning points we said was four.

3:48:10

And so solving that inequality, we get the degree is bigger than equal to five.

3:48:18

Put in some of that information together, we see the degree could be

3:48:23

five, or seven, or nine, or any odd number greater than or equal to five. But it couldn't

3:48:31

be for example, three or six. Because even numbers and and also numbers less than five

3:48:38

are right out.

3:48:41

This video gave a lot of definitions, including the definition of degree

3:48:46

leading coefficients,

3:48:49

turning points,

3:48:51

and and behavior.

3:48:54

We saw that knowing the degree and the leading coefficient can help you make predictions

3:48:59

about the number of turning points and the end behavior, as well as vice versa.

3:49:06

This video is about exponential functions and their graphs.

3:49:11

an exponential function is a function that can be written in the form f of x equals a

3:49:16

times b to the x, where a and b are any real numbers. As long as a is not zero, and B is

3:49:24

positive.

3:49:26

It's important to notice that for an exponential function, the variable x is in the exponent.

3:49:33

This is different from many other functions we've seen, for example, quite a quadratic

3:49:37

function like f of x equals 3x squared has the variable in the base, so it's not an exponential

3:49:44

function.

3:49:45

For exponential functions, f of x equals a times b dx. We require that A is not equal

3:49:53

to zero, because otherwise, we would have f of x equals zero times b dx.

3:50:00

Which just means that f of x equals zero. And this is called a constant function, not

3:50:04

an exponential function. Because f of x is always equal to zero

3:50:11

in an exponential function, but we require that B is bigger than zero, because otherwise,

3:50:17

for example, if b is equal to negative one, say, we'd have f of x equals a times negative

3:50:27

one to the x. Now, this would make sense for a lot of values of x. But if we try to do

3:50:34

something like f of one half, with our Bs, negative one, that would be the same thing

3:50:40

as a times the square root of negative one, which is not a real number,

3:50:45

we'd get the same problem for other values of b, if the values of b were negative. And

3:50:52

if we tried b equals zero, we'd get a kind of ridiculous thing a times zero to the x,

3:50:59

which again is always zero. So that wouldn't count as an exponential either. So we can't

3:51:03

use any negative basis, and we can't use zero basis, if we want an exponential function.

3:51:09

The number A in the expression f of x equals a times b to the x is called the initial value.

3:51:18

And the number B is called the base.

3:51:21

The phrase initial value comes from the fact that if we plug in x equals zero, we get a

3:51:28

times b to the zero, well, anything to the zero is just one. So this is equal to A. In

3:51:34

other words, f of zero equals a. So if we think of starting out,

3:51:40

when x equals zero, we get the y value

3:51:44

of a, that's why it's called the initial value.

3:51:51

Let's start out with this example, where y equals a times b to the x next special function,

3:51:59

and we've set a equal to one and B equals a two. Notice that the y intercept, the value

3:52:06

when x is zero, is going to be one.

3:52:09

If I change my a value, my initial value, the y intercept changes, the function is stretched

3:52:20

out.

3:52:21

If I make the value of a go to zero, and then negative,

3:52:27

then my initial value becomes negative, and my graph flips across the x axis. Let's go

3:52:35

back to an a value of say one, and see what happens when we change B. Right now, the B

3:52:41

value the basis two, if I increase B, my y intercept sticks at one, but my graph becomes

3:52:52

steeper and steeper. If I put B back down close to one, my graph becomes more flat at

3:53:03

exactly one, my graph is just a constant.

3:53:05

As B gets into fractional territory, point 8.7 point six, my graph starts to slope the

3:53:14

other way, it's decreasing now instead of increasing, but notice that the y intercept

3:53:20

still hasn't changed, I can get it more and more steep as my B gets farther and farther

3:53:26

away from one of course, when B goes to negative territory, my graph doesn't make any sense.

3:53:33

So a changes the y intercept, and B changes the steepness of the graph. So that it's added

3:53:44

whether it's increasing for B values bigger than one, and decreasing for values of the

3:53:50

less than one.

3:53:55

we'll summarize all these observations on the next slide.

3:53:58

We've seen that for an exponential function, y equals a times b to the x, the parameter

3:54:04

or number a gives the y intercept, the parameter B tells us how the graph is increasing or

3:54:13

decreasing. Specifically, if b is greater than one, the graph is increasing.

3:54:19

And if b is less than one, the graph is decreasing.

3:54:23

The closer B is to the number one, the flatter the graph.

3:54:29

So for example, if I were to graph y equals point two five to the x and y equals point

3:54:37

four to the x, they would both be decreasing graphs, since the base for both of them is

3:54:44

less than one. But point two five is farther away from one and point four is closer to

3:54:52

one. So point four is going to be flatter. And point two five is going to be more steep.

3:54:58

So in this picture, This red graph would correspond to point two

3:55:03

five to the x, and the blue graph would correspond to point four to the x. For all these exponential

3:55:11

functions, whether the graphs are decreasing or increasing, they all have a horizontal

3:55:17

asymptote along the x axis. In other words, at the line at y equals zero, the domain is

3:55:25

always from negative infinity to infinity, and the range is always from zero to infinity,

3:55:34

because the range is always positive y values. Actually, that's true. If a is greater than

3:55:43

zero, if a is less than zero, then our graph flips over the x axis, our domain stays the

3:55:52

same, but our range becomes negative infinity to zero. The most famous exponential function

3:55:59

is f of x equals e to the x. This function is also sometimes written as f of x equals

3:56:04

x of x. The number E is Oilers number as approximately 2.7. This function has important applications

3:56:13

to calculus and to some compound interest problems. In this video, we looked at exponential

3:56:21

functions, functions of the form a times b to the x, where the variable is in the exponent,

3:56:28

we saw that they all have the same general shape, either increasing like this, or decreasing

3:56:37

like this, unless a is negative, in which case they flip over the x axis. They all have

3:56:45

a horizontal asymptote at y equals zero, the x axis. In this video, we'll use exponential

3:56:55

functions to model real world examples. Let's suppose you're hired for a job. The starting

3:57:02

salary is $40,000. With a guaranteed annual raise of 3% each year, how much will your

3:57:10

salary be after one year, two years, five years. And in general after two years, let

3:57:16

me chart out the information. The left column will be the number of years since you're hired.

3:57:23

And the right column will be your salary. When you start work, at zero years after you're

3:57:30

hired, your salary will be $40,000. After one year, you'll have gotten a 3% raise. So

3:57:40

your salary will be the original 40,000 plus 3% of 40,000.03 times 40,000. I can think

3:57:52

of this first number as one at times 40,000. And I can factor out the 40,000 from both

3:58:00

terms, to get 40,000 times one plus 0.03. we rewrite this as 40,000 times 1.03. This

3:58:15

is your original salary multiplied by a growth factor of 1.03. After two years, you'll get

3:58:26

a 3% raise from your previous year salary, your previous year salary was 40,000 times

3:58:33

1.03.

3:58:34

But you'll add

3:58:36

3% of that, again, I can think of the first number is one times 40,000 times 1.03. And

3:58:45

I can factor out the common factor of 40,000 times 1.03 from both terms, to get 40,000

3:58:53

times 1.03 times one plus 0.03. Let me rewrite that as 40,000 times 1.03 times 1.03 or 40,000

3:59:08

times 1.3 squared. We can think of this as your last year salary multiplied by the same

3:59:19

growth factor of 1.03. After three years, a similar computation will give you that your

3:59:26

new salary is your previous year salary times that growth factor of 1.03 that can be written

3:59:37

as 40,000 times 1.03 cubed. And in general, if you're noticing the pattern, after two

3:59:45

years, your salary should be 40,000 times 1.03 to the t power. In other words, your

3:59:53

salary after two years is your original salary multiplied By the growth factor of 1.03, taken

4:00:06

to the t power, let me write this as a formula s of t, where S of t is your salary is equal

4:00:15

to 40,000 times 1.03 to the T. This is an exponential function, that is a function of

4:00:24

the form a times b to the T, where your initial value a is 40,000 and your base b is 1.03.

4:00:38

Notice that your base is the amount that your salary gets multiplied by each year. Given

4:00:45

this formula, we can easily figure out what your salary will be after, for example, five

4:00:49

years by plugging in five for T. I worked that out on my calculator to be $46,370.96

4:01:01

to the nearest cent. exponential functions are also useful in modeling population growth.

4:01:08

The United Nations estimated that the world population in 2010 was 6.7 9 billion growing

4:01:15

at a rate of 1.1% per year. Assuming that the growth rate stayed the same and will continue

4:01:20

to stay the same, we'll write an equation for the population at t years after the year

4:01:26

2010 1.1%, written as a decimal is 0.011. So if we work out a chart as before, we see

4:01:39

that after zero years since 2010, we have our initial population 6.7 9 billion, after

4:01:46

one year, we'll take that 6.7 9 billion and add 1.1% of it. That is point 011 times 6.79.

4:01:59

This works out to 6.79 times one plus point 011 or 6.79 times 1.011. Here we have our

4:02:15

initial population of 6.7 9 billion, and our growth factor of 1.011. That's how much the

4:02:27

population got multiplied by in one year. As before, we can work out that after two

4:02:35

years, our population becomes 6.79 times 1.011 squared, since they got multiplied by 1.011,

4:02:45

twice, and after two years, it'll be 6.79 times 1.011 to the t power. So our function

4:02:56

that models population is going to be 6.79 times 1.011 to the T. Here, t represents time

4:03:09

in years, since 2010. Just for fun, I'll plug in t equals 40. That's 40 years since 2010.

4:03:21

So that's the year 2050. And I get 6.79 times 1.011 to the 40th power, which works out to

4:03:31

10 point 5 billion. That's the prediction based on this exponential model.

4:03:40

The previous two examples were examples of exponential growth. This last example is an

4:03:45

example of exponential decay. The drugs Seroquel is metabolized and eliminated from the body

4:03:51

at a rate of 11% per hour. If 400 milligrams are given how much remains in the body. 24

4:03:58

hours later, I'll chart out my information where the left column will be time in hours

4:04:09

since the dose was given, and the right column will be the number of milligrams of Seroquel

4:04:15

still on the body. zero hours after the dose is given, we have the full 400 milligrams

4:04:22

in the body. One hour later, we have the formula 100 milligrams minus 11% of it, that's minus

4:04:30

point one one times 400. If I factor out the 400 from both terms, I get 400 times one minus

4:04:40

0.11 or 400 times point eight nine. The 400 represents the initial amount. The point eight

4:04:53

nine I'll call the growth factor, even though in this case the quantity is decreasing, not

4:04:58

growing. So really it's kind of a shrink factor. Let's see what happens after two hours. Now

4:05:08

I'll have 400 times 0.89 my previous amount, it'll again get multiplied by point eight,

4:05:15

nine, so that's going to be 400 times 0.89 squared. And in general, after t hours, I'll

4:05:22

have 400 multiplied by this growth or shrinkage factor of point eight nine, raised to the

4:05:29

t power. Since each hour, the amount of Seroquel gets multiplied by point eight, nine, that

4:05:36

number less than one. All right, my exponential decay function as f of t equals 400 times

4:05:44

0.89 to the T, where f of t represents the number of milligrams of Seroquel in the body.

4:05:54

And t represents the number of hours since the dose was given. To find out how much is

4:06:01

in the body after 24 hours, I just plug in 24 for t. This works out to about 24.4 milligrams,

4:06:13

I hope you notice the common form for the functions used to model all three of these

4:06:17

examples. The functions are always in the form f of t equals a times b to the T, where

4:06:26

a represented the initial amount, and B represented the growth factor. To find the growth factor

4:06:36

B, we started with the percent increase or decrease. We wrote it as a decimal. And then

4:06:45

we either added or subtracted it from one, depending on whether the quantity was increasing

4:06:50

or decreasing. Let me show you that as a couple examples. In the first example, we had a percent

4:06:58

increase of 3% on the race as a decimal, I'll call this r, this was point O three. And to

4:07:09

get the growth factor, we added that to one to get 1.03. In the world population example,

4:07:16

we had a 1.1% increase, we wrote this as point 011 and added it to one to get 1.011. In the

4:07:29

drug example, we had a decrease of 11%. We wrote that as a decimal. And we subtracted

4:07:42

the point one one from one to get 0.89. In fact, you can always write the growth factor

4:07:50

B as one plus the percent change written as a decimal. If you're careful to make that

4:07:58

percent change, negative when the quantity is decreasing and positive when the quantity

4:08:04

is increasing. Since here one plus negative point 11

4:08:09

gives us the correct growth factor of point eight nine which is less than one as a good

4:08:14

check. Remember that if your quantity is increasing, the base b should be bigger than one. And

4:08:23

if the quantity is decreasing, then B should be less than one.

4:08:30

exponential functions can also be used to model bank accounts and loans with compound

4:08:35

interest, as we'll see in another video. This video is about interpreting exponential functions.

4:08:42

An antique car is worth $50,000 now, and its value increases by 7%. Each year, let's write

4:08:49

an equation to model its value x years from now. After one year, it's value is the 50,000

4:09:00

plus 0.07 times the 50,000. That's because its value has grown by 7% or point oh seven

4:09:10

times 50,000. this can be written as 50,000 times one plus point O seven. Notice that

4:09:20

adding 7% to the original value is the same as multiplying the original value by one plus

4:09:28

point oh seven or by 1.07. After two years, the value will be 50,000 times one plus 0.07

4:09:41

squared, or 50,000 times 1.07 squared. That's because the previous year's value is multiplied

4:09:51

again by 1.7. In general, after x years We have x, the antique cars value will be 50,000

4:10:06

times 1.07 to the x. That's because the original value of 50,000 gets multiplied by 1.07x times

4:10:16

one time for each year. If we dissect this equation, we see that the number of 50,000

4:10:24

comes from the original value of the car. The one point is seven, which I call the growth

4:10:29

factor comes from one plus point 707. The point O seven being the, the percent increase

4:10:41

written as a decimal. So the form of this equation is an exponential equation, V of

4:10:50

x equals a times b to the x, where A is the initial value, and B is the growth factor.

4:10:57

But we could also write this as a times one plus r to the x, where A is still the initial

4:11:06

value,

4:11:07

but

4:11:08

R is the percent increase written as a decimal. This same equation will come up in the next

4:11:15

example, here, my Toyota Prius is worth only 3000. Now, and its value is decreasing by

4:11:22

5% each year. So after one year, its value will be 3000 minus 0.05 times 3000. That is

4:11:34

3000 times one minus 0.05. I can also write that as 3000 times 0.95. Decreasing the value

4:11:48

by 5% is like multiplying the value by one minus point oh five, or by point nine, five.

4:12:00

After two years, the value will be multiplied by point nine five again. So the value will

4:12:08

be 3000 times point nine, five squared. And after x years, the value will be 3000 times

4:12:17

point nine five to the x. So my equation for the value is 3000 times point nine five to

4:12:26

the x. This is again, an equation of the form V of x equals a times b to the x, where here,

4:12:36

a is 3000, the initial value, and B is point nine, five, I'll still call that the growth

4:12:47

factor, even though we're actually declining value not growing. Now remember where this

4:12:53

point nine five came from, it came from taking one and subtracting point 05, because of the

4:13:01

5%, decrease in value, so I can again, write my equation in the form a times well, this

4:13:09

time times one minus r to the x, where R is our point 05. That's our percent decrease

4:13:19

written as a decimal. Please take a moment to study this equation and the previous one.

4:13:26

They say that when you have an exponential function, the number here is the initial value.

4:13:35

If it's written in this form, B is your growth factor. But you can think of B as being either

4:13:42

one minus r, where r is the percent decrease, or one plus r, where r is the percent increase.

4:13:50

In this example, we're given a function f of x to model the number of bacteria in a

4:13:55

petri dish x hours after 12 o'clock noon, we want to know what was the number of bacteria

4:14:03

at noon, and by what percent, the number of bacteria is increasing every hour, we can

4:14:09

see from the equation that the number of bacteria is increasing and not decreasing, because

4:14:14

the base of the exponential function 1.45 is bigger than one. Notice that our equation

4:14:21

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

4:14:30

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

4:14:43

this familiar form, we can recognize that the initial amount of bacteria is going to

4:14:49

be 12 12,000. Since those are our units and this value 1.45 is our growth factor. What

4:15:00

the number of bacteria is multiplied by each hour? Well are the point four five is the

4:15:08

the rate of increase, in other words, a 45% increase each hour. So our answers to the

4:15:18

questions are 12,045%. In this example, the population of salamanders is modeled by this

4:15:29

exponential function, where x is the number of years since 2015. Notice that the number

4:15:37

of salamanders is decreasing, because the base of our exponential function point seven,

4:15:42

eight is less than one. So if we recognize the form of our exponential function, a times

4:15:50

b to the x, or we can think of this as a times one minus r to the x, where A is our initial

4:15:59

value, and r is our percent decrease written as a decimal.

4:16:08

Our initial value is 3000. So that's the number of salamanders zero years after 2015. Our

4:16:18

growth factor B is 0.78. But if I write that as one minus r, I see that R has to be one

4:16:27

minus 0.78, or 0.22. In other words, our population is decreasing by 22% each year. In this video,

4:16:40

we saw that exponential functions can be written in the form f of x equals a times b to the

4:16:47

x, where A is the initial value. And B is the growth factor. We also saw that they can

4:16:58

be written in the form a times one plus r to the x when the amount is increasing. And

4:17:05

as a times one minus r to the x when the amount is decreasing. In this format, R is the percent

4:17:16

increase or the percent decrease written as a decimal. So 15% increase will be an R value

4:17:28

of 0.15 and a growth factor B of 1.15. Whereas a 12% decrease will be an R value of point

4:17:43

one, two, and a B value of one minus point one, two, or 0.7. Sorry, 0.88. These observations

4:17:55

help us quickly interpret exponential functions. For example, here, we have an initial value

4:18:03

of 100 and a 15% increase. And here, we have an initial value of 50 and a 40%. decrease.

4:18:17

exponential functions can be used to model compound interest for loans and bank accounts.

4:18:23

Suppose you invest $200 in a bank account that earns 3% interest every year. If you

4:18:30

make no deposits or withdrawals, how much money will you have accumulated after 10 years,

4:18:37

because 3% of the money that's in the bank is getting added each year, the money in the

4:18:41

bank gets multiplied by 1.03 each year. So after one year, the amount will be 200 times

4:18:51

1.3, after two years 200 times one point O three squared, and and after t years 200 times

4:19:02

one point O three to the t power. So the function modeling the amount of money, I'll call it

4:19:08

P of t is given by 200 times 1.03 to the T. More generally, if you invest a dollars

4:19:20

at an annual interest rate of our for t years. In the end, you'll have P of t equals a times

4:19:32

one plus r to the T

4:19:37

here r needs to be written as a decimal, so 0.03 In our example, for the 3% annual interest

4:19:45

rate. Going back to our specific example, After 10 years, the amount of money is going

4:19:51

to be P of 10 which is 200 times 1.03 to the 10 which works out to 206 $68.78 to the nearest

4:20:03

cent. In this problem, we've assumed that the interest accumulates once per year. But

4:20:10

in the next few examples, we'll see what happens when the interest rate accumulates more frequently,

4:20:14

twice a year, or every month. For example, let's deposit $300 in an account that earns

4:20:23

4.5% annual interest compounded semi annually, this means two times a year or every six months.

4:20:31

A 4.5% annual interest rate compounded two times a year means that we're actually getting

4:20:40

4.5 over 2% interest, every time the interest is compounded. That is every six months. That's

4:20:53

2.25% interest every half a year. Note that 2.25% is the same as 0.02 to five as a decimal.

4:21:05

So every time we earn interest, our money gets multiplied by 1.02 to five. Let me make

4:21:17

a chart of what happens. After zero years, which is also zero half years, we have our

4:21:25

original $300. for half a year, that's one half year, our money gets earns interest one

4:21:33

time, so we multiply the 300 by 1.02 to five, after one year, that's two half years, our

4:21:42

money earns interest two times. So we multiply 300 by 1.0225 squared. continuing this way,

4:21:52

after 1.5 years, that's three half years, we have 300 times 1.02 to five cubed. And

4:22:01

after two years or four half years, we have 300 times 1.02 to five to the fourth. In general,

4:22:09

after t years, which is to T half years, our money will grow to 300 times 1.02 to five

4:22:20

to the two t power. Because we've compounded interest to tee times our formula for our

4:22:28

amount of money is P of t equals 300 times one point O two to five to the two t where

4:22:35

t is the number of years. To finish the problem, after seven years, we'll have P of $7, which

4:22:45

is 300 times 1.02 to five to the two times seven or 14 power. And that works out to $409.65

4:22:59

to the nearest cent. In this next example, we're going to take out a loan for $1,200

4:23:06

in annual interest rate of 6% compounded monthly. Although loans and bank accounts might feel

4:23:14

different, they're mathematically the same. It's like from the bank's point of view, they're

4:23:19

investing money in you and getting interest on their money from you. So we can work them

4:23:25

out with the same kind of math 6% annual interest rate compounded monthly means you're compounding

4:23:36

12 times a year. So each time you compound interest, you're just going to get six over

4:23:42

12% interest. That's point 5% interest. And as a decimal, that's 0.005. Let's try it out

4:23:56

again, what happens. Time is zero, of course, you'll have the original loan amount of 1200.

4:24:05

After one year, that's 12 months, your loan has had interest added to it 12 times so it

4:24:15

gets multiplied by 1.05 to the 12th.

4:24:22

After two years, that's 24 months, it's had interest added to it 24 times, so it gets

4:24:30

multiplied by 1.05 to the 24th power. Similarly, after three years or 36 months, your loan

4:24:39

amount will be 1200 times 1.05 to the 36th power. And in general, after two years, that's

4:24:49

12 t months. So the interest will be compounded 12 t times and so we have to raise the 1.05

4:24:59

To the 12 t power. This gives us the general formula for the money owed is P of t equals

4:25:08

1200 times 1.05 to the 12 T, where T is the number of years. In particular, after three

4:25:20

years, we'll have to pay back a total of 1200 times 1.05 to the 12 times three or 36 power,

4:25:31

which works out to $1,436.02 to the nearest cent. These last two problems follow a general

4:25:42

pattern. If A is the initial amount of the loan or bank account, and r is the annual

4:25:51

interest rate, compounded n times per year, then our formula for the amount of money is

4:26:03

going to be a times one plus r over n to the n t. This formula exactly matches what we

4:26:13

did in this problem. First, we took the interest rate our which was 6%, and divided it by the

4:26:20

number of compounding periods each year 12. We wrote that as a decimal and added one to

4:26:25

it. That's where we got the 1.05 from. We raised this to not to the number of years,

4:26:35

but to 12 times the number of years. That's the number of compounding periods per year,

4:26:39

times the number of years. And we multiplied all that by the initial amount of money, which

4:26:46

was 1200. This formula for compound interest is a good one to memorize. But it's also important

4:26:52

to be able to reason your way through it, like we did in this chart. There's one more

4:26:57

type of compound interest. And that's interest compounded continuously. You can think of

4:27:02

continuous compounding as the limit of compounding more and more frequently 10 times a year,

4:27:09

100 times a year 1000 times a year a million times a year. In the limit, you get continuous

4:27:14

compounding. The formula for continuous compounding is P of t equals a times e to the RT, where

4:27:24

p of t is the amount of money, t is the time in years. A is the initial amount of money.

4:27:37

And R is the annual interest rate written as a decimal. So 0.025 in this problem from

4:27:49

the 2.5% annual interest rate. He represents the famous constant Oilers constant, which

4:27:57

is about 2.718. So in this problem, we have P of t is 4000 times e to the 0.025 T. And

4:28:12

after five years, we'll have P of five, which is 4000 times e to the 0.0 to five times five,

4:28:21

which works out to $4,532.59 to the nearest cent. To summarize, if R represents the annual

4:28:32

interest rate written as a decimal, that is 2% would be 0.02. And t represents the number

4:28:43

of years and a represents the initial amount of money, then for just simple annual interest

4:28:50

compounded once a year. Our formula is P of t is a times one plus r to the T for compound

4:29:00

interest compounded n times per year. Our formula is P of t is a times one plus r over

4:29:08

n to the n T. And for compound interest compounded continuously, we get P of t is a times e to

4:29:16

the RT.

4:29:18

In this video, we looked at three kinds of compound interest problems, simple annual

4:29:24

interest, interest compounded and times per year, and continuously compounded interest.

4:29:33

This video introduces logarithms. logarithms are a way of writing exponents. The expression

4:29:42

log base a of B equals c means that a to the C equals b. In other words, log base a of

4:29:52

B is the exponent that you raise a to to get BE THE NUMBER A It is called the base of the

4:30:02

logarithm. It's also called the base when we write it in this exponential form. Some

4:30:07

students find it helpful to remember this relationship. log base a of B equals c means

4:30:14

a to the C equals b, by drawing arrows,

4:30:17

a to the C equals b.

4:30:20

Other students like to think of it in terms of asking a question, log base a of B, asks,

4:30:30

What power do you raise a to in order to get b? Let's look at some examples. log base two

4:30:41

of eight is three, because two to the three equals eight. In general, log base two of

4:30:50

y is asking you the question, What power do you have to raise to to to get y? So for example,

4:30:59

log base two of 16 is four, because it's asking you the question to what power equals 16?

4:31:08

And the answer is four. Please pause the video and try some of these other examples. log

4:31:15

base two of two is asking, What power do you raise to two to get two? And the answer is

4:31:22

one. Two to the one equals two. log base two of one half is asking two to what power gives

4:31:34

you one half? Well, to get one half, you need to raise two to a negative power. So that

4:31:40

would be two to the negative one. So the answer is negative one. log base two of 1/8 means

4:31:49

what power do we raise to two in order to get 1/8. Since 1/8, is one over two cubed,

4:31:58

we have to raise two to the negative three power to get one over two cubed. So our exponent

4:32:06

is negative three. And that's our answer to our log expression. Finally, log base two

4:32:13

of one is asking to what power equals one. Well, anything raised to the zero power gives

4:32:21

us one, so this log expression evaluates to zero. Notice that we can get positive negative

4:32:28

and zero answers for our logarithm expressions. Please pause the video and figure out what

4:32:36

these logs evaluate to. to work out log base 10 of a million. Notice that a million is

4:32:46

10 to the sixth power. Now we're asking the question, What power do we raise tend to to

4:32:54

get a million? So that is what power do we raise 10 to to get 10 to the six? Well, of

4:32:59

course, the answer is going to be six. Similarly, since point O one is 10 to the minus three,

4:33:11

this log expression is the same thing as asking, what's the log base 10 of 10 to the minus

4:33:16

three? Well, what power do you have to raise 10? to to get 10 to the minus three? Of course,

4:33:22

the answer is negative three. Log base 10 of zero is asking, What power do we raise

4:33:31

10 to to get zero. If you think about it, there's no way to raise 10 to an exponent

4:33:38

get zero. Raising 10 to a positive exponent gets us really big positive numbers. Raising

4:33:43

10 to a negative exponent is like one over 10 to a power that's giving us tiny fractions,

4:33:49

but they're still positive numbers, we're never going to get zero. Even if we raise

4:33:55

10 to the zero power, we'll just get one. So there's no way to get zero and the log

4:33:59

base 10 of zero does not exist. If you try it on your calculator using the log base 10

4:34:05

button, you'll get an error message. Same thing happens when we do log base 10 of negative

4:34:11

100. We're asking 10 to what power equals negative 100. And there's no exponent that

4:34:17

will work. And more generally, it's possible to take the log of numbers that are greater

4:34:23

than zero, but not for numbers that are less than or equal to zero. In other words, the

4:34:29

domain of the function log base a of x, no matter what base you're using for a, the domain

4:34:36

is going to be all positive numbers. A few notes on notation. When you see ln of x, that's

4:34:44

called natural log, and it means the log base e of x where he is that famous number that's

4:34:50

about 2.718. When you see log of x with no base at all, by convention, that means log

4:34:59

base 10 of x And it's called the common log. Most scientific calculators have buttons for

4:35:05

natural log, and for common log. Let's practice rewriting expressions with logs in them. log

4:35:13

base three of one nine is negative two, can be rewritten as the expression three to the

4:35:20

negative two equals 1/9.

4:35:26

Log of 13 is shorthand for log base 10 of 13. So that can be rewritten as 10 to the

4:35:34

1.11394 equals 13. Finally, in this last expression, ln means natural log, or log base e, so I

4:35:45

can rewrite this equation as log base e of whenever E equals negative one. Well, that

4:35:52

means the same thing as e to the negative one equals one over e, which is true. Now

4:36:00

let's go the opposite direction. We'll start with exponential equations and rewrite them

4:36:05

as logs. Remember that log base a of B equals c means the same thing as a to the C equals

4:36:14

b, the base stays the same in both expressions. So for this example, the base of three in

4:36:23

the exponential equation, that's going to be the same as the base in our log. Now I

4:36:29

just have to figure out what's in the argument of the log. And what goes on the other side

4:36:33

of the equal sign. Remember that the answer to a log is an exponent. So the thing that

4:36:40

goes in this box should be my exponent for my exponential equation. In other words, you.

4:36:48

And I'll put the 9.78 as the argument of my log. This works, because log base three of

4:36:57

9.78 equals view means the same thing as three to the U equals 9.78, which is just what we

4:37:06

started with. In the second example, the base of my exponential equation is E. So the base

4:37:13

of my log is going to be the answer to my log is an exponent. In this case, the exponent

4:37:22

3x plus seven. And the other expression, the four minus y becomes my argument of my log.

4:37:32

Let me check, log base e of four minus y equals 3x plus seven means e to the 3x plus seven

4:37:41

equals four minus y, which is just what I started with. I can also rewrite log base

4:37:47

e as natural log. This video introduced the idea of logs, and the fact that log base a

4:37:56

of B equal c means the same thing as a to the C equals b. So log base a of B is asking

4:38:07

you the question, What power exponent Do you raise a to in order to get b. In this video,

4:38:16

we'll work out the graph, so some log functions and also talk about their domains. For this

4:38:22

first example, let's graph a log function by hand by plotting some points. The function

4:38:29

we're working with is y equals log base two of x, I'll make a chart of x and y values.

4:38:36

Since we're working this out by hand, I want to pick x values for which it's easy to compute

4:38:41

log base two of x. So I'll start out with the x value of one because log base two of

4:38:47

one is zero, log base anything of one is 02 is another x value that's easy to compute

4:38:56

log base two of two, that's asking, What power do I raise to two to get to one? And the answer

4:39:03

is one. Power other powers of two are easy to work with. So for example, log base two

4:39:11

of four that saying what power do I raise to two to get four, so the answer is two.

4:39:18

Similarly, log base two of eight is three and log base two of 16 is four. Let me also

4:39:26

work with some fractional values for X. If x is one half, then log base two of one half

4:39:33

that saying what power do I raise to two to get one half? Well, that needs a power of

4:39:39

negative one. It's also easy to compute by hand, the log base two of 1/4 and 1/8. log

4:39:48

base two of 1/4 is negative two since two to the negative two is 1/4. And similarly,

4:39:57

log base two of one eight is negative f3 I'll put some tick marks on my x and y axes. Please

4:40:04

pause the video and take a moment to plot these points. Let's see, I have the point,

4:40:10

one zero, that's here to one that's here, for two,

4:40:17

that is here, and then eight, three, which is here. And the fractional x values, one

4:40:26

half goes with negative one, and 1/4 with negative two 1/8 with negative three. And

4:40:33

if I connect the dots, I get a graph that looks something like this. If I had smaller

4:40:39

and smaller fractions, I would keep getting more and more negative answers when I took

4:40:44

log base two of them, so my graph is getting more and more negative, my y values are getting

4:40:51

more and more negative, as x is getting close to zero. Now I didn't draw any parts of the

4:40:56

graph over here with negative X values, I didn't put any negative X values in my chart,

4:41:02

that omission is no accident. Because if you try to take the log base two or base anything

4:41:08

of a negative number, like say negative four or something, there's no answer. This doesn't

4:41:15

exist because there's no power that you can raise to to to get a negative number. So there

4:41:24

are no points on the graph for negative X values. And similarly, there are no points

4:41:29

on the graph where x is zero, because you can't take log base two of zero, there's no

4:41:35

power you can raise to two to get zero. I want to observe some key features of this

4:41:42

graph. First of all, the domain is x values greater than zero. In interval notation, I

4:41:51

can write that as a round bracket because I don't want to include zero to infinity,

4:41:58

the range is going to be the y values, while they go all the way down into the far reaches

4:42:05

of the negative numbers. And the graph gradually increases y value is getting bigger and bigger.

4:42:11

So the range is actually all real numbers are an interval notation negative infinity

4:42:17

to infinity. Finally, I want to point out that this graph has a vertical asymptote at

4:42:25

the y axis, that is at the line x equals zero. I'll draw that on my graph with a dotted line.

4:42:35

A vertical asymptote is a line that our functions graph gets closer and closer to. So this is

4:42:42

the graph of y equals log base two of x. But if I wanted to graph say, y equals log base

4:42:50

10 of x, it would look very similar, it would still have a domain of X values greater than

4:42:57

zero, a range of all real numbers and a vertical asymptote at the y axis, it will still go

4:43:02

through the point one zero, but it would go through the point 10 one instead, because

4:43:10

log base 10 of 10 is one, it would look pretty much the same, just a lot flatter over here.

4:43:20

But even though it doesn't look like it with the way I've drawn it, it's still gradually

4:43:24

goes up to n towards infinity. In fact, the graph of y equals log base neaa of X for a

4:43:34

bigger than one looks pretty much the same, and has the same three properties. Now that

4:43:42

we know what the basic log graph looks like, we can plot at least rough graphs of other

4:43:46

log functions without plotting points. Here we have the graph of natural log of X plus

4:43:52

five. And again, I'm just going to draw a rough graph. If I did want to do a more accurate

4:43:57

graph, I probably would want to plot some points. But I know that roughly a log graph,

4:44:02

if it was just like y equals ln of x, that would look something like this, and it would

4:44:09

go through the point one zero, with a vertical asymptote along the y axis. Now if I want

4:44:17

a graph, ln of x plus five, that just shifts our graph by five units, it'll still have

4:44:23

the same vertical asymptote. Since the vertical line shifted up by five units, there's still

4:44:27

a vertical line, but instead of going through one zero, it'll go through the point one,

4:44:33

five. So I'll draw a rough sketch here. Let's compare our starting function y equals ln

4:44:42

x and the transformed version y equals ln x plus five in terms of the domain, the range

4:44:51

and the vertical asymptote. Our original function y equals ln x had a domain of zero to infinity

4:44:59

Since adding five on the outside affects the y values, and the domain is the x values,

4:45:08

this transformation doesn't change the domain. So the domain is still from zero to infinity.

4:45:16

Now the range of our original y equals ln x was from negative infinity to infinity.

4:45:22

Shifting up by five does affect the y values, and the range is talking about the y values.

4:45:28

But since the original range was all real numbers, if you add five to all set of all

4:45:33

real numbers, you still get the set of all real numbers. So in this case, the range doesn't

4:45:38

change either. And finally, we already saw that the original vertical asymptote of the

4:45:43

y axis x equals zero, when we shift that up by five units, it's still the vertical line

4:45:49

x equals zero. In this next example, we're starting with a log base 10 function. And

4:45:57

since the plus two is on the inside, that means we shift that graph left by two. So

4:46:02

I'll draw our basic log function. Here's our basic log function. So I'll think of that

4:46:09

as y equals log of x going through the point one, zero,

4:46:15

here's its vertical asymptote. Now I need to shift everything left by two. So my vertical

4:46:23

asymptote shifts left, and now it's at the line x equals negative two, instead of at

4:46:28

x equals zero, and my graph, let's see my point, one zero gets shifted to, let's see

4:46:37

negative one zero, since I'm subtracting two from the axis, and here's a rough sketch of

4:46:45

the resulting graph. Let's compare the features of the two graphs drawn here. We're talking

4:46:54

about domains, the original had a domain of from zero to infinity. But now I've shifted

4:47:01

that left. So I've subtracted two from all my x values. And here's my new domain, which

4:47:07

I can also verify just by looking at the picture. My range was originally from negative infinity

4:47:15

to infinity, well, shifting left only affects the x value. So it doesn't even affect the

4:47:20

range. So my range is still negative infinity to infinity. My vertical asymptote was originally

4:47:26

at x equals zero. And since I subtract two from all x values, that shifts it to x equals

4:47:32

negative two. In this last problem, I'm not going to worry about drawing this graph. I'll

4:47:38

just use algebra to compute its domain. So let's think about what's the issue, when you're

4:47:44

taking the logs of things? Well, you can't take the log of a negative number is zero.

4:47:50

So whatever's inside the argument of the log function, whatever is being fed into log had

4:47:56

better be greater than zero. So I'll write that down, we need to minus 3x to be greater

4:48:03

than zero. Now it's a matter of solving an inequality to it's got to be greater than

4:48:08

3x. So two thirds is greater than x. In other words, x has to be less than two thirds. So

4:48:15

our domain is all the x values from negative infinity to two thirds, not including two

4:48:23

thirds. It's a good idea to memorize the basic shape of the graph of a log function. It looks

4:48:31

something like this, go through the point one zero, and has a vertical asymptote on

4:48:36

the y axis. Also, if you remember that you can't take the log of a negative number, or

4:48:44

zero, then that helps you quickly compute domains for log functions. Whatever's inside

4:48:52

the log function, you set that greater than zero, and solve. This video is about combining

4:49:01

logs and exponents. Please pause the video and take a moment to use your calculator to

4:49:07

evaluate the following four expressions. Remember, that log base 10 on your calculator is the

4:49:15

log button. While log base e on your calculator is the natural log button, you should find

4:49:24

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

4:49:34

the log base 10 of 1000 is 1000. And eat the log base e of 9.6 is 9.6. In each case, the

4:49:45

log and the exponential function with the same base undo each other and we're left with

4:49:51

the exponent. In fact, it's true that for any base a the log base a of a to the x is

4:49:59

equal to x, the same sort of cancellation happens if we do the exponential function

4:50:05

in the log function with the same base in the opposite order. For example, we take 10

4:50:10

to the power of log base 10 of 1000, the 10 to the power and the log base 10 undo each

4:50:15

other, and we're left with the 1000s. This happens for any base, say, a data log base

4:50:21

a of x is equal to x. We can describe this by saying that an exponential function and

4:50:31

a log function with the same base undo each other. If you're familiar with the language

4:50:40

of inverse functions, the exponential function and log function are inverses. Let's see why

4:50:46

these roles hold for the first log role. log base a of a dx is asking the question, What

4:50:55

power do we raise a two in order to get a to the x? In other words, a two what power

4:51:03

is a dx? Well, the answer is clearly x. And that's why log base a of a to the x equals

4:51:11

x. For the second log rule,

4:51:15

notice that the log base a of x means the power we raise a two to get x. But this expression

4:51:25

is saying that we're supposed to raise a to that power. If we raise a to the power, we

4:51:30

need to raise a two to get x, then we'll certainly get x. Now let's use these two rules. In some

4:51:37

examples. If we want to find three to the log base three of 1.43 to the power and log

4:51:44

base three undo each other, so we're left with 1.4. If we want to find ln of e iliacs.

4:51:51

Remember that ln means log base e. So we're taking log base e of e to the x, well, those

4:51:59

functions undo each other, and we're left with x. If we want to take 10 to the log of

4:52:05

three z, remember that log without a base written implies that the base is 10. So really,

4:52:12

we want to take 10 to the log base 10 of three z will tend to a power and log base 10 undo

4:52:18

each other. So we're left with a three z. Finally, this last statement hold is ln of

4:52:26

10 to the x equal to x, well, ln means log base e. So we're taking log base e of 10 to

4:52:36

the x, notice that the base of the log and the base of the exponential function are not

4:52:41

the same. So they don't undo each other. And in fact, log base e of 10 to the x is not

4:52:47

usually equal to x, we can check with one example. Say if x equals one, then log base

4:52:53

e of 10 to the one, that's log base e of 10. And we can check on the calculator that's

4:52:59

equal to 2.3. And some more decimals, which is not the same thing as one. So this statement

4:53:06

is false, it does not hold. We need the basis to be the same for logs and exponents to undo

4:53:13

each other. In this video, we saw that logs and exponents with the same base undo each

4:53:23

other. Specifically,

4:53:26

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

4:53:33

x

4:53:34

for any values of x and any base a. This video is about rules or properties of logs. The

4:53:42

log rules are closely related to the exponent rules. So let's start by reviewing some of

4:53:46

the exponent rules. To keep things simple, we'll write everything down with a base of

4:53:51

two. Even though the exponent rules hold for any base. We know that if we raise two to

4:53:57

the zero power, we get one, we have a product rule for exponents, which says that two to

4:54:04

the M times two to the n is equal to two to the m plus n. In other words, if we multiply

4:54:12

two numbers, then we add the exponents. We also have a quotient rule that says that two

4:54:18

to the M divided by n to the n is equal to two to the m minus n. In words, that says

4:54:26

that if we divide two numbers, then we subtract the exponents. Finally, we have a power rule

4:54:34

that says if we take a power to a power, then we multiply the exponents. Each of these exponent

4:54:43

rules can be rewritten as a log rule. The first rule, two to the zero equals one can

4:54:50

be rewritten in terms of logs as log base two of one equals zero. That's because log

4:54:57

base two of one equals zero mean To the zero equals one. The second rule, the product rule

4:55:05

can be rewritten in terms of logs by saying log of x times y equals log of x plus log

4:55:14

of y. I'll make these base two to agree with my base that I'm using for my exponent rules.

4:55:21

In words, that says the log of the product is the sum of the logs. Since logs really

4:55:28

represent exponent, this is saying that when you multiply two numbers together, you add

4:55:34

their exponents, which is just what we said for the exponent version. The quotient rule

4:55:41

for exponents can be rewritten in terms of logs by saying the log of x divided by y is

4:55:48

equal to the log of x minus the log of y. In words, we can say that the log of the quotient

4:55:58

is equal to the difference of the logs. Since logs are really exponents, another way of

4:56:04

saying the same thing is that when you divide two numbers, you subtract their exponents.

4:56:11

That's how we described the exponent rule above. Finally, the power rule for exponents

4:56:18

can be rewritten in terms of logs by saying the log of x to the n is equal to n times

4:56:27

log of x. Sometimes people describe this rule by saying when you take the log of an expression

4:56:34

with an exponent, you can bring down the exponent and multiply. If we think of x as being some

4:56:40

power of two, this is really saying when we take a power to a power, we multiply their

4:56:45

exponents. That's exactly how we described the power rule above. It doesn't really matter

4:56:50

if you multiply this exponent on the left side, or on the right side. But it's more

4:56:56

traditional to multiply it on the left side. I've given these rules with the base of two,

4:57:03

but they actually work for any base. To help you remember them, please take a moment to

4:57:07

write out the log roles using a base of a you should get the following chart. Let's

4:57:17

use the log rules to rewrite the following expressions as a sum or difference of logs.

4:57:23

In the first expression, we have a log base 10 of a quotient. So we can rewrite the log

4:57:30

of the quotient as the difference of the logs. Now we still have the log of a product, I

4:57:39

can rewrite the log of a product as the sum of the logs. So that is log of y plus log

4:57:47

of z. When I put things together, I have to be careful because here I'm subtracting the

4:57:55

entire log expression. So I need to subtract both terms of this song on the make sure I

4:58:02

do that by putting them in parentheses. Now I can simplify a little bit by distributing

4:58:08

the negative sign. And here's my final answer. In my next expression, I have the log of a

4:58:16

product. So I can rewrite that as the sum of two logs.

4:58:21

I can also use my power rule to bring down the exponent T and multiply it in the front.

4:58:31

That gives me the final expression log of five plus t times log of to one common mistake

4:58:38

on this problem is to rewrite this expression as t times log of five times two. In fact,

4:58:47

those two expressions are not equal. Because the T only applies to the two, not to the

4:58:54

whole five times two, we can't just bring it down in front using the power wall. After

4:58:59

all, the power rule only applies to a single expression raised to an exponent, and not

4:59:08

to a product like this. And these next examples, we're going to go the other direction. Here

4:59:14

we're given sums and differences of logs. And we want to wrap them up into a single

4:59:18

log expression. By look at the first two pieces, that's a difference of logs. So I know I can

4:59:24

rewrite it as the log of a quotient. Now I have the sum of two logs. So I can rewrite

4:59:34

that as the log of a product. I'll clean that up a little bit and rewrite it as log base

4:59:43

five of a times c over B. In my second example, I can rewrite the sum of my logs as the log

4:59:54

of a product now, I will Like to rewrite this difference of logs as the log of a quotient.

5:00:03

But I can't do it yet, because of that factor of two multiplied in front. But I can use

5:00:09

the power rule backwards to put that two back up in the exponent. So I'll do that first.

5:00:15

So I will copy down the ln of x plus one times x minus one, and rewrite this second term

5:00:23

as ln of x squared minus one squared. Now I have a straightforward difference of two

5:00:31

logs, which I can rewrite as the log of a quotient. I can actually simplify this some

5:00:40

more. Since x plus one times x minus one is the same thing as x squared minus one. I can

5:00:50

cancel factors to get ln of one over x squared minus one. In this video, we saw four rules

5:01:00

for logs that are related to exponent rules. First, we saw that the log with any base of

5:01:07

one is equal to zero. Second, we saw the product rule, the log of a product is equal to the

5:01:15

sum of the logs. We saw the quotient rule, the log of a quotient is the difference of

5:01:23

the logs. And we saw the power rule. When you take a log of an expression with an exponent

5:01:30

in it, you can bring down the exponent and multiply it. It's worth noticing that there's

5:01:37

no log rule that helps you split up the log of a psalm. In particular, the log of a psalm

5:01:44

is not equal to the sum of the logs. If you think about logs and exponent rules going

5:01:52

together, this kind of makes sense, because there's also no rule for rewriting the sum

5:02:00

of two exponential expressions.

5:02:04

Log rules will be super handy, as we start to solve equations using logs. Anytime you

5:02:10

have an equation like this one that has variables in the exponent logs are the tool of choice

5:02:17

for getting those variables down where you can solve for them. In this video, I'll do

5:02:23

a few examples of solving equations with variables in the exponent. For our first example, let's

5:02:29

solve for x the equation five times two to the x plus one equals 17. As my first step,

5:02:37

I'm going to isolate the difficult spot, the part that has the variables and the exponent.

5:02:42

In this example, I can do that by dividing both sides by five. That gives me two to the

5:02:47

x plus one equals 17 over five. Next, I'm going to take the log of both sides, it's

5:02:56

possible to take the log with any base, but I prefer to take the log base 10, or the log

5:03:00

base e for the simple reason that my calculator has buttons for those logs. So in this example,

5:03:06

I'll just take the log base 10. So I can omit the base because it's a base 10 is implied

5:03:12

here. And that gives me this expression. As my next step, I'm going to use log roles to

5:03:19

bring down my exponent and multiply it on the front. It's important to use parentheses

5:03:24

here because the entire x plus one needs to get multiplied by log two. So that was my

5:03:31

third step using the log roles. Now all my variables are down from the exponent where

5:03:37

I can work with them, but I still need to solve for x. Right now x is trapped in the

5:03:42

parentheses. So I'm going to free it from the parentheses by distributing. So I get

5:03:47

x log two plus log two equals log of 17 fifths. Now I will try to isolate x by moving all

5:03:55

my terms with x's in them to one side, and all my terms without x's in them to the other

5:04:01

side. Finally, I factor out my x. Well, it's already kind of factored out, and I divide

5:04:08

to isolate it. So I read out what I did. So far I distributed. I moved all the x terms

5:04:17

on one side, and the terms without x's on the other side. And then I isolated the x

5:04:24

by factoring out and dividing. We have an exact solution for x. This is correct, but

5:04:31

maybe not so useful if you want a decimal answer. At this point, you can type in everything

5:04:36

into your calculator using parentheses liberally to get a decimal answer about 0.765. It's

5:04:46

always a good idea to check your work by typing in this decimal answer and seeing if the equation

5:04:51

checks out. This next example is trickier because there are variables in the exponent

5:04:57

in two places with two different bases. First step is normally to clean things up and simplify

5:05:05

by isolating the tricky stuff. But in this example, there's nothing really to clean up

5:05:10

or simplifier, or no way to isolate anything more than it already is. So we'll go ahead

5:05:15

to the next step and take the log of both sides. Again, I'll go ahead and use log base

5:05:20

10. But we couldn't use log base e instead. Next, we can use log roles to bring down the

5:05:27

exponents. This gives me 2x minus three in parentheses, log two equals x minus two, log

5:05:35

five. Now I'm going to distribute things out to free the excess from the parentheses. That

5:05:43

gives me 2x log two minus three log two equals x log five minus two log five. Now I need

5:05:52

to group the x terms on one side, and the other terms that don't involve x on the other

5:05:59

side. So I'll keep the 2x log two on the left, put the minus log x log five on the left,

5:06:08

and that gives me the minus two log five that stays on the right and a plus three log two

5:06:13

on the right. Finally, I need to isolate x by factoring and dividing by factoring I just

5:06:22

mean I factor out the x from all the terms on the left, so that gives me x times the

5:06:27

quantity to log two minus log five. And that equals this mess on the right. Now I can divide

5:06:37

the right side by the quantity on the left side. When you type this into your calculator,

5:06:48

I encourage you to type the whole thing in rather than type bits and pieces in because

5:06:53

if you round off, you'll get a less accurate answer than if you type the whole thing in

5:06:58

at once. In this example, when I type it in, I get a final answer of about 5.106.

5:07:06

In this equation, we have the variable t in the exponent in two places. The letter E is

5:07:12

not a variable, it represents the number e whose decimal approximation is about 2.7.

5:07:19

Because there's already an E and the expression, it's going to be handy to use natural log

5:07:24

in this problem instead of log base 10. But before we take the log of both sides, let's

5:07:29

clean things up. by isolating the tricky parts, we could at least divide both sides by five.

5:07:35

And that will give us either the negative 0.05 t is equal to three fifths it is 0.2

5:07:42

T. One way to proceed would now be to clean things out further by dividing both sides

5:07:47

by either the point two t, but I'm going to take a different approach and go ahead and

5:07:51

take the natural log of both sides. That gives me ln of e to the minus 0.05 t is equal to

5:08:01

ln of three fifths e to the 0.2 t. Now on the left side, I can immediately use my log

5:08:08

rule to bring down my exponent and get minus 0.05 t, L and E. But on the right side, I

5:08:18

can't bring down the exponent yet because this e to the point two t is multiplied by

5:08:23

three fifths. So before I can bring down the exponent, I need to split up this product

5:08:28

using the product rule. So I can rewrite this as ln of three fifths plus ln e to the 0.2

5:08:36

T. And now I can bring down the exponent. So that third step was using log roles to

5:08:47

ultimately bring down my exponents. Now ln of E is a really nice expression, ln IV means

5:08:56

log base e of E. So that's asking what power do I raise E to in order to get e? And the

5:09:02

answer is one. So anytime I have ln of E, I can just replace that with one. That's why

5:09:10

using natural log is a little bit handier here than using log base 10. You can make

5:09:14

that simplification. Next, I'm ready to solve for t. So I don't need to distribute here.

5:09:23

But I do need to bring my T terms to one side of my terms without t to the other side. So

5:09:29

let's see. I'll put my T terms on the left and my team turns without T on the right.

5:09:37

And finally I'm going to isolate t by factoring and dividing. So by factoring I mean I factor

5:09:45

out my T and now I can divide. Using my calculator,

5:09:57

I can get a decimal answer of 2.04 Three, three. This video gave some examples of solving

5:10:04

equations with variables in the exponent. And the key idea was to take the log of both

5:10:10

sides and use the log properties to bring the exponents down. This video, give some

5:10:21

examples of equations with logs in them like this one. In order to solve the equations

5:10:26

like this one, we have to free the variable from the log. And we'll do that using exponential

5:10:32

functions. My first step in solving pretty much any kind of equation is to simplify it

5:10:38

and isolate the tricky part. In this case, the tricky part is the part with the log in

5:10:44

it. So I can isolate it by first adding three to both sides. That gives me two ln 2x plus

5:10:51

five equals four, and then I can divide both sides by two. Now that I've isolated the tricky

5:11:01

part, I still need to solve for x, but x is trapped inside the log function. to free it,

5:11:07

I need to somehow undo the log function. Well log functions and exponential functions undo

5:11:13

each other. And since this is a log base e, I need to use the exponential function base

5:11:18

e also. So I'm going to take e to the power of both sides. In other words, I'll take e

5:11:26

to the ln 2x plus five, and that's going to equal e to the power of two.

5:11:33

Now e to the ln of anything, I'll just write a for anything. That means e to the log base

5:11:40

e of a, that you the power and log base e undo each other. So we just get a, I'm going

5:11:46

to use that principle over here, e to the ln of 2x plus five, the E and the log base

5:11:53

e undo each other. And we're left with 2x plus five on the left side. So 2x plus five

5:11:59

is equal to E squared. And from there, it's easy to finish the problem and solve for x

5:12:04

by subtracting five from both sides and then dividing by two. So I'll just write the third

5:12:11

step here is to just say finish solving for x. There's one last step we need to do when

5:12:17

solving equations with logs in them. And that's to check our answers. Because we may get extraneous

5:12:25

solutions. an extraneous solution is a solution that comes out of the solving process, but

5:12:31

doesn't actually satisfy the original equation. And those can happen for equations with logs

5:12:36

in them, because we might get a solution that makes the argument of the log negative or

5:12:42

zero, and we can't take the log of a negative numbers zero. So let's check and plug in the

5:12:52

solution of E squared minus five over two. We'll plug that in for x in our original equation

5:13:01

and see if that works. So let's see the twos cancel here. So I get two ln e squared minus

5:13:09

five plus five, minus three, I want that to equal one. And now my fives cancel. And so

5:13:17

I have two ln e squared minus three that I want to equal one. Well, I think this is going

5:13:23

to work out because let's see, ln is log base e. and log base e of E squared that's asking

5:13:34

the question, What power do I raise e two to get e squared? Well, I have to raise it

5:13:38

to the power of two to get e squared. So this becomes two times two minus three. And does

5:13:46

that equal one, four minus three does equal one. So that all checks out. And we didn't

5:13:53

have any problem with taking the log of negative numbers zero, we didn't get any extraneous

5:13:58

solutions. So this is our solution. The second equation is a little bit trickier, because

5:14:04

there's a log into places. Now notice that this is a log where there's no base written.

5:14:10

So a base 10 is implied. So I'm already thinking in order to undo a log base 10, I'm going

5:14:16

to want to take a 10 to the power of both sides. Of course, it's still a good idea to

5:14:23

isolate the tricky part, but there's nothing really to isolate here. So I'm going to say,

5:14:29

can't do it here. So we'll just jump right to straight step two, and take 10 to the power

5:14:37

of both sides. Okay, so that's going to give me 10 to the whole thing, log x plus three

5:14:46

plus log x, that whole thing is in the exponent equals 10 to the One Power. Well, I know what

5:14:52

to do with the right side 10 to the one is just 10. But what do I do with the 10 to these

5:14:58

two things added

5:14:59

up Up,

5:15:00

while remembering my exponent rules, I know that when you add up the exponent, that's

5:15:06

what happens when you multiply two things. So this is the same thing as 10 to the log

5:15:12

x plus three times 10 to the log x, right, because when you multiply two things, you

5:15:18

add the exponent, so these are the same. Okay, now we're in business, because 10 to the log

5:15:24

base 10, those undo each other. And so this whole expression here simplifies to x plus

5:15:31

three. Similarly, 10 to the log base, 10 of x is just x. So I'm multiplying x plus three

5:15:38

by x, that's equal to 10. Now I have an equation I can deal with, it's a quadratic, so I'm

5:15:46

going to first multiply out to make it look more like a quadratic, get everything to one

5:15:52

side. So is equal to zero. And, and now I can either factor or use the quadratic formula.

5:15:59

I think this one factors, it looks like X plus five times x minus two, so I'm going

5:16:06

to get x is negative five, or x is two. So that was all the third step to finish solving

5:16:14

for x. Finally, we need to check our solutions to make sure we haven't gotten some extraneous

5:16:20

ones. So let's see if x equals negative five. If I plug that into my original equation,

5:16:27

that says, I'm checking that log of negative five plus three, plus log of negative five,

5:16:36

checking that's equal to one. Well, this is giving me a queasy feeling. And I hope it's

5:16:40

giving you a queasy feeling too, because log of negative two does not exist, right, you

5:16:48

can't take the log of a negative number. Same thing with log of negative five. So x equals

5:16:53

negative five is an extraneous solution, it doesn't actually solve our original equation.

5:17:00

Let's check the other solution, x equals two. So now we're checking to see if log of two

5:17:08

plus three plus log of two is equal to one. Since there's no problem with taking logs

5:17:15

of negative numbers, or zero here, this should work out fine. And just to finish checking,

5:17:20

we can see let's see this is log of five plus log of two, we want that equal one. But using

5:17:27

my log rules, the sum of two logs is the log of the product. So log of five times two,

5:17:34

we want that to equal one. And that's just log base 10 of 10. And that definitely equals

5:17:41

one because log base 10 of 10 says, What power do I raise tend to to get 10. And that power

5:17:46

is one. So the second solution x equals two does check out. And that's our final answer.

5:17:56

Before I leave this problem, I do want to mention that some people have an alternative

5:18:00

approach. Some people like to start with the original equation, and then use log roles

5:18:05

to combine everything into one log expression. So since we have the sum of two logs, we know

5:18:12

that's the same as the log of a product, right, so we can rewrite the left side as log of

5:18:18

x plus three times x, that equals one, then we do the same trick of taking 10 to the power

5:18:25

of both sides. And as before, the 10 to the power and the log base 10 undo each other,

5:18:32

and we get x plus three times x equals 10, just like we did before. In the first solution,

5:18:40

we ended up using exponent rules to rewrite things. In the second alternative method,

5:18:45

we use log rules to rewrite things. So the methods are really pretty similar, pretty

5:18:50

equivalent, and they certainly will get us to the same answer. So we've seen a couple

5:18:54

examples of equations with logs in them and how to solve them. And the key step is to

5:19:02

use exponential functions to undo the log. In other words, take e to the power of both

5:19:13

sides to undo natural log, and take 10 to the power of both sides. To undo log base

5:19:25

10.

5:19:27

In this video, we'll use exponential equations in some real life applications, like population

5:19:33

growth and radioactive decay. I'll also introduce the ideas of half life and doubling time.

5:19:40

In this first example, let's suppose we invest $1,600 in a bank account that earns 6.5% annual

5:19:47

interest compounded once a year. How many years will it take until the account has $2,000

5:19:52

in it if you don't make any further deposits or withdrawals since our money is earning

5:19:59

six point 5% interest each year, that means that every year the money gets multiplied

5:20:05

by 1.065. So after t years, my 1600 gets multiplied by 1.065 to the t power. I'll write this in

5:20:18

function notation as f of t equals 1600 times 1.065 to the T, where f of t is the amount

5:20:30

of money after t years. Now we're trying to figure out how long it will take to get $2,000

5:20:41

$2,000 is a amount of money. So that's an amount for f of t. And we're trying to solve

5:20:48

for T the amount of time. So let me write out my equation and make a note that I'm solving

5:20:55

for t. Now to solve for t, I want to first isolate the tricky part. So I'm going to the

5:21:04

tricky part is the part with the exponential in it. So I'm going to divide both sides by

5:21:08

1600. That gives me 2000 over 1600 equals 1.065. To the tee, I can simplify this a little

5:21:18

bit further as five fourths. Now that I've isolated the tricky part, my next step is

5:21:24

going to be to take the log of both sides. That's because I have a variable in the exponent.

5:21:29

And I know that if I log take the log of both sides, I can use log roles to bring that exponent

5:21:33

down where I can solve for it. I think I'll use log base e this time. So I have ln fi

5:21:40

force equals ln of 1.065 to the T. Now by the power rule for logs, on the right side,

5:21:48

I can bring that exponent t down and multiply it in the front. Now it's easy to isolate

5:21:56

t just by dividing both sides by ln of 1.065. Typing that into my calculator, I get that

5:22:07

t is approximately 3.54 years. And the next example, we have a population of bacteria

5:22:17

that initially contains 1.5 million bacteria, and it's growing by 12% per day, we want to

5:22:24

find the doubling time, the doubling time means the amount of time it takes for a quantity

5:22:34

to double in size. For example, the amount of time it takes to get from the initial 1.5

5:22:41

million bacteria to 3 million bacteria would be the doubling time. Let's start by writing

5:22:47

an equation for the amount of bacteria. So if say, P of t represents the number of bacteria

5:22:56

in millions, then my equation and T represents time in days, then P of t is going to be given

5:23:08

by the initial amount of bacteria times the growth factor 1.12 to the T. That's because

5:23:19

my population of bacteria is growing by 12% per day. That means the number of bacteria

5:23:25

gets multiplied by one point 12. Since we're looking for the doubling time, we're looking

5:23:30

for the t value when P of D will be twice as big. So I can set P of t to be three and

5:23:39

solve for t. As before, I'll start by isolating the tricky part, taking the log

5:23:56

bringing the T down. And finally solving for T. Let's see three over 1.5 is two so I can

5:24:08

write this as ln two over ln 1.12. Using my calculator, that's about 6.12 days. It's an

5:24:19

interesting fact that doubling time only depends on the growth rate, that 12% growth, not the

5:24:32

initial population. In fact, I could have figured out the doubling time without even

5:24:37

knowing how many bacteria were in my initial population. Let me show you how that would

5:24:41

work. If I didn't know how many I started with, I could still write P of t equals a

5:24:48

times 1.12 to the t where a is our initial population that I don't know what it is.

5:24:55

Then if I want to figure out how long it takes for my population, to double Well, if I start

5:25:01

with a and double that I get to a. So I'll set my population equal to two a. And I'll

5:25:08

solve for t. Notice that my A's cancel. And so when I take the log of both sides and bring

5:25:19

the T down and solve for t, I get the exact same thing as before, it didn't matter what

5:25:32

the initial population was, I didn't even have to know what it was. In this next example,

5:25:38

we're told the initial population, and we're told the doubling time, we're not told by

5:25:42

what percent the population increases each minute, or by what number, we're forced to

5:25:49

multiply the population by each minute. So we're gonna have to solve for that, I do know

5:25:53

that I want to use an equation of the form y equals a times b to the T, where T is going

5:25:59

to be the number of minutes, and y is going to be the number of bacteria. And I know that

5:26:04

my initial amount a is 350. So I can really write y equals 350 times b to the T. Now the

5:26:11

doubling time tells me that when 15 minutes have elapsed, my population is going to be

5:26:19

twice as big, or 700. plugging that into my equation, I have 700 equals 350 times b to

5:26:28

the 15. Now I need to solve for b. Let me clean this up a little bit by dividing both

5:26:34

sides by 350. That gives me 700 over 350 equals b to the 15th. In other words, two equals

5:26:43

b to the 15th. To solve for B, I don't have to actually use logs here, because my variables

5:26:51

in the base not in the exponent, so I don't need to bring that exponent down. Instead,

5:26:57

the easiest way to solve this is just by taking the 15th root of both sides or equivalently.

5:27:03

The 1/15 power. That's because if I take B to the 15th to the 1/15, I multiply by exponents,

5:27:12

that gives me B to the one is equal to two to the 1/15. In other words, B is two to the

5:27:19

1/15, which as a decimal is approximately 1.047294. I like to use a lot of decimals

5:27:29

if I'm doing a decimal approximation and these kind of problems to increase accuracy. But

5:27:34

of course, the most accurate thing is just to leave B as it is. And I'll do that and

5:27:38

rewrite my equation is y equals 350 times two to the 1/15 to the T. Now I'd like to

5:27:50

work this problem one more time. And this time, I'm going to use the form of the equation

5:27:56

y equals a times e to the RT. This is called a continuous growth model. It looks different,

5:28:04

but it's actually an equivalent form to this growth model over here. And I'll say more

5:28:09

about why these two forms are equivalent at the end, I can use the same general ideas

5:28:14

to solve in this form. So I know that my initial amount is 350. And I know again that when

5:28:21

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.

5:28:33

Again, I'm going to simplify things by dividing both sides by 350. That gives me two equals

5:28:44

e to the r times 15. This time my variable does end up being in the exponent. So I do

5:28:51

want to take the log of both sides, I'm going to use natural log, because I already have

5:28:56

an E and my problem. So natural log and E are kind of more harmonious the other than

5:29:01

then common log with base 10 and E. So I take the log of both sides. Now I can pull the

5:29:09

exponent down. So that's our times 15 times ln of E. Well Elena V is just one, right?

5:29:14

Because Elena V is asking what power do I raise E to to get e that's just one. And so

5:29:19

I get 15 r equals ln two. So r is equal to ln two divided by 15. Let me plug that back

5:29:27

in to my original equation as e to the ln two over 15 times t. Now I claimed that these

5:29:35

two equations were actually the same thing just looking different. And the way to see

5:29:40

that is if I start with this equation, and I rewrite as e to the ln two over 15 like

5:29:49

that

5:29:50

to the tee. Well, I claim that this quantity right here is the same thing as my two to

5:29:58

the 1/15. And in fact One way to see that is e to the ln two already to the 115. The

5:30:06

Right, that's the same, because every time I take a power to a power, I multiply exponents.

5:30:10

But what's Ed Oh, and two, E and ln undo each other, so that's just 350 times two to the

5:30:18

1/15 to the T TA, the equations are really the same. And these, you'll always be able

5:30:24

to find two different versions of an exponential equation, sort of the standard one, a times

5:30:30

b to the T, or the continuous growth one, a times e to the RT. In this last example,

5:30:38

we're going to work with half life, half life is pretty much like doubling time, it just

5:30:43

means the amount of time that it takes for a quantity to decrease to half as much as

5:30:55

we originally started with, we're told that the half life of radioactive carbon 14 is

5:31:02

5750 years. So that means it takes that long for a quantity of radioactive carbon 14 to

5:31:11

decay, so that you just have half as much left and the rest is nonradioactive form.

5:31:16

So we're told a sample of bone that originally contained 200 grams of radioactive carbon

5:31:21

14 now contains only 40 grams, we're supposed to find out how old the sample is, is called

5:31:29

carbon dating. Let's use the continuous growth model this time. So our final amount, so this

5:31:36

is our amount of radioactive c 14 is going to be the initial amount

5:31:49

times e to the RT, we could have used the other model to we could have used f of t equals

5:31:54

a times b to the T, but I just want to use the continuous model for practice. So we know

5:31:59

that our half life is 5750. So what that means is when t is 5750, our amount is going to

5:32:13

be one half of what we started with. Let me see if I can plug that into my equation and

5:32:19

figure out use that to figure out what r is. That's ours called the continuous growth rate.

5:32:26

So I plug in one half a for the final amount, a is still the initial amount, e to the r

5:32:35

and I have 5750 I can cancel my A's. And now I want to solve for r, r is in my exponent.

5:32:46

So I do need to take the log of both sides to solve for it. I'm going to use log base

5:32:50

e since I already have an E and my problem, log base e is more compatible with E then

5:32:57

log base 10 is okay. Now, on the left side, I still have log of one half and natural log

5:33:04

of one half on the right side, ln n e to a power those undo each other. So I'm left with

5:33:09

R times 5750. Now I can solve for r, it's ln one half over 5750 could work that as a

5:33:20

decimal, but it's actually more accurate just to keep it in exact form. So now I can rewrite

5:33:25

my equation, I have f of t equals a times e to the ln one half over five 750 t. Now

5:33:40

I can use that to figure out my problem. And my problem the bone originally contained 200

5:33:46

grams, that's my a, I want to figure out when it's going to contain only 40 grams, that's

5:33:53

my final amount. And so I need to solve for t. I'll clean things up by dividing both sides

5:34:00

by 200. Let's say 40 over 200 is 1/5. Now I'm going to take the ln of both sides. And

5:34:10

ln and e to the power undo each other. So I'm left with Elena 1/5 equals ln of one half

5:34:19

divided by five 750 T. And finally I can solve for t but super messy but careful use of my

5:34:27

calculator gives me an answer of 13,351 years approximately.

5:34:35

That kind of makes sense in terms of the half life because to get from 200 to 40 you have

5:34:41

to decrease by half a little more than two times right or increasing by half once would

5:34:45

get you to 100 decreasing to half again would get you to 50 a little more than 40 and two

5:34:50

half lives is getting pretty close to 13,000 years. This video introduced a lot of new

5:34:57

things it introduced continue Less growth model, which is another equivalent way of

5:35:04

writing an exponential function. The relationship is that the B in this example, is the same

5:35:16

thing as EDR. In that version, it also introduced the ideas of doubling time. And HalfLife the

5:35:27

amount of time it takes a quantity to double or decreased to half in an exponential growth

5:35:32

model. Recall that a linear equation is equation like for example, 2x minus y equals one. It's

5:35:42

an equation without any x squared or y squared in it, something that could be rewritten in

5:35:47

the form y equals mx plus b, the equation for a line a system of linear equations is

5:35:55

a collection of two or more linear equations. For example, I could have these two equations.

5:36:03

A solution to a system of equations is that an x value and a y value that satisfy both

5:36:10

of the equations. For example, the ordered pair two three, that means x equals two, y

5:36:18

equals three is a solution to this system. Because if I plug in x equals two, and y equals

5:36:26

three into the first equation, it checks out since two times two minus three is equal to

5:36:33

one. And if I plug in x equals two and y equals three into the second equation, it also checks

5:36:40

out two plus three equals five. However, the ordered pair one for that is x equals one,

5:36:49

y equals four is not a solution to the system. Even though this X and Y value work in the

5:36:55

second equation, since one plus four does equal five, it doesn't work in the first equation,

5:37:02

because two times one minus four is not equal to one. In this video, we'll use systematic

5:37:10

methods to find the solutions to systems of linear equations. In this first example, we

5:37:17

want to solve this system of equations, there are two main methods we could use, we could

5:37:22

use the method of substitution, or we could use the method of elimination. If we use the

5:37:29

method of substitution, the main idea is to isolate one variable in one equation, and

5:37:38

then substitute it in to the other equation. For example, we can start with the first equation

5:37:44

3x minus two y equals four, and isolate the x by adding two y to both sides, and then

5:37:53

dividing both sides by three. Think I'll rewrite that a little bit by breaking up the fraction

5:38:01

into two fractions four thirds plus two thirds y. Now, I'm going to copy down the second

5:38:07

equation 5x plus six y equals two. And I'm going to substitute in my expression for x.

5:38:16

That gives me five times four thirds plus two thirds y plus six y equals two. And now

5:38:24

I've got an equation with only one variable in it y. So I can solve for y as a number.

5:38:30

First, I'm going to distribute the five so that gives me 20 thirds plus 10 thirds y plus

5:38:38

six y equals two. And now I'm going to keep all my y terms, my terms with y's in them

5:38:45

on the left side, but I'll move all my terms without y's in them to the right side. At

5:38:51

this point, I could just add up all my fractions and solve for y. But since I don't really

5:38:56

like working with fractions, I think I'll do the trick of clearing the denominators

5:39:00

here. So I'm going to actually multiply both sides by my common denominator of three just

5:39:05

to get rid of the denominators and not have to work with fractions. So let me write that

5:39:10

down. Distributing the three, I get 10 y plus eight t and y equals six minus 20. Now add

5:39:20

things together. So that's 28 y equals negative 14. So that means that y is going to be negative

5:39:28

14 over 28, which is negative one half.

5:39:32

So I've solved for y. And now I can go back and plug y into either of my equations to

5:39:38

solve for x, I plug it into my first equation. So I'm plugging in negative one half for y.

5:39:47

That gives me 3x plus one equals four. So 3x equals three, which means that x is equal

5:39:55

to one. I've solved my system of equations and gotten x equals one On y equals minus

5:40:01

one half, I can also write that as an ordered pair, one, negative one half for my solution.

5:40:09

Now let's go back and solve the same system, but use a different method, the method of

5:40:14

elimination, the key idea to the method of elimination is to multiply each equation by

5:40:23

a constant to make the coefficients of one variable match. Let me start by copying down

5:40:34

my two equations. Say I'm trying to make the coefficient of x match. One way to do that

5:40:42

is to multiply the first equation by five, and the second equation by three. That way,

5:40:48

the coefficient of x will be 15 for both equations, so let me do that. So for the first equation,

5:40:55

I'm going to multiply both sides by five. And for the second equation, I'm going to

5:41:00

multiply both sides by three. That gives me for the first equation 15x minus 10, y equals

5:41:09

20. And for the second equation, 15x plus 18, y is equal to six. Notice that the equate

5:41:19

the coefficients of x match. So if I subtract the second equation from the first, the x

5:41:25

term will completely go away, it'll be zero times x, and I'll be left with let's see negative

5:41:32

10 y minus 18, y is going to give me minus 28 y. And if I do 20 minus six, that's going

5:41:44

to give me 14. solving for y, I get y is 14 over minus 28, which is minus one half just

5:41:52

like before. Now we can continue, like we did in the previous solution, and substitute

5:41:59

that value of y into either one of the equations. I'll put it again in here. And my solution

5:42:06

proceeds as before. So once again, I get the solution that x equals one and y equals minus

5:42:13

one half. Before we go on to the next problem, let me show you graphically what this means.

5:42:20

Here I've graphed the equations 3x minus two, y equals four, and 5x plus six y equals two.

5:42:29

And we can see that these two lines intersect in the point with coordinates one negative

5:42:36

one half, just like we predicted by solving equations algebraically. Let's take a look

5:42:42

at another system of equations. I'm going to rewrite the first equation, so the x term

5:42:49

is on the left side with the y term, and the constant term stays on the right. And I'll

5:42:55

rewrite or copy down the second equation. Since the coefficient of x in the first equation

5:43:02

is minus four, and the second equation is three, I'm going to try using the method of

5:43:08

elimination and multiply the first equation by three and the second equation by four.

5:43:14

That'll give me a coefficient of x of negative 12. And the first equation, and 12. And the

5:43:19

second equation, those are equal and opposite, right, so I'll be able to add together my

5:43:23

equations to cancel out access. So let's do that. My first equation becomes negative 12x

5:43:31

plus 24, y equals three, and I'll put everything by three. And my second equation, I'll multiply

5:43:39

everything by four. So that's 12x minus 24, y equals eight. Now something kind of funny

5:43:46

has happened here, not only do the x coefficients match has in with opposite signs, but the

5:43:53

Y coefficients do also. So if I add together my two equations, in order to cancel out the

5:44:00

x term, I'm also going to cancel out the y term, and I'll just get zero plus zero is

5:44:06

equal to three plus eight is 11. Well, that's a contradiction, we can't have zero equal

5:44:12

to 11. And that shows that these two equations actually have no solution.

5:44:19

Let's look at this situation graphically. If we graph our two equations, we see that

5:44:25

they're parallel lines with the same slope. This might be more clear, if I isolate y in

5:44:31

each equation, the first equation, I get y equals, let's see, dividing by eight that's

5:44:36

the same thing as four eighths or one half x plus one eight. And the second equation,

5:44:43

if I isolate y, let's say minus six y equals minus 3x plus two divided by minus six, that's

5:44:50

y equals one half x minus 1/3. So indeed, they have the same slope. And so they're parallel

5:44:59

with different In intercepts, and so they can have no intersection. And so it makes

5:45:04

sense that we have no solution to our system of linear equations. This kind of system that

5:45:10

has no solution is called an inconsistent system. In this third example, yet a third

5:45:18

behavior happens. This time, I think I'm going to solve by substitution because I already

5:45:24

have X with a coefficient of one. So it's really easy to just isolate X in the first

5:45:30

equation, and then plug in to the second equation to get three times six minus five y plus 15,

5:45:39

y equals 18. If I distribute out, I get the strange phenomenon that the 15 y's cancel

5:45:46

and I just get 18 equals 18, which is always true. This is what's called a dependent system

5:45:54

of linear equations. If you look more closely, you can see that the second equation is really

5:46:00

just a constant multiple, the first equation is just three times every term is three times

5:46:06

as big as the corresponding term and the first equation. So there's no new information in

5:46:10

the second equation, anything, any x and y values that satisfy the first one will satisfy

5:46:16

the second one. So this system of equations has infinitely many solutions. Any ordered

5:46:23

pair x y, where X plus five y equals six, or in other words, X's minus five y plus six

5:46:32

will satisfy this system of equations. That would include a y value of zero corresponding

5:46:40

x value of six or a y value of one corresponding to an x value of one, or a y value of 1/3.

5:46:49

Corresponding to an x value of 13 thirds just by plugging into this equation will work.

5:46:55

Graphically, if I graph both of these equations, the lines will just be on top of each other,

5:47:00

so I'll just see one line. In this video, we've solved systems of linear equations,

5:47:06

using the method of substitution and the method of elimination. We've seen that systems of

5:47:13

linear equations can have one solution. When the lines that the equations represent intersect

5:47:19

in one point, they can be inconsistent, and have no solutions that corresponds to parallel

5:47:26

lines, or they can be dependent and have infinitely many solutions that corresponds to the lines

5:47:33

lying on top of each other. In this video, I'll work through a problem involving distance

5:47:39

rate and time. The key relationship to keep in mind is that the rate of travel is the

5:47:46

distance traveled divided by the time it takes to travel again. For example, if you're driving

5:47:51

60 miles an hour, that's your rate. And that's because you're going a distance of 60 miles

5:47:57

in one hour. Sometimes it's handy to rewrite that relationship by multiplying both sides

5:48:02

by T time. And that gives us that R times T is equal to D. In other words, distance

5:48:10

is equal to rate times time. There's one more important principle to keep in mind. And that's

5:48:15

the idea that rates add. For example, if you normally walk at three miles per hour, but

5:48:23

you're walking on a moving sidewalk, that's going at a rate of two miles per hour, then

5:48:29

your total speed of travel with respect to you know something stationary is going to

5:48:34

be three plus two, or five miles per hour. All right, that is a formula as our one or

5:48:43

the first rate per second rate is equal to the total rate. Let's use those two key ideas.

5:48:51

distance equals rate times time, and rates add in the following problem. else's boat

5:48:59

has a top speed of six miles per hour and still water. While traveling on a river at

5:49:04

top speed. She went 10 miles upstream in the same amount of time she went 30 miles downstream,

5:49:10

we're supposed to find the rate of the river currents. I'm going to organize the information

5:49:15

in this problem into a chart.

5:49:17

During the course of Elsa stay, there were two situations we need to keep in mind. For

5:49:23

one period of time she was going upstream. And for another period of time she was going

5:49:27

downstream. For each of those, I'm going to chart out the distance you traveled. The rate

5:49:35

she went at and the time it took when she was going upstream. She went a total distance

5:49:42

of 10 miles. When she was going downstream she went a longer distance of 30 miles. But

5:49:49

the times to travel those two distances were the same. Since I don't know what that time

5:49:55

was, I'll just give it a variable I'll call it T now Think about her rate of travel, the

5:50:02

rate she traveled upstream was slower because she was going against the current and faster

5:50:07

when she was going downstream with the current. We don't know what the speed of the current

5:50:11

is, that's what we're trying to figure out. So maybe I'll give it a variable R. But we

5:50:17

do know that in still water also can go six miles per hour. And when she's going downstream,

5:50:23

since she's going with the direction of the current rates should add, and her rate downstream

5:50:29

should be six plus R, that's her rate, and still water plus the rate of the current.

5:50:40

On the other hand, when she's going upstream, then she's going against the current, so her

5:50:46

rate of six miles per hour, we need to subtract the rate of the current from that. Now that

5:50:53

we've charted out our information, we can turn it into equations using the fact that

5:50:57

distance equals rate times time, we actually have two equations 10 equals six minus R times

5:51:04

T, and 30 is equal to six plus R times T. Now that we've converted our situation into

5:51:12

a system of equations, our next job is to solve the system of equations. In this example,

5:51:17

I think the easiest way to proceed is to isolate t in each of the two equations. So in the

5:51:23

first equation, I'll divide both sides by six minus r, and a second equation R divided

5:51:28

by six plus R. That gives me 10 over six minus r equals T, and 30 over six plus r equals

5:51:36

t. Now if I set my T variables equal to each other, I get, I get 10 over six minus r is

5:51:45

equal to 30 over six plus R. I'm making progress because now I have a single equation, the

5:51:52

single variable that I need to solve. Since the variable R is trapped in the denominator,

5:51:58

I'm going to proceed by clearing the denominator. So I'm going to multiply both sides by the

5:52:03

least common denominator, that is six minus r times six plus R. Once I cancel things out,

5:52:14

I get that the six plus r times 10 is equal to 30 times six minus r, if I distribute,

5:52:23

I'm going to get 60 plus xR equals 180 minus 30 ar, which I can now solve, let's see, that's

5:52:36

going to be 40 r is equal to 120. So our, the speed of my current is going to be three

5:52:43

miles per hour. This is all that the problem asked for the speed of the current. If I also

5:52:51

wanted to solve for the other unknown time, I could do so by plugging in R into one of

5:52:57

my equations and solving for T. In this video, I saw the distance rate and time problem by

5:53:04

charting out my information for the two situations and my problem using the fact that rates add

5:53:12

to fill in some of my boxes, and then using the formula distance equals rate times time

5:53:17

to build a system of equations. In this video, I'll do a standard mixture problem in which

5:53:24

we have to figure out what quantity of two solutions to mix together. household bleach

5:53:30

contains 6% sodium hypochlorite. The other 94% is water. How much household bleach should

5:53:39

be combined with 70 liters of a weaker 1% hypochlorite solution in order to form a solution,

5:53:50

that's 2.5% sodium hypochlorite. I want to turn this problem into a system of equations.

5:53:59

So I'm asking myself what quantities are going to be equal to each other? Well, the total

5:54:05

amount of sodium hypochlorite that has symbol and a co o before mixing,

5:54:14

it should equal the total amount of sodium hypochlorite after mixing. Also, the total

5:54:23

amount of water before mixing should equal the total amount of water after. And finally,

5:54:30

there's just the total amount of solution. In my two jugs, sodium hypochlorite together

5:54:37

with water should equal the total amount of solution after that gives me a hint for what

5:54:47

I'm looking for. But before I start reading out equations, I find it very helpful to chart

5:54:51

out my quantities. So I've got the 6% solution. The household leech, I've got the 1% solution.

5:55:05

And I've got my Desired Ending 2.5% solution. Now on each of those solutions, I've got a

5:55:16

certain volume of sodium hypochlorite. I've also got a volume of water. And I've got a

5:55:26

total volume of solution. Let me see which of these boxes I can actually fill in, I know

5:55:37

that I'm adding 70 liters of the 1% solution. So I can put a 70 in the total volume of solution

5:55:45

here. I don't know what volume of the household bleach, I want to add, that's what I'm trying

5:55:53

to find out. So I'm going to just call that volume x. Now since my 2.5% solution is made

5:56:02

by combining my other two solutions, I know its volume is going to be the sum of these

5:56:08

two volumes, so I'll write 70 plus x in this box. Now, the 6% solution means that whatever

5:56:16

the volume of solution is, 6% of that is the sodium hypochlorite. So the volume of the

5:56:22

sodium hypochlorite is going to be 0.06 times x, the volume of water and that solution is

5:56:31

whatever's left, so that's going to be x minus 0.06x, or point nine four times X with the

5:56:40

following the same reasoning for the 1% solution 1% of the 70 liters is a sodium hypochlorite.

5:56:46

So that's going to be 0.01 times 70. Or point seven, the volume of water and that solution

5:56:56

is going to be 99% or point nine, nine times seven day. That works out to 69.3. Finally,

5:57:05

for the 2.5% solution, the volume of the sodium hypochlorite is going to be 0.0 to five times

5:57:15

the volume of solution 70 plus x and the volume of water is going to be the remainder. So

5:57:22

that's 0.975 times 70 plus x. Now I've already used the fact that the volume of solutions

5:57:32

before added up is the volume of solution after in writing a 70 plus x in this box.

5:57:39

But I haven't yet used the fact that the volume of the sodium hypochlorite is preserved before

5:57:43

and after. So I can write that down as an equation. So that means 0.06x plus 0.7 is

5:57:53

equal to 0.025 times 70 plus x. Now I've got an equation with a variable. I'll try to solve

5:58:01

it. Since I don't like all these decimals, I'm going to multiply both sides of my equation

5:58:06

by let's see, 1000 should get rid of all the decimals. After distributing, I get 60x plus

5:58:15

700 equals 25 times 70 plus x. Distributing some more, I get 60x plus 700 is equal to

5:58:27

1750 plus 25x. So let's see 60 minus 25 is 35x is equal to 1050. Which gives me that

5:58:42

x equals 30 liters of the household bleach.

5:58:49

Notice that I never actually had to use the fact that the water quantity of water before

5:58:54

mixing is equal to the quantity of water after that is I never use the information in this

5:59:00

column. In fact, that information is redundant. Once I know that the quantities of sodium

5:59:08

hypochlorite add up, and the total volume of solutions add up. The fact that that volume

5:59:15

of waters add up is just redundant information. The techniques to use to solve this equation

5:59:21

involving solutions can be used to solve many many other equations involving mixtures of

5:59:28

items. My favorite method is to first make a chart involving the types of mixtures and

5:59:36

the types of items in your mixture. Fill in as many boxes size I can and then use the

5:59:43

fact that the quantities add. This video is about rational functions and their graphs.

5:59:53

Recall that a rational function is a function that can be written as a ratio or quotient

5:59:59

of two power. No Here's an example. The simpler function, f of x equals one over x is also

6:00:07

considered a rational function, you can think of one and x as very simple polynomials. The

6:00:14

graph of this rational function is shown here. This graph looks different from the graph

6:00:20

of a polynomial. For one thing, its end behavior is different. The end behavior of a function

6:00:27

is the way the graph looks when x goes through really large positive, or really large negative

6:00:33

numbers, we've seen that the end behavior of a polynomial always looks like one of these

6:00:38

cases. That is why marches off to infinity or maybe negative infinity, as x gets really

6:00:44

big or really negative. But this rational function has a different type of end behavior.

6:00:50

Notice, as x gets really big, the y values are leveling off

6:00:54

at about a y value of three. And similarly, as x values get really negative, our graph

6:01:00

is leveling off near the line y equals three, I'll draw that line, y equals three on my

6:01:07

graph, that line is called a horizontal asymptote. A horizontal asymptote is a horizontal line

6:01:16

that our graph gets closer and closer to as x goes to infinity, or as X goes to negative

6:01:21

infinity, or both. There's something else that's different about this graph from a polynomial

6:01:26

graph, look at what happens as x gets close to negative five. As we approach negative

6:01:32

five with x values on the right, our Y values are going down towards negative infinity.

6:01:37

And as we approach the x value of negative five from the left, our Y values are going

6:01:42

up towards positive infinity. We say that this graph has a vertical asymptote at x equals

6:01:49

negative five. A vertical asymptote is a vertical line that the graph gets closer and closer

6:01:56

to. Finally, there's something really weird going on at x equals two, there's a little

6:02:02

open circle there, like the value at x equals two is dug out. That's called a hole. A hole

6:02:10

is a place along the curve of the graph where the function doesn't exist. Now that we've

6:02:16

identified some of the features of our rational functions graph,

6:02:19

I want to look back at the equation and see how we could have predicted those features

6:02:24

just by looking at the equation. To find horizontal asymptotes, we need to look at what our function

6:02:31

is doing when x goes through really big positive or really big negative numbers. Looking at

6:02:36

our equation for our function, the numerator is going to be dominated by the 3x squared

6:02:42

term when x is really big, right, because three times x squared is going to be absolutely

6:02:47

enormous compared to this negative 12. If x is a big positive or negative number, in

6:02:53

the denominator, the denominator will be dominated by the x squared term. Again, if x is a really

6:02:59

big positive or negative number, like a million, a million squared will be much, much bigger

6:03:04

than three times a million or negative 10. For that reason, to find the end behavior,

6:03:10

or the horizontal asymptote, for our function, we just need to look at the term on the numerator

6:03:18

and the term on the denominator that have the highest exponent, those are the ones that

6:03:22

dominate the expression in size. So as x gets really big, our functions y values are going

6:03:29

to be approximately 3x squared over x squared, which is three. That's why we have a horizontal

6:03:37

asymptote at y equals three. Now our vertical asymptotes, those tend to occur where our

6:03:45

denominator of our function is zero. That's because the function doesn't exist when our

6:03:51

denominator is zero. And when we get close to that place where our denominator is zero,

6:03:56

we're going to be dividing by tiny, tiny numbers, which will make our Y values really big in

6:04:01

magnitude. So to check where our denominators zero, let's factor our function. In fact,

6:04:06

I'm going to go ahead and factor the numerator and the denominator. So the numerator factors,

6:04:11

let's see, pull out the three, I get x squared minus four, factor in the denominator, that

6:04:18

factors into X plus five times x minus two, I can factor a little the numerator a little

6:04:25

further, that's three times x minus two times x plus two over x plus 5x minus two. Now,

6:04:34

when x is equal to negative five, my denominator will be zero, but my numerator will not be

6:04:41

zero. That's what gives me the vertical asymptote at x equals negative five. Notice that when

6:04:49

x equals two, the denominator is zero, but the numerator is also zero. In fact, if I

6:04:57

cancelled the x minus two factor from the numerator, and in nominator, I get a simplified

6:05:01

form for my function that agrees with my original function as long as x is not equal to two.

6:05:11

That's because when x equals two, the simplified function exists, but the original function

6:05:16

does not, it's zero over zero, it's undefined. But for every other x value, including x values

6:05:23

near x equals to our original functions just the same as this function. And that's why

6:05:29

our function only has a vertical asymptote at x equals negative five, not one at x equals

6:05:36

two, because the x minus two factor is no longer in the function after simplifying,

6:05:41

it does have a hole at x equals two, because the original function is not defined there,

6:05:45

even though the simplified version is if we want to find the y value of our hole, we can

6:05:52

just plug in x equals two into our simplified version of our function, that gives a y value

6:05:59

of three times two plus two over two plus seven, or 12 ninths, which simplifies to four

6:06:06

thirds. So our whole is that to four thirds. Now that we've been through one example in

6:06:15

detail, let's summarize our findings. We find the vertical asymptotes and the holes by looking

6:06:22

where the denominator is zero. The holes happen where the denominator and numerator are both

6:06:29

zero and those factors cancel out. The vertical asymptotes are all other x values where the

6:06:35

denominator is zero, we find the horizontal asymptotes. By considering the highest power

6:06:42

term on the numerator and the denominator, I'll explain this process in more detail in

6:06:47

three examples. In the first example, if we circle the highest power terms, that simplifies

6:06:55

to 5x over 3x squared, which is five over 3x. As x gets really big, the denominator

6:07:03

is going to be huge. So I'm going to be dividing five by a huge, huge number, that's going

6:07:09

to be going very close to zero. And therefore we have a horizontal asymptote

6:07:15

at y equals zero. In the second example, the highest power terms, 2x cubed, over 3x cubed

6:07:24

simplifies to two thirds. So as x gets really big, we're going to be heading towards two

6:07:30

thirds, and we have a horizontal asymptote at y equals two thirds. In the third example,

6:07:36

the highest power terms, x squared over 2x simplifies to x over two. As x gets really

6:07:45

big, x over two is getting really big. And therefore, we don't have a horizontal asymptote

6:07:51

at all. This is going to infinity, when x gets through go through big positive numbers,

6:07:59

and is going to negative infinity when x goes through a big negative numbers. So in this

6:08:07

case, the end behavior is kind of like that of a polynomial, and there's no horizontal

6:08:12

asymptote. In general, when the degree of the numerator is smaller than the degree of

6:08:18

the denominator, we're in this first case where the denominator gets really big compared

6:08:23

to the numerator and we go to zero. In the second case, where the degree of the numerator

6:08:29

and the degree of the dominant are equal, things cancel out, and so we get a horizontal

6:08:34

asymptote at the y value, that's equal to the ratio of the leading coefficients. Finally,

6:08:42

in the third case, when the degree of the numerator is bigger than the degree of the

6:08:46

denominator, then the numerator is getting really big compared to the denominator, so

6:08:52

we end up with no horizontal asymptote. Final Finally, let's apply all these observations

6:08:57

to one more example. Please pause the video and take a moment to find the vertical asymptotes,

6:09:03

horizontal asymptotes and holes for this rational function. To find the vertical asymptotes

6:09:09

and holes, we need to look at where the denominator is zero. In fact, it's going to be handy to

6:09:15

factor both the numerator and the denominator. Since there if there are any common factors,

6:09:20

we might have a whole instead of a vertical asymptote. The numerator is pretty easy to

6:09:25

factor. Let's see that's 3x times x plus one for the denominator or first factor out an

6:09:31

x. And then I'll factor some more using a guess and check method. I know that I'll need

6:09:40

a 2x and an X to multiply together to the 2x squared and I'll need a three and a minus

6:09:47

one or alpha minus three and a one. Let's see if that works. If I multiply out 2x minus

6:09:56

one times x plus three that does get me back to x squared plus 5x minus three, so that

6:10:02

checks out. Now I noticed that I have a common factor of x in both the numerator and the

6:10:08

denominator. So that's telling me I'm going to have a hole at x equals zero. In fact,

6:10:14

I could rewrite my rational function by cancelling out that common factor. And that's equivalent,

6:10:21

as long as x is not equal to zero. So the y value of my whole is what I get when I plug

6:10:28

zero into my simplified version, that would be three times zero plus one over two times

6:10:35

zero minus one times zero plus three, which is three over negative three or minus one.

6:10:42

So my whole is at zero minus one. Now all the remaining places in my denominator that

6:10:49

make my denominator zero will get me vertical asymptotes. So I'll have a vertical asymptote,

6:10:55

when 2x minus one times x plus three equals zero, that is, when 2x minus one is zero,

6:11:04

or x plus three is zero. In other words, when x is one half, or x equals negative three.

6:11:13

Finally, to find my horizontal asymptotes, I just need to consider the highest power

6:11:20

term in the numerator and the denominator. That simplifies to three over 2x, which is

6:11:28

bottom heavy, right? When x gets really big, this expression is going to zero. And that

6:11:35

means that we have a horizontal asymptote at y equals zero. So we found the major features

6:11:41

of our graph, the whole, the vertical asymptotes and the horizontal asymptotes. Together, this

6:11:48

would give us a framework for what the graph of our function looks like. horizontal asymptote

6:11:54

at y equals zero, vertical asymptotes at x equals one half, and x equals minus three

6:12:02

at a hole at the point zero minus one. plotting a few more points, or using a graphing calculator

6:12:09

of graphing program, we can see that our actual function will look something like this.

6:12:17

Notice that the x intercept when x is negative one, corresponds to where the numerator of

6:12:24

our rational function or reduced rational function is equal to zero. That's because

6:12:29

a zero on the numerator that doesn't make the denominator zero makes the whole function

6:12:33

zero. And an X intercept is where the y value of the whole function is zero. In this video,

6:12:40

we learned how to find horizontal asymptotes have rational functions. By looking at the

6:12:44

highest power terms, we learned to find the vertical asymptotes and holes. By looking

6:12:50

at the factored version of the functions. The holes correspond to the x values that

6:12:56

make the numerator and denominator zero, his corresponding factors cancel. The vertical

6:13:03

asymptotes correspond to the x values that make the denominator zero, even after factoring

6:13:09

any any common and in common factors in the numerator denominator.

6:13:13

This video is about combining functions by adding them subtracting them multiplying and

6:13:19

dividing them. Suppose we have two functions, f of x equals x plus one and g of x equals

6:13:26

x squared. One way to combine them is by adding them together. This notation, f plus g of

6:13:34

x means the function defined by taking f of x and adding it to g of x. So for our functions,

6:13:43

that means we take x plus one and add x squared, I can rearrange that as the function x squared

6:13:50

plus x plus one. So f plus g evaluated on x means x squared plus x plus one. And if

6:14:01

I wanted to evaluate f plus g, on the number two, that would be two squared plus two plus

6:14:08

one, or seven. Similarly, the notation f minus g of x means the function we get by taking

6:14:18

f of x and subtracting g of x. So that would be x plus one minus x squared. And if I wanted

6:14:25

to take f minus g evaluated at one, that would be one plus one minus one squared, or one,

6:14:35

the notation F dot g of x, which is sometimes also written just as f g of x. That means

6:14:44

we take f of x times g of x. In other words, x plus one times x squared, which could be

6:14:52

simplified as x cubed plus x squared. The notation f divided by g of x means I take

6:15:02

f of x and divided by g of x. So that would be x plus one divided by x squared. In this

6:15:11

figure, the blue graph represents h of x. And the red graph represents the function

6:15:19

p of x, we're asked to find h minus p of zero.

6:15:26

We don't have any equations to work with, but that's okay. We know that for any x, h

6:15:32

minus p of x is defined as h of x minus p of x. So for x equals zero, H minus p of zero

6:15:40

is going to be h of zero minus p of zero. Using the graph, we can find h of zero by

6:15:48

finding the value of zero on the x axis, and finding the corresponding y value for the

6:15:55

function h of x. So that's about 1.8. Now P of zero, we can find similarly by looking

6:16:04

for zero on the x axis, and finding the corresponding y value for the function p of x, and that's

6:16:10

a y value of one 1.8 minus one is 0.8. So that's our approximate value for H minus p

6:16:18

of zero. If we want to find P times h of negative three, again, we can rewrite that as P of

6:16:28

negative three times h of negative three. And using the graphs, we see that for an x

6:16:34

value of negative three, the y value for P is two. And the x value of negative three

6:16:45

corresponds to a y value of negative two for H.

6:16:50

Two times negative two is negative four. So that's our value for P times h of negative

6:16:55

three.

6:16:56

In this video, we saw how to add two functions, subtract two functions, multiply two functions,

6:17:03

and divide two functions in the following way. When you compose two functions, you apply

6:17:13

the first function, and then you apply the second function to the output of the first

6:17:19

function. For example, the first function might compute population size from time in

6:17:27

years. So its input would be time in years, since a certain date, as output would be number

6:17:35

of people in the population. The second function g, might compute health care costs as a function

6:17:45

of population size. So it will take population size as input, and its output will be healthcare

6:17:53

costs. If you put these functions together, that is compose them, then you'll go all the

6:18:00

way from time in years to healthcare costs. This is your composition, g composed with

6:18:07

F. The composition of two functions, written g with a little circle, f of x is defined

6:18:16

as follows. g composed with f of x is G evaluated on f of x, we can think of it schematically

6:18:26

and diagram f x on a number x and produces a number f of x, then g takes that output

6:18:36

f of x and produces a new number, g of f of x. Our composition of functions g composed

6:18:45

with F is the function that goes all the way from X to g of f of x. Let's work out some

6:18:51

examples where our functions are defined by tables of values. If we want to find g composed

6:18:58

with F of four, by definition, this means g of f of four. To evaluate this expression,

6:19:08

we always work from the inside out. So we start with the x value of four, and we find

6:19:15

f of four, using the table of values for f of x, when x equals four, f of x is seven,

6:19:23

so we can replace F of four with the number seven. Now we need to evaluate g of seven,

6:19:33

seven becomes our new x value in our table of values for G, the x value of seven corresponds

6:19:39

to the G of X value of 10. So g of seven is equal to 10. We found that g composed with

6:19:47

F of four is equal to 10. If instead we want to find f composed with g of four, well, we

6:19:56

can rewrite that as f of g of four Again work from the inside out. Now we're trying to find

6:20:04

g of four. So four is our x value. And we use our table of values for G to see that

6:20:10

g of four is one. So we replaced you a four by one. And now we need to evaluate f of one.

6:20:20

Using our table for F values, f of one is eight. Notice that when we've computed g of

6:20:29

f of four, we got a different answer than when we computed f of g of four. And in general,

6:20:36

g composed with F is not the same thing as f composed with g. Please pause the video

6:20:43

and take a moment to compute the next two examples. We can replace f composed with F

6:20:49

of two by the equivalent expression, f of f of two. Working from the inside out, we

6:20:56

know that f of two is three, and f of three is six. If we want to find f composed with

6:21:06

g of six, rewrite that as f of g of six, using the table for g, g of six is eight. But F

6:21:17

of eight, eight is not on the table as an x value for the for the f function. And so

6:21:25

there is no F of eight, this does not exist, we can say that six is not in the domain,

6:21:35

for F composed with g. Even though it was in the domain of g, we couldn't follow all

6:21:42

the way through and get a value for F composed with g of six. Next, let's turn our attention

6:21:49

to the composition of functions that are given by equations.

6:21:54

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.

6:22:03

As usual, I can rewrite this as Q of P of one and work from the inside out. P of one

6:22:14

is one squared plus one, so that's two. So this is the same thing as Q of two. But Q

6:22:21

of two is negative two times two or negative four. So this evaluates to negative four.

6:22:28

In this next example, we want to find q composed with P of some arbitrary x, or rewrite it

6:22:35

as usual as Q of p of x and work from the inside out. Well, p of x, we know the formula

6:22:44

for that. That's x squared plus x. So I can replace my P of x with that expression. Now,

6:22:52

I'm stuck with evaluating q on x squared plus x. Well, Q of anything is negative two times

6:22:59

that thing. So q of x squared plus x is going to be negative two times the quantity x squared

6:23:08

plus x, what I've done is I've substituted in the whole expression x squared plus x,

6:23:13

where I saw the X in this formula for q of x, it's important to use the parentheses here.

6:23:20

So that will be multiplying negative two by the whole expression and not just by the first

6:23:24

piece, I can simplify this a bit as negative 2x squared minus 2x. And that's my expression

6:23:32

for Q composed with p of x. Notice that if I wanted to compute q composed with P of one,

6:23:40

which I already did in the first problem, I could just use this expression now, negative

6:23:45

two times one squared minus two and I get negative four, just like I did before. Let's

6:23:52

try another one. Let's try p composed with q of x. First I read write this P of q of

6:24:00

x. Working from the inside out, I can replace q of x with negative 2x. So I need to compute

6:24:07

P of negative 2x. Here's my formula for P. to compute P of this expression, I need to

6:24:15

plug in this expression everywhere I see an x in the formula for P. So that means negative

6:24:23

2x squared plus negative 2x. Again, being careful to use parentheses to make sure I

6:24:29

plug in the entire expression in forex. let me simplify. This is 4x squared minus 2x.

6:24:40

Notice that I got different expressions for Q of p of x, and for P of q of x. Once again,

6:24:50

we see that q composed with P is not necessarily equal to P composed with Q. Please pause the

6:24:57

video and try this last example yourself. rewriting. And working from the inside out,

6:25:06

we're going to replace p of x with its expression x squared plus x. And then we need to evaluate

6:25:13

p on x squared plus x. That means we plug in x squared plus x, everywhere we see an

6:25:22

x in this formula, so that's x squared plus x quantity squared plus x squared plus x.

6:25:29

Once again, I can simplify by distributing out, that gives me x to the fourth plus 2x

6:25:36

cubed plus x squared plus x squared plus x, or x to the fourth plus 2x cubed plus 2x squared

6:25:45

plus x. In this last set of examples, we're asked to go backwards, we're given a formula

6:25:52

for a function of h of x. But we're supposed to rewrite h of x as a composition of two

6:25:57

functions, F and G. Let's think for a minute, which of these two functions gets applied

6:26:04

first, f composed with g of x, let's see, that means f of g of x. And since we evaluate

6:26:13

these expressions from the inside out, we must be applying g first, and then F. In order

6:26:21

to figure out what what f and g could be, I like to draw a box around some thing inside

6:26:27

my expression for H, so I'm going to draw a box around x squared plus seven, then whatever's

6:26:32

inside the box, that'll be my function, g of x, the first function that gets applied,

6:26:38

whatever happens to the box, in this case, taking the square root sign, that becomes

6:26:43

my outside function, my second function f.

6:26:46

So here, we're gonna say g of x is equal to x squared plus seven, and f of x is equal

6:26:54

to the square root of x, let's just check and make sure that this works. So I need to

6:27:00

check that when I take the composition, f composed with g, I need to get the same thing

6:27:08

as my original h. So let's see, if I do f composed with g of x, well, by definition,

6:27:17

that's f of g of x, working from the inside out, I can replace g of x with its formula

6:27:23

x squared plus seven. So I need to evaluate f of x squared plus seven. That means I plug

6:27:30

in x squared plus seven, into the formula for for F. So that becomes the square root

6:27:37

of x squared plus seven to the it works because it matches my original equation. So we found

6:27:44

a correct answer a correct way of breaking h down as a composition of two functions.

6:27:48

But I do want to point out, this is not the only correct answer. I'll write down my formula

6:27:54

for H of X again, and this time, I'll put the box in a different place, I'll just box

6:27:59

the x squared. If I did that, then my inside function of my first function, g of x would

6:28:08

be x squared. And my second function is what happens to the box. So my f of x is what happens

6:28:17

to the box, and the box gets added seven to it, and taking the square root. So in other

6:28:23

words, f of x is going to be the square root of x plus seven. Again, I can check that this

6:28:32

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

6:28:39

I'm taking f of x squared. When I plug in x squared for x, I do in fact, get the square

6:28:46

root of x squared plus seven. So this is that alternative, correct solution. In this video,

6:28:53

we learn to evaluate the composition of functions. by rewriting it and working from the inside

6:29:00

out. We also learn to break apart a complicated function into a composition of two functions

6:29:10

by boxing one piece of the function and letting the first function applied in the composition.

6:29:16

Let that be the inside of the box, and the second function applied in the composition

6:29:21

be whatever happens to the box.

6:29:33

The inverse of a function undoes what the function does. So the inverse of tying your

6:29:38

shoes would be to untie them. And the inverse of the function that adds two to a number

6:29:46

would be the function that subtracts two from a number. This video introduces inverses and

6:29:53

their properties. Suppose f of x is a function defined by this chart. In other words, Have

6:30:00

two is three, f of three is five, f of four is six, and f of five is one, the inverse

6:30:08

function for F written f superscript. Negative 1x undoes what f does. Since f takes two to

6:30:17

three, F inverse takes three, back to two. So we write this f superscript, negative one

6:30:26

of three is to. Similarly, since f takes three to five, F inverse takes five to three. And

6:30:38

since f takes four to six, f inverse of six is four. And since f takes five to one, f

6:30:46

inverse of one is five. I'll use these numbers to fill in the chart. Notice that the chart

6:30:54

of values when y equals f of x and the chart of values when y equals f inverse of x are

6:31:00

closely related. They share the same numbers, but the x values for f of x correspond to

6:31:07

the y values for f inverse of x, and the y values for f of x correspond to the x values

6:31:14

for f inverse of x. That leads us to the first key fact inverse functions reverse the roles

6:31:21

of y and x. I'm going to plot the points for y equals f of x in blue. Next, I'll plot the

6:31:29

points for y equals f inverse of x in red. Pause the video for a moment and see what

6:31:35

kind of symmetry you observe in this graph. How are the blue points related to the red

6:31:40

points, you might have noticed that the blue points and the red points are mirror images

6:31:46

over the mirror line, y equals x. So our second key fact is that the graph of y equals f inverse

6:31:55

of x can be obtained from the graph of y equals f

6:31:58

of x

6:31:59

by reflecting over the line y equals x. This makes sense, because inverses, reverse the

6:32:06

roles of war annex. In the same example, let's compute f inverse of f of two, this open circle

6:32:15

means composition. In other words, we're computing f inverse of f of two, we compute this from

6:32:23

the inside out. So that's f inverse of three. Since F of two is three, and f inverse of

6:32:33

three, we see as to similarly, we can compute f of f inverse of three. And that means we

6:32:44

take f of f inverse of three. Since f inverse of three is two, that's the same thing as

6:32:52

computing F of two, which is three. Please pause the video for a moment and compute these

6:33:02

other compositions. You should have found that in every case, if you take f inverse

6:33:11

of f of a number, you get back to the very same number you started with. And similarly,

6:33:16

if you take f of f inverse of any number, you get back to the same number you started

6:33:20

with. So in general, f inverse of f of x is equal to x, and f of f inverse of x is also

6:33:29

equal to x. This is the mathematical way of saying that F and n f inverse undo each other.

6:33:37

Let's look at a different example. Suppose that f of x is x cubed. Pause the video for

6:33:43

a moment, and guess what the inverse of f should be. Remember, F inverse undoes the

6:33:48

work that F does. You might have guessed that f inverse of x is going to be the cube root

6:33:57

function, we can check that this is true by looking at f of f inverse of x, that's F of

6:34:05

the cube root of function, which means the cube root function

6:34:09

cubed, which gets us back to x. Similarly, if we compute f inverse of f of x, that's

6:34:16

the cube root of x cubed.

6:34:19

And we get back to excellence again. So the cube root function really is the inverse of

6:34:25

the cubing function. When we compose the two functions, we get back to the number that

6:34:30

we started with. It'd be nice to have a more systematic way of finding inverses of functions

6:34:37

besides guessing and checking. One method uses the fact that inverses reverse the roles

6:34:44

of y and x. So if we want to find the inverse of the function, f of x equals five minus

6:34:50

x over 3x. We can write it as y equals five minus x over 3x. Reverse the roles of y and

6:34:59

x To get x equals five minus y over three y, and then solve for y. To solve for y, let's

6:35:09

multiply both sides by three y. Bring all terms with y's in them to the left side, and

6:35:19

alternate without wizened them to the right side, factor out the why. And divide to isolate

6:35:28

why this gives us f inverse of x as five over 3x plus one. Notice that our original function

6:35:40

f and our inverse function, f inverse are both rational functions, but they're not the

6:35:46

reciprocals of each other. And in general, f inverse of x is not usually equal to one

6:35:53

over f of x. This can be confusing, because when we write two to the minus one, that does

6:36:01

mean one of our two, but f to the minus one of x means the inverse function and not the

6:36:08

reciprocal. It's natural to ask us all functions have inverse functions. That is for any function

6:36:16

you might encounter. Is there always a function that its is its inverse? In fact, the answer

6:36:23

is no. See, if you can come up with an example of a function that does not have an inverse

6:36:30

function. The word function here is key. Remember that a function is a relationship between

6:36:37

x values and y values, such that for each x value in the domain, there's only one corresponding

6:36:47

y value. One example of a function that does not have an inverse function is the function

6:36:57

f of x equals x squared.

6:37:01

To see that,

6:37:02

the inverse of this function is not a function. Note that for the x squared function, the

6:37:09

number two and the number negative two, both go to number four. So if I had an inverse,

6:37:17

you would have to send four to both two and negative two, the inverse would not be a function,

6:37:27

it might be easier to understand the problem, when you look at a graph of y equals x squared.

6:37:34

Recall that inverse functions reverse the roles of y and x and flip the graph over the

6:37:40

line y equals x. But when I flipped the green graph over the line y equals x, I get this

6:37:47

red graph. This red graph is not the graph of a function because it violates the vertical

6:37:53

line test. The reason that violates the vertical line test is because the original green function

6:37:59

violates the horizontal line test, and has 2x values with the same y value. In general,

6:38:10

a function f has an inverse function if and only if the graph of f satisfies the horizontal

6:38:14

line test, ie every horizontal line intersects the graph. In it most one point, pause the

6:38:21

video for a moment and see which of these four graphs satisfy the horizontal line test.

6:38:26

In other words, which of the four corresponding functions would have an inverse function?

6:38:34

You may have found that graphs A and B violate the horizontal line test. So their functions

6:38:42

would not have inverse functions. But graph C and D satisfy the horizontal line test.

6:38:48

So these graphs represent functions that do have inverses. functions that satisfy the

6:38:54

horizontal line test are sometimes called One to One functions. Equivalently a function

6:39:02

is one to one, if for any two different x values, x one and x two, the y value is f

6:39:10

of x one and f of x two are different numbers. Sometimes, as I said, f is one to one, if,

6:39:18

whenever f of x one is equal to f of x two, then x one has to equal x two. As our last

6:39:27

example, let's try to find P inverse of x, where p of x is the square root of x minus

6:39:32

two drawn here. If we graph P inverse on the same axis as p of x, we get the following

6:39:40

graph simply by flipping over the line y equals x. If we try to solve the problem algebraically

6:39:49

we can write y equal to a squared of x minus two, reverse the roles of y and x and solve

6:39:56

for y by squaring both sides adding two. Now if we were to graph y equals x squared plus

6:40:07

two, that would look like a parabola, it would look like the red graph we've already drawn

6:40:13

together with another arm on the left side. But we know that our actual inverse function

6:40:22

consists only of this right arm, we can specify this algebraically by making the restriction

6:40:31

that x has to be bigger than or equal to zero. This corresponds to the fact that on the original

6:40:40

graph for the square root of x, y was only greater than or equal to zero. Looking more

6:40:47

closely at the domain and range of P and P inverse, we know that the domain of P is all

6:40:55

values of x such that x minus two is greater than or equal to zero. Since we can't take

6:41:01

the square root of a negative number. This corresponds to x values being greater than

6:41:07

or equal to two, or an interval notation, the interval from two to infinity. The range

6:41:14

of P, we can see from the graph is all y value is greater than or equal to zero, or the interval

6:41:20

from zero to infinity. Similarly, based on the graph, we see the domain of P inverse

6:41:29

is x values greater than or equal to zero, the interval from zero to infinity. And the

6:41:34

range of P inverse is Y values greater than or equal to two, or the interval from two

6:41:40

to infinity. If you look closely at these domains and ranges, you'll notice that the

6:41:45

domain of P corresponds exactly to the range of P inverse, and the range of P corresponds

6:41:53

to the domain of P inverse.

6:41:57

This makes sense, because inverse functions reverse the roles of y and x. The domain of

6:42:04

f inverse of x is the x values for F inverse, which corresponds to the y values or the range

6:42:10

of F. The range of f inverse is the y values for F inverse, which correspond to the x values

6:42:18

or the domain of f. In this video, we discussed five key properties of inverse functions.

6:42:28

inverse functions, reverse the roles of y and x. The graph of y equals f inverse of

6:42:36

x is the graph of y equals f of x reflected over the line y equals x. When we compose

6:42:47

F with F inverse, we get the identity function y equals x. And similarly, when we compose

6:42:56

f inverse with F, that brings x to x. In other words, F and F inverse undo each other. The

6:43:04

function f of x has an inverse function if and only if the graph of y equals f of x satisfies

6:43:18

the horizontal line test. And finally, the domain of f is the range of f inverse and

6:43:31

the range of f is the domain of f inverse. These properties of inverse functions will

6:43:39

be important when we study exponential functions and their inverses logarithmic functions.

College Algebra - Full Course

More from freeCodeCamp.org

Trending Transcripts