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Calculus 1 - Full College Course

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This video is about working with rational expressions. A rational expression is a fraction,

usually with variables in it, something like x plus two over x squared minus three is a

rational expression. In this video, we'll practice adding, subtracting, multiplying

and dividing rational expressions, and simplifying them to lowest terms. We'll start with simplifying

to lowest terms. Recall, if you have a fraction with just numbers in it something like 21

over 45, we can reduce it to lowest terms by factoring the numerator and factoring the

denominator and then canceling common factors. So in this example, the three is cancel, and

our fraction reduces to seven over 15. If we want to reduce rational expression with

variables and add to lowest terms, we proceed the same way. First, we'll factor the numerator,

that's three times x plus two, and then factor the denominator. In this case of factors 2x

plus two times x plus two, we could also write that as x plus two squared. Now we cancel

the common factors. And we're left with three over x plus two. Definitely a simpler way

of writing that rational expression. Now next, let's practice multiplying and dividing. Recall

that if we multiply two fractions with just numbers in them, we simply multiply the numerators

and multiply the denominators. So in this case, we would get four times two over three

times five or 8/15. If we want to divide two fractions, like in the second example, then

we can rewrite it as multiplying by the reciprocal of the fraction on the denominator. So here,

we get four fifths times three halves, and that gives us 12 tenths. But actually, we

could reduce that fraction to six fifths, we use the same rules when we compute the

product or quotient of two rational expressions with variables on them. Here, we're trying

to divide to rational expressions. So instead, we can multiply by the reciprocal, I call

this flipping and multiplying. And now we just multiply the numerators. And multiply

the denominators. It might be tempting at this point to multiply out to distribute out

the numerator and the denominator. But actually, it's better to leave it in this factored form

and factored even more completely. That way, we'll be able to reduce the rational expression

to cancel the common factors. So let's factor even more the x squared plus x factors as

x times x plus one, and x squared minus 16. And that's a difference of two squares, that's

x plus four times x minus four, the denominator is already fully factored, so we'll just copy

it over. And now we can cancel common factors here and here, and we're left with x times

x minus four. This is our final answer. Adding and subtracting fractions is a little more

complicated because we first have to find a common denominator. A common denominator

is an expression that both denominators divided into, it's usually best of the long run to

use the least common denominator, which is the smallest expression that both denominators

divided into.

In this example, if we just want a common denominator, we could use six times 15, which

is 90 because both six and 15 divided evenly into 90. But if we want the least common denominator,

the best way to do that is to factor the two denominators. So six is two times 315 is three

times five, and then put together only the factors we need for both six and 15 to divide

our numbers, so if we just use two times three times five, which is 30. We know that two

times three will divide it and three times five will also divide it and we won't be able

to get a denominator any smaller because we need the factor of two three and five, in

order to ensure both these numbers divided, once we have our least common denominator,

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

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

by the factors that are missing from the current denominator in order to get my least common

denominator of 30. For the second fraction, for 1515 times two is 30. So I'm going to

multiply by two over two, I can rewrite this as 3530 s minus 830 s. And now that I have

a common denominator, I can just subtract my two numerators. And I get 27/30. If I factor,

I can reduce this to three squared over two times five, which is nine tenths. The process

for finding the sum of two rational expressions with variables in them follows the exact same

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

the two denominators. So 2x plus two factors as two times x plus 1x squared minus one,

that's a difference of two squares. So that's x plus one times x minus one. Now for the

least common denominator, I'm going to take all the factors, I need to get an expression

that each of these divides into, so I need the factor two, I need the factor x plus one,

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

need to have at one time. And so I will get my least common denominator two times x plus

one times x minus one, I'm not going to bother multiplying this out, it's actually better

to leave it in factored form to help me simplify later. Now I can rewrite each of my two rational

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

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

the 2x plus two is two times x plus one, I'll write it in factored form. And then I noticed

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

I multiply the numerator and the denominator by x minus one I just needed in the denominator,

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

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

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

the second rational expression. I'll write the denominator in factored form to make it

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

to my least common denominator is just the factor two. So I multiply the numerator and

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

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

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

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

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

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

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

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

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

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

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

canceling common factors. We also saw how to multiply rational expressions by multiplying

the numerator and multiplying the denominator had divide rational expressions by flipping

and multiplying, and how to add and subtract rational expressions by writing them in terms

of the least common denominator. This video is about the difference quotient and the average

rate of change. These are topics that are related to the concept of derivative and calculus.

For function y equals f of x, like the function whose graph right here, a secant line is a

lie. that stretches between two points on the graph of the function. I'm going to label

this x value has a, and this x value as B. So this point here on the graph is going to

have an x value of a, and a y value given by f of a, the second point will have x value

B and Y value, f of b. Now the average rate of change for a function on the interval from

a to b can be defined as the slope have the secret line between the two points A, F of

A and B, F of B. In symbols, that slope m is the rise over the run, or the change in

y over the change in x, which is the difference in Y coordinates f of b minus F of A over

the difference in x coordinates b minus a. So this is the average rate of change. To

put this in context, if for example, f of x represents the height of a tree, and x represents

time in years, then f of b minus f of a represents a difference in height, or the amount the

tree grows. And B minus A represents a difference in in years, so a time period. So this average

rate of change is the amount the tree grows in a certain time period. For example, if

it grows 10 inches in two years, that would be 10 inches per two years, or five inches

per year would be its average rate of change its average rate of growth, let's compute

the average rate of change for the function f of x equals square root of x on the interval

from one to four.

the average rate of change is f of four minus f of one over four minus one, well, f of four

is the square root of four of one's the square root of one. So that's going to be two minus

one over three or 1/3. Instead of calling these two locations on the x axis, a and b,

this time, I'm going to call the first location, just x and the second location, x plus h.

The idea is that h represents the horizontal distance between these two locations on the

x axis. In this notation, if I want to label this point on the graph of y equals f of x,

it'll have an x coordinate of x and a y coordinate of f of x. The second point will have an x

coordinate of x plus h and a y coordinate of f of x plus h. a difference quotient is

simply the average rate of change using this x x plus h notation. So a different version

represents the average rate of change of a function, f of x on the interval from x to

x plus h. Equivalent way, the difference quotient represents the slope of the secant line for

the graph of y equals f of x between the points with coordinates x, f of x, and x plus h,

f of x plus h. Let's work out a formula for the difference quotient. Remember that the

formula for the average rate of change could be written as f of b minus F of A over B minus

A, where a and b are the two locations on the x axis. But now I'm calling instead of

a I'm using x instead of B, I'm using x plus h. So I can rewrite this average rate of change

as f of x plus h minus f of x over x plus h minus x. That simplifies a little bit on

the denominator. Because x plus h minus x, I can cancel the access and I get the difference

quotient formula f of x plus h minus f of x over h. The quantity h on denominator It

looks like a single entity, but it still represents a difference in x values. Let's find and simplify

the difference quotient for this function given first or write down the general formula

for the difference quotient. That's an F of x plus h minus f of x over h. I'm going to

compute f of x plus h first, I do this by shoving in x plus h, everywhere I see an x

in the formula for the function. So that's going to give me two times x plus h squared

minus x plus h plus three. Notice how I use parentheses here. That's important, because

I need to make sure I shove in the entire x plus h for x. So the entire x plus h needs

to be subtracted, not just the x part, so the parentheses are mandatory. Similarly,

the parentheses here signal that the entire x plus h is squared as it needs to be, I'm

going to go ahead and simplify a bit right now, I can multiply out the x plus h squared,

I can go ahead and distribute the negative sign. So if I multiply out, I'm gonna get

x squared plus XH plus h x plus h squared. Now I can distribute the two to get to x squared

plus 2x H plus two h x plus two h squared minus x minus h plus three, these two terms

are actually the same, I can add them up to get 4x H. And I think that's as simple as

I can get that part. Now, I'm going to write out F of x plus h minus f of x. So that's

going to be this thing right here, minus f of x. Again, I need to put the F of X formula

in parentheses to make sure I subtract the whole thing. I'll distribute the negative.

And now I noticed that a bunch of things cancel out. So the 2x squared and the minus 2x squared

add to zero, the minus x and the x add to zero,

and the three and the minus three, add to zero. So I'm left with 4x H plus two h squared

minus h. Finally, I'll write out the whole difference quotient by dividing everything

by H. I can simplify this further, because notice that there's an H in every single term

of the numerator. If I factor out this H, H times 4x, plus two h minus one divided by

h, these two H's cancel, and I'm left with a difference quotient of 4x plus two h minus

one. This difference quotient will become important in calculus, when we calculate the

difference quotient for smaller and smaller values of h, eventually letting h go to zero,

and ending up with an expression that has no H's in it, and represents the derivative

or slope of the function itself. In this video, we use the formula f of b minus F of A over

B minus A to calculate an average rate of change, and the related formula f of x plus

h minus f of x over h to calculate and simplify a difference quotient. This video will introduce

the idea of limits through some graphs and examples. When I worked in San Francisco,

I used to eat at Julia sushi and salad bar that had an interesting price structure, the

food there cost $10 a pound. But if you happen to load up your plate to exactly one pound,

you got a free lunch. So let's let y equals f of x represent the price of your lunch in

dollars as a function of its weight x in pounds. And I want to write an equation to describe

f of x as a piecewise defined function. It makes sense to use a piecewise defined function.

Because there are two situations, the weight x could be exactly one pound, or it could

be different from one pound. If the weight is exactly one pound, then f of x The price

is zero. But if x is different from a pound, then the price is 10 times x as a graph, my

function is going to follow the line y equals 10x. But when x is exactly one, my function

is going to have a value of zero and not 10. The open circle here represents a hole or

a place where a point is missing on the graph. Now when x is near one, but not equal to one,

then the F of X values that is the y values are very close to 10. In the language of limits,

we say that the limit as x approaches one of F have x is equal to Ted. More informally,

we can write, as x approaches one, f of x approaches 10. Notice that the value of f

at one is actually equal to zero, not 10. And so the limit of f of x as x goes to one,

and the value of f at one, are not equal. This illustrates the important fact that the

limit here as x goes to one doesn't care about the value of f, at one, it only cares about

the value of f, when x is near one, in general, for any function f of x, and for real numbers,

a and L, the limit as x goes to a of f of x equals L means that f of x gets arbitrarily

close to L, as x gets arbitrarily close to a, in other words, as x heads towards a, f

of x heads towards L. Let me draw this as a picture. In this picture, I can say that

the limit as x goes to a of f of x is L. Because by taking a sufficiently small interval around

a of x values, I can guarantee that my y values, my f of x values lie in an arbitrarily small

interval around L. Now the limit doesn't care about s value at exactly a. So if I change

the function, if I change the functions value at a, or even leave the function undefined

at a, the limit is still L. But the limit does care about what happens for x values

on both sides of a.

If, for example, the function didn't even exist for x values bigger than a, then we

could no longer say that the limit, as x approached a was out, the limit would not exist. In other

words, the limit of f of x is L, only if the y values are approaching now, as x approaches

a from both the left and the right. For the function g of x graph below, the limit as

x goes to two of g of x does not exist. Because the y values approach one as the x values

approach two from the left and the y values approach three, as the x values approach two

from the right. Although the limit doesn't exist, we can say that the left sided limit

exists. And we write this as limit as x goes to two from the left of g of x is equal to

one. The superscript minus sign here means that x is going to two from the left side.

In other words, the x values are less than two and approaching two. Similarly, we can

talk about right sided limits. In this example, the limit as x goes to two from the right

side of g of x is three. And here, the superscript plus side means that we're approaching two

from the right side. In other words, our x values are greater than to and approaching

to. In general, the limit as x goes to a minus of f of x equals L means that f of x approaches

L, as x approaches a from the left, and the limit as x goes to a plus of f of x equals

our means that f of x approaches our as x approaches a from the right limits from the

left or from the right are also called one sided limits. In this last example, let's

look at the behavior of y equals h of x graph below when x is near negative two. As x approaches

negative two from the right, our Y values are getting arbitrarily large, larger than

any real number we might choose. We can write this in terms of limits by saying the limit

as x goes to negative two from the right of h of x is equal to infinity. As x approaches

negative two from the left, the y values are getting smaller. In fact, they go below any

negative real number we might choose. In terms of limits, we can say that the limit as x

goes to negative two from the left of h of x is negative infinity. In this example, the

limit of h of x as x goes to negative two, not specifying from the left or the right

means we have to approach negative two from both sides. And so this limit does not exist,

because the limits from the left and the limit from the right are heading in opposite directions.

I want to mention that some people say that the limits from the right and the limits from

the left also do not exist. Because the functions don't approach any finite number. I prefer

to say that these limits do not exist as a finite number. But they do exist as infinity

and negative infinity. This video introduced the idea of limits, and one sided limits and

infinite limits. This video gives some examples of when limits fail to exist. For this function,

f of x graph below, let's look at the behavior of f of x in terms of limits, as x approaches

negative one, one, and two. Let's start with x approaching negative one. When x approaches

negative one from the last, the y value is going to be approaching about one half. When

x approaches negative one from the right, the y values sandvi, approaching one. So when

x approaches negative one, and we don't specify from either the left or the right, we can

only say that the limit does not exist, because these two limits from the left and right are

not equal.

Now let's look at the limit as x is approaching one. This time, we approach from the left,

we get a limiting y value of two, we approach from the right, the y values are going towards

two. So both of these left and right limits are equal to

and therefore the limit as x goes to one of f of x equals two. That's true, even though

f of one f of one itself does not exist. The limit doesn't care what happens at exactly

x equals one just what happens when x is near one.

Finally, let's look at the limit as x goes to two. So here on the left side, the limit

is going to negative infinity. And on the right side, it's negative infinity. So we

can say the limit as x goes to two is negative infinity. Or we can also say that the limit

as x goes to two does not exist. This is the correct answer. This is a better answer because

it carries more information. What values of x does a limit of f of x fail to exist? Well,

let's see. Negative one and two are the only two values. Let's talk about the ways that

limits can fail to exist, we've seen at least a couple different ways. So we've seen examples

where the limit from the left is not equal to the limit from the right. Here's our number

a where we're calculating the limit at. So that's one example we've seen. We've also

seen examples where they're vertical asymptotes. There's a vertical asymptote here at a, that

limit fails to exist because of the unbounded behavior because the y values are going off

to infinity. There's one other way that limits can fail to exist that comes up, sometimes

not quite as frequently. And that's wild behavior. Not a technical term,

just to descriptive term.

Let's look at an example that has this wild behavior, forcing a limit not to exist. And

the one of the most classic examples as the limit as x goes to zero of sine pi over x

or sometimes you'll see sine one over x. If you graph this on your graphing calculator

and zoom in near x equals zero, you're gonna see something that looks roughly like this.

It just keeps oscillating up and down and up and down, as x goes towards zero As x goes

towards zero, pi over x is getting bigger and bigger. And you're going to go through

these oscillations between one and negative one for the faster and faster.

From the other side, when x is negative, you'll see a similar kind of behavior just oscillating

faster and faster. As x goes to zero,

here, this top values up here at one, and the bottom value of these are all supposed

to hit it negative one. Now, when you try to decide what the limit is, as x goes to

zero, well, the y values are going through all possible real numbers and between negative

one and one infinitely often as x goes to zero, so there's no single number that the

limit can settle at. And so the limit as x goes to zero, of sine pi over x does not exist.

In this video, we saw three types of examples when limits fail to exist, they can fail to

exist because the one sided limits on the left and the right are not equal. Or they

can fail to exist because of vertical asymptotes. Also, limits can fail to exist when there's

wild behavior. And the function fails to settle down at any single value. This video is about

limit laws, rules for finding the limits of sums, differences, products, and quotients

and functions. Let's start with an example. Suppose that the limit as x goes to seven

of f of x is 30. And the limit as x goes to seven of g of x is two. What's the limit as

x goes to seven of f of x divided by negative three times g of

Wilson's f of x is heading towards 30. And g of x is heading towards two, it makes sense

that the quotient should head towards 30 divided by negative three times two, or negative five.

in calculating this limit, by plugging in numbers for the components, we were actually

using the limit loss, which are now state. Suppose that c is a constant, just some number,

and that the limits as x goes to a of f of x and g of x exist as finite numbers that

is not as limits that are infinity or negative infinity. Then the limit of the sum f of x

plus g of x is equal to the limit of f of x plus the limit of g of x. In other words,

the limit of the sum is the sum of the limits. Similarly, the limit of the difference is

the difference of the limits. The limit of C times f of x is just c times the limit of

f of x. And the limit of the product is the product of the limits. The limit of the quotient

is the quotient of the limits, provided that the limit of g of x is not equal to zero.

Since we can't divide by zero. Let's use these limit laws in an example, to find the limit

as x goes to two of x squared plus 3x plus six divided by x plus nine. Well, the limit

rule about quotients allows us to rewrite this limit of a quotient as a quotient of

limits, provided that the limits of the numerator and denominator exist, and that the limit

of the denominator is not zero. But we'll see in a moment that these conditions do in

fact hold. Next, we can use the limit rule about sums to rewrite the numerator as a sum

of limits. And we can rewrite the limit in the denominator also as a sum of limits. Next,

we can use the limit rule about products to rewrite the limit of x squared as the square

of the limit. That's because x squared is really x times x. And so the limit of x squared

is really the limit of x times x, which by the product rule is a limit of x times the

limit of x, which is the limit of x quantity squared. Going back to the original problem,

we can now use the limit rule about multiplying by a constant to rewrite the limit of three

times x as three times the limit of x. And I'll just carry the rest forward. Now that

we've got things broken down into bite sized pieces, we can start evaluating some limits.

Notice that the limit as x goes to two of x is just two, because as x heads towards

2x, heads towards two, so we can replace all these limits of access by just the number

two. So we get two squared plus three times two. Now notice that the limit as x goes to

two of six, well, six doesn't have any x's in it. So as x two heads towards to six days

at six, and the limit here is just six. So in my original problem, I can replace the

set limit of six with six. And on the denominator here, I get to plus nine. And after a little

arithmetic, this simplifies to 16/11. Notice that we could have gotten this answer a lot

faster, just by substituting in the value of two into our original expression. And in

fact, that's the beauty of the lemon laws, they allow us to evaluate limits of rational

functions, just by plugging in the number that x is going towards, as long as plugging

in that number doesn't make the denominator zero. It's not nearly so simple. When plugging

in the value does make the done at zero. And in the future, we'll build up a bunch of algebraic

techniques for handling the situation. In this video, we've talked about the limit loss.

It's important to note, these limit laws only apply if the limits of the component functions

actually exist as finite numbers.

If the limit of one or both of the component functions don't exist, then a limit rules

just simply don't apply. And instead, we have to use other techniques to try to evaluate

the limit of a sum, difference product or quotient. This video is about the squeeze

theorem, which is another method for finding limits. Let's start with an example. Suppose

we have a function g of x. We don't know much about it. But we do know that for x value

is near one, g of x is greater than or equal to four times the square root of x drawn here

in red, and less than or equal to 3x squared minus 4x plus five, drawn here in blue. So

on the picture, G has to lie between the red and the blue curves for x values near one,

it could look something like this. What can we say about the limit of g of x as x goes

to one? Well, if you notice, the red curve and the blue curve have the exact same limit

of four as x goes to one. And since the green curve is squeezed in between the red and the

blue curve, its limit must also be for this example as a special case of the squeeze theorem.

Now, let's say the squeeze theorem in general, suppose that we have three functions, f of

x g of x and h of x. And let's suppose that f of x is less than or equal to g of x, which

is less than or equal to h of x, at least for x values near some number A. this inequality

doesn't necessarily have to hold for x equal to a, because we're going to be talking about

limits. And limits don't care what happens when x is exactly a just when x is near a.

Let's suppose also, that like in the previous example, f of x and h of x have the exact

same limit as x approaches a. So we're going to suppose that the limit as x goes to a of

f of x is equal to the limit as x goes to a of h of x. And we'll call this limit. Now.

The picture looks a lot like the previous example. Again, it doesn't matter exactly

what happens at x equals a, for example, g of x could have a hole there. And its value

could be for example, way up here. Since g of x is trapped here in between f of x and

h of x, which both have the same limit l at a, we can conclude that the limit as x goes

to a of g of x is equal to l also. And that's the squeeze theorem, also known as the pinching

theorem, and the sandwich theorem, three very descriptive names that capture the idea of

geovax being trapped here, in between lower and upper bounds. Now let's use the squeeze

there. To find the limit as x goes to zero of x squared sine one over x. Now, you might

remember that sine one over x by itself has this crazy oscillating behavior. In fact,

the limit as x goes to zero of sine one over x does not exist, because the function never

settles down to a single finite value. At first glance, you might think the limit of

x squared sine of one over x also wouldn't exist. In fact, it's tempting to try to use

the product rule and say that the limit of the product is the product of the limits.

But in fact, the product rule only applies when the component limits both exist. And

since the second limit doesn't exist, the product rule tells us absolutely nothing about

whether the limit that we're interested in exists or doesn't. So we can't use the product

rule. But it turns out, we can use the squeeze theorem. Now this example is a little trickier

than the first example, because in the first example, we were told what the upper and lower

bounding functions should be. And in this example, we have to come up with. But if we

look at a graph of x squared sine one over x, we can see that it does seem to be trapped

in an envelope here.

Let's use algebra to see what those two bounding functions might be. Now we know that sine

of one over x is always between one and negative one, just because sine of anything has between

one and negative one. And if we multiply this whole inequality by x squared, we get minus

x squared is less than or equal to x squared, sine one over x, which is less than or equal

to x squared. Notice that x squared is always positive, so we don't have to worry about

flipping any of the inequality signs when we multiply by this positive number. So x

squared and minus x squared are good bounding functions. And if we notice that the limit

as x goes to zero of x squared is zero, and the limit as x goes to zero of negative x

squared is also zero, we can conclude by the squeeze theorem, that the limit as x goes

to zero of x squared, sine of one over x is also zero, because it squeezed in between

these two functions with the same limit. the squeeze theorem is a great trick for evaluating

limits when you happen to have a function that you're interested in, trapped in between

two other functions with the same limit. The example in this video is a classic example,

where we have a crazy oscillating trig function, multiplied by a power of x, that x is one

of our bounding functions. In this video, we'll compute a bunch of limits using algebraic

tricks. All these limits are of the zero over zero indeterminate form kind. Recall that

this means that the limits are of the form the limit of f of x over g of x where the

limit of f of x equals zero and the limit of g of x also equals zero. For a first example,

let's look at the limit as x goes to one of x cubed minus one over x squared minus one.

Notice that the numerator and the denominator are both going to zero as x goes to one. To

calculate this limit, we want to simplify this expression. And one way to simplify it

is to factor it. So let's rewrite this as the limit as x goes to one of x minus one

times x squared plus x plus one. We're factoring the numerator here is a difference of cubes.

We can also factor the denominator as a difference of squares, x minus one times x plus one.

Now, as long as x is not equal to one, we can cancel out these two factors of x minus

one. And so the limit of this expression is just the same as the limit of this expression.

Now we can just plug in one, because plugging in one gives us a numerator of three and a

denominator of two. And that way, we've evaluated our limits. Next, let's look at the limit

of five minus z quantity squared minus 25. All divided by z. Once again, when we plug

in x equals zero in the numerator, we get zero and then nominator we also get zero.

This time instead of factoring, the trick is going to be to multiply out. So let's rewrite

this limit as the limit as t goes to zero

25 minus 10 z plus z squared minus 25. I just distributed to get this expression divided

by z. Since 25 minus 25 is zero, I just have the limit as t goes to zero of negative 10

z plus c squared over

Now what? Well, I could factor out the Z here. One z is not zero, I can cancel here. So my

original limit is the same as this limit. Plugging in z equals zero, I just get negative

10 as my answer. This third example is also a situation where the numerator is going to

zero, and the denominator is also going to zero. In this case, I'm going to try to simplify

the expression by adding together the fractions and the numerator. So I'll need a common denominator,

which is r plus three, times three. So rewriting, I get the limit of one over r plus three,

I multiply that by three over three, in order to get the appropriate common denominator,

minus 1/3, which I have to multiply by R plus three over r plus three, all that over are

continuing to rewrite, I have in the numerator, adding together these fractions mature, have

the denominator of R plus three times three, I get three minus quantity r plus three. And

then this entire fraction is still divided by our distributing the negative sign, I have

three minus r minus three divided by R plus three times three, all divided by R. three

minus three is zero. So I can rewrite this as negative r over r plus three, times three.

And now instead of dividing by r, which is r of r one, I can multiply by the reciprocal,

one over R. R divided by R is one. So this expression simplifies to negative one, over

r plus three, times three. So finally, I'm in a good position, because now I can just

go ahead and let our go to zero. And by plugging in R equals zero, I have a limit of negative

one over zero plus three times three, or negative one, nine. This example is a little tricky,

because involve square roots can be hard to deal with. Well, there's one nice trick for

dealing with square roots that works here, which is the conjugate. So I'm going to take

the expression that we're given, and multiply by the conjugate of the numerator, in this

case, because the numerator is the place where the square root is by the conjugate of A minus

B, I just mean a plus b conjugate of A plus B is a minus b. Well, of course, if I multiply

on the numerator, by something, I also have to multiply the denominator by the same thing,

so that I won't alter the value of the expression. So this limit here is equal to the limit of

the expression down here that looks more complicated, than in a moment, if we're lucky, things will

clear up and become simpler. So Multiplying the numerators across, I get the square root

of x plus three squared.

I get

plus two times the square root of x plus three minus two times the square root of x plus

three minus four. On the denominator, I get x times the square root of x plus three

plus

2x minus the square root of x plus three minus two. Let's see what simplifies here. So the

square root of x plus three squared is just x plus three. And I see that this expression

and this expression are opposites, so they subtract to zero here. So on the numerator,

I just have x plus three, and then I still have the minus four on the denominator, nominator

looks a little messy. I'll just copy it over for now. Okay, so on my numerator, I'm getting

x minus one. Notice that when x goes to one, that numerator is still going to zero. And

in fact, as I let x go to one, that denominator, if I plug in here, everything is gonna also

cancel out to zero. So I still got a zero over zero indeterminate form. But maybe I

can use one of the previous tricks of factoring. Because look at here, if I factor an X out

of these two expressions as a square root of x plus three out of these two expressions,

and I could maybe factor a two out of these two, factoring by grouping, let's try that.

So we have x squared of x plus three times x minus one from these two expressions, and

then I have a plus two times x minus one. From these two expressions, this is looking

promising. So now I've got an x minus one on the top. And if I factor out the x minus

one from each of these two expressions, I'm going to have an x minus one on the bottom

times the square root of x plus three, plus two. Now, for x values near one, but not equal

to one, I can cancel those. And my limit simplifies to just one over square root of x plus three

plus two, plugging in one, I get a one on the numerator, and square root of four, which

is two plus two, four on the denominator, and we have calculated this limit. Last example

here, another zero over zero indeterminate form. Anytime I see an absolute value, I'm

going to want to take cases, because the absolute value of X plus five naturally follows in

the cases, if X plus five is greater than zero, in other words, x is greater than negative

five, then the absolute value of this positive number is just

itself.

On the other hand, if X plus five is less than zero, in other words, x is less than

negative five, than the absolute value of a negative number is its opposite. And we

make the expression x plus five, turn it into its opposite by putting a negative sign in

front. Now let's look at one sided limits. As x goes to negative five from the left,

we have

a situation where x is less than negative five, this situation right here. And so we

can rewrite the absolute value by taking the negative of the expression. We still have

a zero over zero and determinant form. But if we use the old factoring trick,

factor out a two cancel, we got the limit of two over negative one, which is just negative

two.

We do the same exercise on the right side. When we're on the right, then x is greater

than negative five. So we're in this situation here, where we can just replace the absolute

value with the stuff inside and again factoring the numerator and canceling the X plus five

We just get a limit

of two.

So we have a left limit, and a right limit that are different. And so in this example,

the limit does not exist. So we've seen a five different kinds of limits all of the

zero or zero indeterminate form. And we've used five different methods to evaluate them.

For the first example, we use factoring. For the second example, do the opposite of factoring,

we multiplied out. For the third example, we added together our rational expressions

to simplify things. The next example uses the old multiply by the conjugate trick. And

the last example, we used cases and looked at one sided limits.

The limit law about quotients tells us that the limit of the quotient is the question

of the limit, provided that the limits of the component functions actually exist, and

that the limit of the function on the denominator is not equal to zero. But what happens if

the limit of the function on the denominator is equal to zero? This video will begin to

answer that question. In fact, there are two different situations we'll want to consider.

It could be that even though the limit on the denominator is equal to zero,

the limit on the numerator exists and is not equal to zero. Or it could happen that both

limits are zero. We'll focus on the first situation, first starting with an example.

And we'll look at the second situation later on.

In this example, the limit of the numerator, negative 4x is just negative 12, which we

can see by plugging in three for x. But the limit as x goes to three of the denominator

is zero. So we're exactly in one of these situations where the numerator goes to a finite

nonzero number, but the denominator goes to zero. Let's see what happens as we approach

three from the left first. As we approach three from the left, x is going through numbers

that are slightly less than three numbers like 2.9 2.99 2.999, and so on. If we plug

in those numbers into the expression, here on our calculator, we didn't get answers of

116 1196, and 11,996. If even without a calculator, we could approximate these answers pretty

closely by just thinking about the fact that since x is very close to three, the numerator

is about negative four times three, so about negative 12. The denominator 2.9 minus three

is negative 0.1. That quotient of two negative numbers gives us a positive value of 120.

Similarly, we could approximate the value when x is 2.9 times as almost negative 12

divided by 0.01, which is 1200. And approximate the third value as 12,000. Either way, we

do it exact answers on our calculator or approximations in our head when noticing that these values

are positive numbers that are getting larger and larger as x goes towards three from the

left. This makes sense. Because if we look at our expression, as x goes towards three

from the left, the numerator is getting close to negative 12, which is a negative number.

And the denominator, since x is less than three will always be a small negative number,

negative over negative is a positive. And as x is getting really close to three, those

denominators are getting smaller and therefore the fractions are getting bigger and bigger

in magnitude. So we can conclude that our limit is positive infinity. We can make a

similar argument By looking at the limit, as x goes to three from the positive side,

it's supposed to be an X minus three here. So now x is going through a value is slightly

bigger than three 3.1 3.01 3.01. And again, we can plug directly into our calculator and

figure out the answers are negative 124. Negative 1204, negative 12,004. Or we can make a similar

approximating argument. This answer is approximately negative 12 over a positive point one, which

is negative 120, and so on. Like before, if we consider the signs of our numerator and

denominator, we can see that as x goes to three, our numerator is a negative number.

But our denominator is a positive number, since we're approaching three from the right

where x is bigger than three, and therefore, our quotient is a negative number, it's still

getting bigger and bigger in magnitude as x goes towards three, because the denominator

is still getting tinier and tinier, while the numerator stays pretty close to negative

12. So in this case, we're getting a negative number that's bigger and bigger magnitude,

so that makes a limit of negative infinity. Now, since our limit on the left is infinity,

and our limit on the right is negative infinity, the only thing we can say about the limit

as x goes to three

is that it does not exist. Now let's look at another example. The limit as x goes to

negative four of 5x, over the absolute value of x plus four, notice that the limit of the

numerator is just negative 20 by plugging in negative four for x, and the limit of the

denominator is zero. Because we've got an absolute value in our expression, here, it's

screaming out at us to look at cases. Remember that the absolute value of x plus four is

going to equal just x plus four, if x plus four is positive, in other words, if x is

greater than negative four, however, if x is less than negative four, then the expression

x plus four will be negative. So taking the absolute value has to switch at sign in order

to make a negative expression positive. Alright, so that's going to come in handy when we look

at the limit as x goes to negative four, from the left, and from the right. So when we approach

negative four from the left, x is going to be less than negative four. So we're going

to be in this situation here, where the absolute value gives us the opposite sign. That means

that the limit as x goes to negative four minus of this expression, the same as the

limit of 5x over negative x plus four. Now reasoning as before, as x is going to negative

four from the left, the numerator here is a negative number. Pretty close to negative

20. The denominator, since x is less than negative 4x plus four is negative, the negative

of it is positive. So our quotient is negative. And since denominator is getting really tiny,

while the numerators will pretty level at at negative 20. This limit is going to be

bigger and bigger magnitude a limit of negative infinity. Now let's look at the limit as x

goes to negative four from the right, in this case, x is just a little bit bigger than negative

four. So we're in this case where the absolute value doesn't change the expression. So we

can rewrite this As 5x

over x plus four,

now the numerator is still gonna be a negative number, the denominator, since x is slightly

bigger than negative four, slightly to the right, this expression is going to be positive

number. Negative or over a positive is a negative. And again, since the denominator is getting

tiny, the fractions getting huge in magnitude. And so this limit is negative infinity. Now,

in this case, look at what's going on, we've got a negative infinity limit on the left

and a negative infinity limit on the right. So we can conclude that the limit as x goes

to negative four of 5x, over the absolute value of x plus four is equal to negative

infinity. In fact, we can confirm that by looking at a graph. If we look, check out

the graph near x equals negative four, it's gonna look something like this, with a vertical

asymptote at x equals negative four, like expected. So we've seen that if the limit

of f of x is equal to something that's not zero, and the limit of g of x is equal to

zero, then the limit of the quotient could be negative infinity as it was in the past

example. It could also be infinity, or it could just not exist. supposed to say does

not exist, if the one sided limits are infinity on one side, and negative infinity on the

other. Now, what about this second situation that I mentioned the beginning when the limit

of f of x is zero, and the limit of f of g of x is zero? What can we say about the limit

of the quotient in this situation? Well, in fact, in this situation, the limit of the

quotient, it could exist and be any finite number, or infinity, or minus infinity, or

it could not exist at all. In fact, in this sort of situation, which is called a zero,

or zero indeterminate form, anything could happen. Which makes it in some ways the hardest,

but in some ways, the most fun situation of all. So in another video, we'll talk about

techniques for dealing with 00 indeterminate forms, and how to use algebra and other simplification

techniques to evaluate these these mysterious limits. So in this video, we've looked at

the limits of quotients, when the limit of the denominator is zero. We've done some examples

when the limit of the numerator was not zero, but the limit of the denominator was zero.

And we saw that these situations corresponded to vertical asymptotes, and gave us an answer

for the limit of the quotient of infinity, or negative infinity, or sometimes infinity

on one side and negative infinity on the other. We also hinted at fun things to come when

we look at the limits in this situation when the numerator and the denominator are both

heading towards zero. And when anything can happen. This video is about graphs and equations

of lines. Here we're given the graph of a line, we want to find the equation, one standard

format for the equation of a line is y equals mx plus b. here, m represents the slope, and

B represents the y intercept, the y value, where the line crosses the y axis, the slope

is equal to the rise over the run. Or sometimes this is written as the change in y values

over the change in x values. Or in other words, y two minus y one over x two minus x one,

where x one y one and x two y two are points on the line.

While we could use any two points on the line, to find the slope, it's convenient to use

points where the x and y coordinates are integers. That is points where the line passes through

grid points. So here would be one convenient point to use. And here's another convenient

point to use. The coordinates of the first point are one, two, and the next point this

is let's say, five negative one. Now I can find this slope by looking at the rise over

the run. So as I go through a run of this distance, I go through a rise of that distance,

especially gonna be a negative rise or a fall because my line is pointing down. So let's

see counting off squares, this is a run of 1234 squares and a rise of 123. So negative

three, so my slope is going to be negative three, over four. I got that answer by counting

squares. But I could have also gotten it by looking at the difference in my y values over

the difference of my x values. That is, I could have done negative one minus two, that's

from my difference in Y values, and divide that by my difference in x values, which is

five minus one, that gives me negative three over four, as before, so my M is negative

three fourths. Now I need to figure out the value of b, my y intercept, well, I could

just read it off the graph, it looks like approximately 2.75. But if I want to be more

accurate, I can again use a point that has integer coordinates that I know it's exact

coordinates. So either this point or that point, let's try this point. And I can start

off with my equation y equals mx plus b, that is y equals negative three fourths x plus

b. And I can plug in the point one, two, for my x and y. So that gives me two equals negative

three fourths times one plus b, solving for B. Let's see that two equals negative three

fourths plus b. So add three fourths to both sides, that's two plus three fourths equals

b. So b is eight fourths plus three fourths, which is 11 fourths, which is actually just

what I eyeballed it today. So now I can write out my final equation for my line y equals

negative three fourths x plus 11 fourths by plugging in for m and b. Next, let's find

the equation for this horizontal line. a horizontal line has slope zero. So if we think of it

as y equals mx plus b, m is going to be zero. In other words, it's just y equals b, y is

some constant. So if we can figure out what that that constant y value is, it looks like

it's to let's see, this three, three and a half, we can just write down the equation

directly, y equals 3.5. For a vertical line, like this one, it doesn't really have a slope.

I mean, if you tried to do the rise over the run, there's no run. So you'd I guess you'd

be divided by zero and get an infinite slope. But But instead, we just think of it as an

equation of the form x equals something. And in this case, x equals negative two, notice

that all of the points on our line have the same x coordinate of negative two and the

y coordinate can be anything. So this is how we write the equation for a vertical line.

In this example, we're not shown a graph of the line, we're just get told that it goes

through two points. But knowing that I go through two points is enough to find the equation

for the line. First, we can find the slope by taking the difference in Y values over

the difference in x values. So that's negative three minus two over four minus one, which

is negative five thirds. So we can use the standard equation for the line, this is called

the slope intercept form.

And we can plug in negative five thirds. And we can use one point, either one will do will

still get the same final answer. So let's use one two and plug that in to get two equals

negative five thirds times one plus b. And so B is two plus five thirds, which is six

thirds plus five thirds, which is 11 thirds. So our equation is y equals negative five

thirds x plus 11 thirds. This is method one. method two uses a slightly different form

of the equation. It's called the point slope form and it goes y minus y naught is equal

to m times x minus x naught where x naught Why not is a point on the line, and again

is the slope. So we calculate the slope the same way, by taking a difference in Y values

over a difference in x values. But then we can simply plug in any point. For example,

the point one, two will work, we can plug one in for x naught and two in for Y not in

this point slope form, that gives us y minus two is equal to minus five thirds x minus

one. Notice that these two equations, while they may look different, are actually equivalent.

Because if I distribute the negative five thirds, and then add the two to both sides,

I get the same equation

as above.

So we've seen two ways of finding the equation for the line is in the slope intercept form,

and using the point slope form. In this video, we saw that you can find the equation for

a line if you know the slope. And you know one point, you can also find the equation

for the line if you know two points, because you can use the two points to get the slope

and then plug in one of those points. To figure out the rest of the equation. We saw two standard

forms for the equation of a line the slope intercept form y equals mx plus b, where m

is the slope, and B is the y intercept. And the point slope form y minus y naught equals

m times x minus x naught, where m again is the slope and x naught y naught is a point

on the line. This video is about rational functions and their graphs. Recall that a

rational function is a function that can be written as a ratio or quotient of two polynomials.

Here's an example. The simpler function, f of x equals one over x is also considered

a rational function, you can think of one and x as very simple polynomials. The graph

of this rational function is shown here. This graph looks different from the graph of a

polynomial. For one thing, its end behavior is different. The end behavior of a function

is the way the graph looks, when x goes through really large positive, or really large negative

numbers, we've seen that the end behavior of a polynomial always looks like one of these

cases. That is why marches off to infinity or maybe negative infinity, as x gets really

big or really negative. But this rational function has a different type of end behavior.

Notice, as x gets really big, the y values are leveling off at about a y value of three.

And similarly, as x values get really negative, our graph is leveling off near the line y

equals three, I'll draw that line, y equals three on my graph, that line is called a horizontal

asymptote. A horizontal asymptote is a horizontal line that our graph gets closer and closer

to as x goes to infinity, or as X goes to negative infinity or both. There's something

else that's different about this graph from a polynomial graph, look at what happens as

x gets close to negative five. As we approach negative five with x values on the right,

our Y values are going down towards negative infinity. And as we approach the x value of

negative five from the left, our Y values are going up towards positive infinity. We

say that this graph has a vertical asymptote at x equals negative five. vertical asymptote

is a vertical line that the graph gets closer and closer to. Finally, there's something

really weird going on at x equals two, there's a little open circle there, like the value

at x equals two is dug out. That's called a hole. A hole is a place along the curve

of the graph where the function doesn't exist. Now that we've identified some of the features

of our rational functions graph, I want to look back at the equation and see how we could

have predicted those features just by looking at the equation. To find horizontal asymptotes.

We need to look at what our function is doing when x goes through really big positive or

really big negative numbers. Looking at our equation for our function, numerator is going

to be dominated by the 3x squared term when x is really big, right, because three times

x squared is going to be absolutely enormous compared to this negative 12. If x is a big

positive or negative number in the denominator, the denominator will be dominated by the x

squared term. Again, if x is a really big positive or negative number, like a million,

a million squared will be much, much bigger than three times a million or negative 10.

For that reason, to find the end behavior, or the horizontal asymptote, for our function,

we just need to look at the terms on the numerator and the term on the denominator that have

the highest exponent, those are the ones that dominate the expression in size. So as x gets

really big, our functions y values are going to be approximately 3x squared over x squared,

which is three. That's why we have a horizontal asymptote at y equals

three. Now our vertical asymptotes, those tend to occur where our denominator of our

function is zero. That's because the function doesn't exist when our denominator is zero.

And when we get close to that place where our denominator is zero, we're going to be

dividing by tiny, tiny numbers, which will make our Y values really big in magnitude.

So to check where our denominators zero, let's factor our function. In fact, I'm going to

go ahead and factor the numerator and the denominator. So the numerator factors, let's

see, pull out the three, I get x squared minus four, factor in the denominator, that factors

into X plus five times x minus two, I can factor a little the numerator a little further,

that's three times x minus two times x plus two over x plus 5x minus two. Now, when x

is equal to negative five, my denominator will be zero, but my numerator will not be

zero. That's what gives me the vertical asymptote at x equals negative five. Notice that when

x equals two, the denominators zero, but the numerator is also zero. In fact, if I cancelled

the x minus two factor from the numerator and denominator, I get a simplified form for

my function that agrees with my original function, as long as x is not equal to two. That's because

when x equals two, the simplified function exists, but the original function does not

it's zero over zero, it's undefined. But for every other x value, including x values near

x equals two, our original function is just the same as this function. And that's why

our function only has a vertical asymptote at x equals negative five, not one at x equals

two, because the x minus two factor is no longer in the function after simplifying,

it does have a hole at x equals two, because the original function is not defined there,

even though the simplified version is if we want to find the y value of our hole, we can

just plug in x equals two into our simplified version of our function, that gives a y value

of three times two plus two over two plus seven, or 12 ninths, which simplifies to four

thirds. So our whole is that to four thirds. Now that we've been through one example in

detail, let's summarize our findings. We find the vertical asymptotes and the holes by looking

where the denominator is zero. The holes happen where the denominator and numerator are both

zero and those factors cancel out. The vertical asymptotes are all other x values where the

denominator is zero, we find the horizontal asymptotes by considering the highest power

term on the numerator and the denominator, I'll explain this process in more detail in

three examples. In the first example, if we circle the highest power terms, that simplifies

to 5x over 3x squared, which is five over 3x. As x gets really big, the denominator

is going to be huge. So I'm going to be dividing five by a huge, huge number, that's going

to be going very close to zero, and therefore we have a horizontal asymptote at y equals

zero. In the second example, the highest power terms, 2x cubed over 3x cubed simplifies to

two thirds. So as x gets really big, we're going to be heading towards two thirds and

we have a horizontal asymptote at y equals two thirds. In the third example, the highest

power terms x squared over 2x simplifies to x over two. As x gets really big, x over two

is getting really big. And therefore, we don't have a horizontal asymptote at all. This is

going to infinity, when x gets through goes through big positive numbers, and is going

to negative infinity when x goes through a big negative numbers. So in this case, the

end behavior is kind of like that of a polynomial, and there's no horizontal asymptote.

In general, when the degree of the numerator is smaller than the degree of the denominator,

we're in this first case where the denominator gets really big compared to the numerator

and we go to zero. In the second case, where the degree of the numerator and the degree

of the denominator equal, things cancel out, and so we get a horizontal asymptote at the

y value, that's equal to the ratio of the leading coefficients. Finally, in the third

case, when the degree of the numerator is bigger than the degree of the denominator,

then the numerator is getting really big compared to the denominator, so we end up with no horizontal

asymptote. Finally, let's apply all these observations to one more example. Please pause

the video and take a moment to find the vertical asymptotes, horizontal asymptotes and holes

for this rational function. To find the vertical asymptotes and holes, we need to look at where

the denominator is zero. In fact, it's going to be handy to factor both the numerator and

the denominator. Since there if there are any common factors, we might have a hole instead

of a vertical asymptote. The numerator is pretty easy to factor. Let's see, that's 3x

times x plus one for the denominator, first factor out an x. And then I'll factor some

more using a guess and check method. I know that I'll need a 2x and an X to multiply together

to the to x squared, and I'll need a three and a minus one or also minus three and a

one. Let's see if that works. If I multiply out 2x minus one times x plus three, that

does give me back my 2x squared plus 5x minus three, so that checks out. Now, I noticed

that I have a common factor of x in both the numerator and the denominator. So that's telling

me I'm going to have a hole at x equals zero. In fact, I could rewrite my rational function

by cancelling out that common factor, and that's equivalent, as long as x is not equal

to zero. So the y value of my

whole

is what I get when I plug zero into my simplified version, that would be three times zero plus

one over two times zero minus one times zero plus three, which is three over negative three

or minus one. So my whole is at zero minus one. Now all the remaining places in my denominator

that make my denominator zero will get me vertical asymptotes. So I'll have a vertical

asymptote, when 2x minus one times x plus three equals zero, that is, when 2x minus

one is zero, or x plus three is zero. In other words, when x is one half, or x equals negative

three. Finally, to find my horizontal asymptotes, I just need to consider the highest power

term in the numerator and the denominator. That simplifies to three over 2x, which is

bottom heavy, right? When x gets really big, this expression is going to zero. And that

means that we have a horizontal asymptote at y equals zero. So we found the major features

of our graph, the whole, the vertical asymptotes and the horizontal asymptotes. Together, this

would give us a framework for what the graph of our function looks like. horizontal asymptote

at y equals zero, vertical asymptotes at x equals one half, and x equals minus three

at a hole at the point zero minus one. plotting a few more points, or using a graphing calculator

of graphing program, we can see that our actual function will look something like this. Notice

that the x intercept when x is negative one corresponds to where the numerator of our

rational function or reduced rational function is equal to zero. That's because a zero on

the numerator that doesn't make the denominator zero makes the whole function zero. And an

X intercept is where the y value of the whole function is zero. In this video, we learned

how to find horizontal asymptotes of rational functions. By looking at the highest power

terms, we learned to find the vertical asymptotes and holes. By looking at the factored version

of the functions, the holes correspond to the x values that make the numerator and denominator

zero, his corresponding factors cancel. The vertical asymptotes correspond to the x values

that make the denominator zero, even after factoring any any common any common factors

in the numerator denominator. This video focuses on the behavior of functions and graphs, as

x goes through arbitrarily large positive and negative values. You may have touched

on these ideas before in the past when you studied horizontal asymptotes. But in this

video, we'll talk about the same ideas in the language of limits. In this first example,

what happens to the function f of x drawn below as x go through larger and larger positive

numbers? Well, the arrow here on the end is supposed to mean the trend continues. So as

x gets bigger and bigger, the values of y, that is f of x, get closer and closer to one,

we can write this in the language of limits by saying the limit as x goes to infinity

of f of x is equal to one. Now what happens to this function as x goes through larger

and larger negative numbers, by larger and larger negative numbers, I mean numbers that

are negative but are larger and larger in magnitude. So for example, negative five,

negative 10, negative 100, and negative a million and so on. Well, assuming this trend

continues, it looks like f of x, even though it's oscillating, it's settling down at a

value of two. So we say that the limit as x goes to negative infinity, of f of x equals

to limits in which x goes to infinity, or negative infinity are called limits at infinity.

The phrase limits at infinity should be contrasted with the phrase infinite limits. An infinite

limit means that the y values, or the F of X values go to infinity, or negative infinity.

limits and infinity correspond to horizontal asymptotes, as drawn above, while infinite

limits correspond to vertical asymptotes. The only exception to this when we don't have

a horizontal or vertical asymptote is when x and f of x are both going infinite at the

same time, we'll see an example of this on the next page. Let's figure out the limits

of infinity for these two functions, g of x and h of x. The function g of x is actually

the function e to the minus x, and it has a horizontal asymptote, heading right here

at y equals zero. So the limit as x goes to infinity of g of x equals zero. But as we

head to the left, and x goes through larger and larger negative values, our Y values don't

settle down to a particular finite value, they get arbitrarily large. And so we say

that the limit as x goes to minus infinity of g of x is equal to infinity. Now let's

look at the graph of y equals h of x. Please pause the video for a moment and try to figure

out the limits of infinity for this function. The limit as x goes to infinity of h of x

is negative infinity. Because as x goes to infinity, the y values get below or more negative

than any finite number. Now as X goes to negative infinity, the y values oscillate, and never

settle down at a particular number. So we say that the limit as x goes to minus infinity

of h of x does not exist. Finally, let's look at some limits of infinity of functions without

looking at their graphs first, to find the limit as x goes to infinity of one over x.

Let's think about what happens to one over x. As x gets bigger and bigger through positive

numbers. As x gets bigger and bigger, one over x gets smaller and smaller. So the limit

as x goes to infinity of one over x equals zero. To find the limit of one over x as x

goes to negative infinity, let's look at what happens as X goes to negative numbers that

are larger and larger in magnitude. Now one of our x goes through numbers that are negative,

but they're still getting smaller and smaller. magnitude. So the limit as x goes to negative

infinity of one over x is also zero. We can use similar reasoning to find the limit as

x goes to infinity of one over x cubed. As x goes to infinity, x cubed also goes to infinity.

So one of our x cubed has to go to zero. To find the limit as x goes to infinity of one

over the square root of x, notice that as x goes to infinity, the square root of x still

goes to infinity. So one over the square root of x also goes to zero. In other words, both

of these limits are equal to zero. Both of these examples are actually closely related,

because both have the form of the limit as x goes to infinity of one over x to the R,

where R is a number greater than zero. In the second example, the square root of x is

really x to the R, where R is one half. In fact, the limit as x goes to infinity of one

over x to the R is always equal to zero. Whenever R is bigger than zero, we can even say the

same thing about the limit as x goes to negative infinity of one over x to the R. As long as

we avoid exponents, like one half that don't make sense for negative numbers. But for other

values of r, as x goes to negative infinity, x to the R is getting bigger and bigger and

magnitude. And so one over x to the R is getting smaller and smaller and magnitude and heading

towards zero. Notice that this is no longer true. if r is less than zero, for example,

something like r equals negative two, because one over x to the minus two is really x squared.

And the limit as x goes to infinity of x squared is going to be infinity, not zero.

In this video, we looked at examples of limits as x goes to infinity, and x goes to minus

infinity. And we saw that those limits could be zero. Any other number, infinity, negative

infinity, or not exist. This video gives some algebraic techniques for computing the limits

at infinity of rational functions. Let's find the limit as x goes to infinity of this rational

function. The numerator and the denominator of this rational function are each getting

arbitrarily large as x goes to infinity. One way to see this is by estimating the graphs,

the graph of the numerator looks like a parabola pointing upwards. And the graph of the denominator

looks like some kind of cubic. So something like this, for both of these graphs, as x

goes to infinity, y also goes to infinity. So this is an infinity over infinity indeterminate

form. And just like the zero have over zero indeterminate forms we saw earlier, and infinity

over infinity and determinant form could turn out to be absolutely anything. So we're going

to use algebra to rewrite this expression in a different form that makes it easier to

evaluate. Specifically, we're going to factor out the highest power of x that we can find

from the numerator, and then from the denominator. In the numerator, I'm going to factor out

the highest power I see in the numerator, which is x squared. When I factor x squared

out of 5x squared, I get five, when I factor x squared out of negative 4x, that's like

dividing negative 4x by x squared, so I get negative four divided by x. You can check

this works by distributing the x squared and making sure we get back to the original expression.

Now the highest power of x SC, and the denominator is x cubed. So I'll factor out an x cubed

for each from each of those terms, I get a two minus 11 over x plus 12 over x squared.

Because factoring out an x cubed is the same as dividing each term by x cubed and then

writing the x cubed on the side. Now, we can rewrite again by canceling an x squared from

the top and the bottom to get the limit of one over x times five minus four over x over

two minus 11 over x plus 12 over x squared. Now as x goes to infinity, four over x goes

to zero, because I'm dividing for by larger and larger numbers. Similarly 11 over x goes

to zero, and 12 over x squared goes to zero. So I end up with one over x, which is itself

going to zero times something that's going to five halves. So my limit is going to be

zero times five halves, which is just zero. I've actually been applying limit laws to

do these last steps, which is fine, because my component limits exist as finite numbers.

Something that wasn't true from my original expression when I had infinity over infinity.

In this example, we're asked to find the limit as x goes to negative infinity of a different

rational expression. I encourage you to stop the video and try it for yourself first. In

this example, the highest power of x in the numerator is x cubed, and the highest power

in the denominator is also x cubed. Factoring out the x cubed from the numerator, we get

x cubed times three plus six over x plus 10 over x squared, plus two over x cubed. And

factoring out the x cubed from the denominator, we get x cubed times two, plus one over X

plus five over x cubed. Now the x cubes cancel, and all these parts go to zero. So when the

dust clears here, our limit is just three halves. In this next example, the highest

power in the numerator is x to the fourth, and the highest power in the denominator is

x squared. So we factor out the x to the fourth from the numerator and the x squared from

the denominator and cancel as much as we can. A lot of these pieces are going to zero.

So our limit is the same as the limit of x squared times one over negative five. As x

goes to negative infinity, x squared is positive and goes towards positive infinity. multiplying

by negative fifth turns it negative, but doesn't change the fact that the magnitudes are getting

arbitrarily large. Therefore, our final limit is negative infinity. Now let's look at the

same three examples again, more informally, using a heuristic to get the same conclusions.

In the first example, the term 5x squared dominates the numerator, because x squared

is much larger than x when x is large. In the denominator, the highest power of x to

x cubed dominates, because x cubed is much larger than x squared, or x when x is large.

If we ignore all the other terms in the numerator and denominator, and just focus on the important

terms, which have the highest powers, then we can rewrite our limit as the limit of 5x

squared over 2x cubed, which is the same as the limit as x goes to infinity of five over

2x, just by canceling Xs, which is zero as x goes to infinity. Similarly, if we just

focus on the highest power terms in the numerator and denominator in the second example, we

get the limit of 3x cubed over 2x cubed, which simplifies to the limit of three halves, which

is just three halves. In the third example, the highest power terms are x to the fourth

and negative 5x squared. And we rewrite the limit using only these highest power terms

and simplify, and we get the limit as x goes to negative infinity of x squared over negative

five, which is negative infinity as before, for rational functions, in general, looking

at the highest power terms, lets you reliably predict the limits that infinity and negative

infinity when the degree of the numerator is less than the degree of the denominator,

then the limit as x goes to infinity, or negative infinity, is zero. As an example, one above,

when the degree of the numerator is equal to the degree of the denominator, then the

limit is just the quotient of the highest power terms, which is how we got three halves

as the limit in the second example. And finally, if the degree of the numerator is greater

than the degree of the denominator, then the limit is going to be plus or minus infinity.

Like it was in the third example. These shortcut rules are really handy. But it's important

to also understand the technique of factoring out highest power terms. Since this technique

can be used more generally. This video gave two methods for computing limits and infinity

of rational functions. First, there's the formal method of factoring out highest power

terms and simplifying. Second, there's the informal method of looking at the degree of

the numerator and the degree of the denominator, and focusing on the highest power terms. In

the past, you may have heard an informal definition of continuity. Something like a function is

continuous, if you can draw it without ever picking up your pencil. In this video, we'll

develop a more precise definition of continuity based on limits. It can be helpful to look

at some examples of functions that are discontinuous, that is functions that fail to be continuous.

In order to better understand what it means to be continuous. Please pause this video

and try to draw graphs of at least two different functions that fail to be continuous in different

ways. One common kind of discontinuity is called a jump discontinuity. A function has

a jump discontinuity, if its graph separates the two pieces with a jump in between them.

This particular function can be described as a piecewise defined function with two linear

equations, f of x equals to x when x is less than or equal to one, and f of x equals negative

x plus two, when x is greater than one. Another common kind of discontinuity is called a removable

discontinuity.

You may have encountered these before, when you learned about rational functions with

holes in them, for example, the function f of x equals x minus three squared times x

minus four divided by x minus four, which has a hole when x equals four, and otherwise

looks like the graph of x minus three squared. This kind of discontinuity is called removable,

because you could get rid of it by plugging the hole just by defining F to have an appropriate

value when x equals four. So in this case, you'd want f of x to be the same when x is

not equal to four, but you'd want it to have the value of one when x equals four. And that

would amount to plugging the hole and making it continuous. In this original example, our

function had a removable discontinuity because it wasn't defined when x equals four. But

a function could also have a removable discontinuity, because it's defined in the wrong place at

x equals four, for example, too high or too low, to fit the trend of f. a discontinuity

can also occur at a vertical asymptote, where it's called an infinite discontinuity. For

example, the rational function g of x is one over x minus two has an infinite discontinuity

at x equals two. Occasionally, you may encounter a discontinuity is that don't look like any

of these. For example, the graph of the function y equals cosine of one over X has a wild discontinuity

at x equals zero, because of the wild oscillating behavior there. So for a function to be continuous

at x equals a, we need it to avoid all of these problems. To avoid a jump discontinuity,

we can insist that the functions limit has to exist at x equals a way to avoid a hole

or removable discontinuity, we can insist that F has to be defined at x equals A to

avoid the other kind of removable discontinuity in which f of a is defined, but it's in the

wrong place. We can insist that the limit of f of x as x goes to a has to equal f of

a. Sometimes the definition of continuity is written with just the third condition,

and the first two conditions are implied. Notice that these three conditions not only

exclude jump this continuity and removal this continuity is they also exclude infinite discontinuities

and wild discontinuities. For example, in our third example, the function can't be continuous

at x equals two, because it fails to have a limit at x equals two and it fails to have

a value at x equals two. In our wild, this continuity example, the limit also fails to

exist at x equals zero, so the function can't be continuous there. So what are the places

where this function f is not in us and why, please take a moment to think about it for

yourself. The functions not continuous at negative three, because the function is simply

not defined there, the functions not continuous at x equals one, because of that jump discontinuity.

In the language of limits, we say that the limit of f of x does not exist, when x equals

two, the limit of the function exists and equals three, but the value of the function

is down here at negative one. So the function is not continuous, because the limit doesn't

equal the value at x equals three, the function is not continuous, because once again, the

limit doesn't exist. Notice that x equals negative two. Even though the function turns

a corner, the function still continuous. Because the limit exists and equals two, and the value

of the function is also two. The function drawn here is not continuous at x equals negative

two, because the limit doesn't exist at x equals negative two, the limit from the left

is one, while the limit from the right is zero. But it is true that the value of the

function at x equals negative two is equal to the limit of the function from the left

side. That is, f of negative two is equal to the limit as x goes to negative two from

the left of f of x.

Notice that we can't say the same thing about the right limit. The limit from the right

is zero, while the value of the function is one, and those are equal. In this situation,

we say that f is continuous from the left, but not from the right. By the same reasoning

at x equals one, the function is not continuous, but it is continuous from the right, because

the limit from the right is equal to the value of the function. Notice that f is not continuous

from the left here, because the limit from the left is about one and a half, while the

value of the function is one. In general, we say that a function f is continuous from

the left at x equals j. If the limit as x goes to a from the left of f of x is equal

to f of a, and a function is continuous from the right at x equals a. If the limit as x

goes to a from the right of f of x equals f of a, in practical terms of function is

continuous from the left if the endpoint is included on the left piece, and a function

is continuous from the right, if the endpoint is included on the right piece. This video

gave a precise definition of continuity at a point in terms of limits. Namely, a function

is continuous at the point x equals a. If the limit as x goes to a of the function is

equal to the functions value

at A.

In a previous video, we gave a definition for continuity at a point. In this video,

we'll discuss continuity on an interval and continuous functions. We say that a function

f of x is continuous on the open interval BC, if f of x is continuous at every point

in that interval. For x to be continuous on a closed interval BC, we require it to be

continuous on every point in the interior of BC, but we just require it to be continuous

from the right at B. And from the left at sea. We can also talk about f being continuous

on half open intervals. For example, on half open interval, BC, which is open at B and

closed at sea, or the other way around, or the happy to open it all from B to infinity

and so on. In all of these cases, we require F to be continuous on the interior of the

interval, and left or right continuous on the closed endpoints of the interval

as appropriate.

So on what intervals is this function geovax continuous? Well, it's continuous. On this

part, the arrows indicate it keeps on going. So I'd say it's continuous from negative infinity

to negative one, not including the endpoint negative one. It's also continuous here, and

we can include the endpoint this time. So this is from negative one to one and Then

again, on this last section, we can't include one. It's not continuous there. It's not even

defined there. So what kinds of functions are continuous? What kinds of functions are

continuous everywhere? And by everywhere, I mean, on the entire real line negative infinity

infinity? Well, polynomials are a great example. Also, sine x and cosine x, the absolute value

of x is another common example. There are certainly many other functions that are continuous

on the whole real line. I'll let you see if you can come up with some more examples. Now,

if we ask the second question, what kinds of functions are continuous on their domains,

we get a lot more answers, not only polynomials, but also all rational functions, things like

f of x equals 5x minus two over x minus three squared times x plus four is a good example

of a rational function, even though it's not continuous everywhere. Because it's not continuous

when x equals three or negative four, it is continuous on its whole domain, because three

and four are not in the domain of this rational function. In addition, all trig functions,

inverse trig functions, log and either the x functions. And pretty much all the functions

we normally encounter are continuous on their domains, although their domains are not necessarily

the whole real line. For example, for natural log of x, the domain is just zero, infinity,

and that's where the function is continuous. In addition, the sums, differences, products

and quotients of continuous functions are continuous on their domains, so for example,

y equals sine of x plus the natural log of x is continuous where it's defined, the compositions

of continuous functions are continuous on their domains. So for example, the function

y equals ln of sine of x is continuous, where defined turns out to be a bunch of disjoint

intervals where sine is positive. Since continuity is defined in terms of limits, it's sometimes

possible to use our knowledge of which functions are continuous to calculate limits. For example,

if we want to find the limit as x goes to zero of cosine of x, because cosine is continuous,

we can evaluate this limit just by plugging in zero for x. And cosine of zero is one.

We're using the definition of continuity here to say that the limit of the function is equal

to the value of the function. The second example is a little trickier, because the function

inside is not continuous at x equals two. In fact, it's not defined at x equals two.

But as X approaches 2x squared minus four over 2x minus four times pi, can be rewritten

as x plus two times x minus two over two times x minus two times pi, which is the same thing

as x plus two over two pi. For x not equal to two. So as X approaches two, this expression

here, approaches two plus two over two times pi, which is just two pi. In other words,

the limit as x goes to two of x squared minus four over 2x minus four pi is just two pi.

And therefore, the limit as x goes to two of cosine of this expression is just cosine

of two pi, which is again equal to one. We're using here the fact that cosine is continuous,

and a property of continuous functions, which says that the limit as x goes to a of f of

g of x is equal to f of the limit as x goes to a of g of x. If f is a continuous function.

In other words, for continuous functions, you can pass the limit inside the function.

That's all for continuity on intervals and continuous functions. The intermediate value

theorem says that if f is a continuous function, on the closed interval a b and if n is any

number, in between F of A and F of Bay, n f has to achieve that value and somewhere.

In other words, if n is a number between F of A and F of b, then there has to be a number

c in the interval a, b, such that f of c equals n. In our example, there are three such possible

values for C, it could be right here, since f of that number equals n, or it could be

here, or here, I'll just mark the middle one. The intermediate value theorem can only be

applied to continuous functions. If the function is not continuous, then it might jump over

and, and never achieve that value. When application of the intermediate value theorem is to prove

the existence of roots or zeros of equations, we call that a real root of p of x is a real

number c,

such that P of C is zero, we're going to want to apply the intermediate value theorem

with n equal to zero. Our polynomial is defined on the whole real line, not just an interval.

But the trick here is to pick an interval a b, so that P of A is negative, and P of

B is positive, or vice versa. So that the intermediate value theorem will tell us that

P has to pass through zero in between. I'm just going to use trial and error here and

calculate a few values of p. So P of zero is easy to calculate, P of zero is just seven.

P of one is going to be five minus three minus 12 plus seven, which equals negative three.

So in this very lucky example, the first two numbers that we pick will work for our A and

B, so we can just let A be equal 01. Because P of zero is a positive number, and P of one

is a negative number. So actually, the graph should look a little different. The graph

looks more like this. But in any case, by the intermediate value theorem, there has

to be a number c, in between, in this case, zero and one, where our polynomial p achieves

this intermediate value of zero. And that number See, we don't know what it is, but

we know it somewhere in the interval zero to one, that value c gives us a real root

for our polynomial. There may be other real roots, but we've proved there exists at least

one. The intermediate value theorem has lots of other applications besides finding roots.

For example, suppose you have a wall that runs in a circle around the castle, and the

height of the wall varies continuously as a function of the angle. Surprisingly, the

intermediate value theorem can be used to show there must be somewhere two diametrically

opposite places on the wall with exactly the same height. So if you can figure out a way

to show this in this video, we stated the intermediate value theorem, which holds for

continuous functions, and talked about a couple of applications. This video introduces the

trig functions, sine, cosine, tangent, secant, cosecant, and cotangent. For right triangles.

For a right triangle with sides of length A, B and C angle theta as drawn, we define

sine of theta as the length of the opposite side over the hypothesis. The side that's

opposite to our angle theta has measure a and I have partners is this side here with

measure C. So that would be a oversee for this triangle. Cosine of theta is defined

as the length of the adjacent side over the length of high partners. This side here is

the side adjacent to theta. Of course, the high partners is also adjacent to theta, but

it's special as the high partners so we don't think of it as the adjacent side. So that

would be B oversea. tangent of theta is the opposite side length over the adjacent side

length. So that would be a over b. The pneumonic to remember this is so A toa. That's sine

is opposite over hypotenuse. Cosine is adjacent over hypotenuse. And tangent is opposite over

adjacent. In fact, there's a relationship between tangent and sine and cosine. Namely,

tangent of theta is equal to sine of theta over cosine of theta. If you want to see why

that's, that's because sine of theta over cosine of theta is given by sine, which is

opposite over hypotenuse, divided by cosine, which is adjacent over hypotenuse. If we compute

these fractions by flipping and multiplying,

the high partners, length cancels, and we just get opposite over adjacent, which is

by definition, tangent of theta. There are three more trig functions that are defined

in terms of sine, cosine and tangent. First of all, their sequence of theta, by definition,

that's one over cosine of theta. So it's going to be one over the adjacent over hypotenuse,

which is the high partners over the adjacent, which in this triangle is C over B. cosecant

of theta is defined as one over sine theta. So that's one over the opposite over the hypotenuse,

which is the high partners over the opposite. And for this triangle, that's going to be

C over a. Finally, cotangent of theta is defined as one over tan theta, that's going to be

one over opposite over adjacent, flip and multiply, I get adjacent over opposite birth,

which in this case is b over a. So notice that the values for cotangent cosecant and

secant are the reciprocals of the values for tangent, sine and cosine, respectively. Let's

use these definitions to find the exact values of all six trig functions for the angle theta

in this triangle. I'll start with sine of theta. That's the opposite of our hypothesis.

Well, for this angle theta, the opposite side is down here, and as measured to the high

partners has measured five. So sine theta is two over five. Cosine theta is adjacent

over hypotenuse, but I don't know the value of this side length. But fortunately, I can

find it using the Fagor in theorem says I have a right triangle here, I know that a

squared, I'll call this side length A plus B squared, where b is this other leg of the

triangle is equal to c squared, where c is the hypothesis. So here I have a squared plus

two squared equals five squared, which means that a squared plus four equals 25. A squared

is 21. So A is plus or minus the square root of 21. But since I'm talking about the length

of a side of a triangle, I can just use the positive answer. Returning to my computation

of cosine theta, I can write it as adjacent, which is the square root of 21. Over hypotony

News, which is five, tangent theta is the opposite over the adjacent, so that's going

to be two over the square root of 21. To compute secant of theta, that's one over cosine theta,

so that's going to be one over square root of 21 over five, which is five over the square

root of 21. The reciprocal of my cosine value. cosecant theta is one over sine theta, that's

going to be the reciprocal of my sine, so five halves, and cotangent theta is one over

a tan theta, so it's going to be the reciprocal of my 10 value square root of 21 over two.

Finally, we'll do an application. So if we have a kite that's flying at an angle of elevation

as the angle from the horizontal bar of 75 degrees with a kite length string of 100 meters,

we want to find out how high it is. I'll call the height Why? Well, we want to relate the

known quantities this angle and this hypothesis to The unknown quantity, the unknown quantity

is the opposite side of our triangle. So if we use sine of theta equals opposite of our

hypothesis, then we can relate these known amounts sine of 75 degrees to our unknown

amount y, which is the opposite and are known amount of 100 meters. solving for y, this

gives that y is 100 meters times sine of 75 degrees, we can use a calculator to compute

sine of 75 degrees, be sure you use degree mode and not radian mode, when you type in

the 75. When I do the computation, I get a final answer of 96.59 meters up to two decimal

places. Notice that we're ignoring the height of the person in this problem.

To remember the definitions of the trig functions, you can use the pneumonic, so tau, and the

fact that secant is the reciprocal of cosine cosecant the reciprocal of sine and cotangent

the reciprocal of tangent. In this video, I'll use geometry to compute the sine and

cosine of a 30 degree angle, a 45 degree angle and a 60 degree angle. One way to compute

the sine of a 45 degree angle is to use a right triangle with a 45 degree angle. This

particular right triangle has I have partners of length one. Since all the angles of the

triangle have to add up to 180 degrees, and we already have 90 degrees and 45 degrees,

the remaining angle must also be 45 degrees. So we have an isosceles triangle with two

sides the same length, I'll call that side length a. If we want sine of 45 degrees, let's

use this 45 degree angle here, then sine is opposite over hypotenuse. So if I can figure

out how long this side length is, I'll be able to compute sine of 45 degrees. Now the

Pythagorean theorem says this side length squared plus that side land squared equals

I have hotness squared. So we have that a squared plus a squared equals one squared.

All right, that is two A squared equals one. So a squared is one half, and a is plus or

minus the square root of one half. Since we're talking about the length of sides of triangles,

I can just use the positive square root it's customary to rewrite this as the square root

of one over the square root of two, which is one over the square root of two, and then

rationalize the denominator by multiplying the top and the bottom by the square root

of two. That gives me a square root of two in the numerator, and the square root of two

squared in the denominator, which is the square root of two over two. So the side lengths

are the square root of two over two. Now I can figure out the sine of 45 degrees, by

computing the opposite over the hypotenuse. That's, I'm looking at this angle. So opposite

is square root of two over two hypothesis one. So the sine of 45 degrees is the square

root of two over two. Cosine of 45 degrees is adjacent over hypotenuse. That's this side

length over this hypothesis. So that's the square root of two over two over one again,

what would happen if instead of using this triangle with hypotony is one, we use this

triangle, also a 4545 90 triangle with hypotony is five, not quite drawn to scale. Please

pause the video for a moment and repeat the computation with this triangle. This time,

I'll call the side length B. Pythagoras theorem tells me b squared plus b squared equals five

squared. So two, b squared equals 25. And b squared equals 25 over two, B is going to

be the plus or minus the square root of 25 over two again, I can just use the positive

version. And so B is the square root of 25 over the square root of two, which is five

over the square root of two, rationalizing the denominator, I get five root two over

two. Now, sine of my 45 degree angle is opposite overhype hotness,

which is

five square root of two over two divided by five. That simplifies to the square root of

two over two as before, and a similar computation shows that cosine of 45 degrees is also square

root of two over two as before. This makes sense because sine and cosine are based on

ratios of sides. And since these two triangles are similar triangles, they'll have the same

ratios of sides. To find the sine and cosine of 30 degrees, let's use this 30 6090 right

triangle with hypotony is one. If we double the triangle, we get an angle of 30 here,

so a total angle of 60 degrees here, and this angle is also 60 degrees. So we have a 60

6060 triangle, that's an equilateral triangle, all side lengths are the same. Since this

side length has length one, this side length is also one. This entire side length is one,

which means this short side of our original triangle has length one half. Going back to

my original triangle, let's use the Pythagorean Theorem to find the length of its longer side

x. But agrin theorem says x squared plus one half squared equals one squared. So x squared

plus a fourth equals 1x squared is three fourths. And so x is plus or minus the square root

of three fourths is the positive version, and get the square root of three over the

square root of four, which is the square root of three over two. Now using our original

triangle, again, let's compute the sine of this 30 degree angle here. We know that sine

of 30 degrees is opposite of our hypothesis, the opposite of this angle is one half and

the hypothesis is one. So we get a sine of one half over one, which is one half cosine

of 30 degrees is adjacent over hypothesis. So that's the square root of three over two

divided by one. To find sine of 60 degrees and cosine of 60 degrees, we can actually

use this same green triangle and just focus on this upper corner angle of 60 degrees instead.

So sine of 60 degrees. Opposite overhype hotness, but this time the opposite to this angle is

the square root of three over two. Cosine of 60 degrees adjacent over hypotenuse gives

us one half. I'll summarize the results in this table below. Notice that a 30 degree

angle corresponds to pi over six radians since 30 degrees times pi over 180 is pi over six.

Similarly, 45 degrees corresponds to pi over four radians and 60 degrees corresponds to

pi over three radians. I recommend that you memorize the three numbers one half, root

two over two, and root three over two. And the fact that one half and root three over

two go together. And root two over two goes with itself. From that information, it's not

hard to reconstruct the triangles, you know that a 4545 90 triangle is my sauciest triangle.

So it must have the side lengths were the same number goes with itself. And a 30 6090

triangle has one side length smaller than the other. So the smaller side must be one

half and the larger side must be root three over two since root three is bigger than one

and so root three over two is bigger than one half. Doing a visual check, you can easily

fill in the angles, the smaller angle must be the 30 degree angle, and the larger one

must be the 60 degree one. In this video, we computed the sine and cosine of three special

angles 30 degrees, 45 degrees, and 60 degrees. This video defined sine and cosine in terms

of points on the unit circle. a unit circle is a circle with radius one.

Up to now we defined sine and cosine and tangent in terms of right triangles. For example,

to find sine of 14 degrees in theory, you could draw a right triangle with an angle

of 14 degrees and then calculate the sign as the length of the opposite side over the

length of pi partners. But if we use this method to try to compute sine of 120 degrees,

things go horribly wrong. When we draw this 120 degree angle, and this right angle, there's

no way to complete this picture to get a right triangle. So instead, we're going to use a

unit circle, that is a circle of radius one. The figure below illustrates how right triangles

and a unit circle are related. If you draw right triangles, with larger of hypotenuse

one with larger and larger angles, then the top vertex sweeps out part of a unit circle.

Let's look at this relationship in more detail. In this figure, I've drawn a right triangle

inside a unit circle, the high partners of the triangle is the radius of the circle,

which is one, one vertex of the right triangle is at the origin, another vertex of the right

triangle is at the edge of the circle, I'm going to call the coordinates of that vertex

A, B. Now the base of this right triangle has length a, the x coordinate, and the height

of the right triangle is B, the y coordinate. If I use the right triangle definition of

sine and cosine of theta, this right here is the angle theta, then cosine of theta is

adjacent over hypotenuse, so that's a over one, or a. Notice that a also represents the

x coordinate of this point on the unit circle. At angle theta from the x axis. I'll write

that down. For sine of theta, if I use the right triangle definition, that's opposite

overhead partners, so B over one, which is just B. But B also represents the y coordinate

of this point in the on the unit circle at angle theta. For tangent theta, if we use

the right side triangle definition, its opposite over adjacent. So that's B over A, I can think

of that as the y coordinate of the point over the x coordinate of the point. Now for angles

theta, that can't be part of a right triangle, because they're too big, they're bigger than

90 degrees, like now I'll call this angle here theta, I can still use this idea of x

and y coordinates to calculate the sine and cosine of theta. So if I just mark this point,

on the end of this line at angle theta, if I mark that to have coordinates x and y, then

cosine theta, I'm still going to define as the x coordinate of this point, sine theta

as the y coordinate, and tangent theta as the ratio of the y coordinate over the x coordinate.

When we use this unit circle definition, we always draw theta starting from the positive

x axis and going counterclockwise. Let's use this unit circle definition to calculate sine,

cosine and tangent of this angle fee. In our figure, we have a unit circle. And these numbers

are supposed to represent the x&y coordinates of this point on the unit circle at the endpoint

of this line segment, that lies at angle fee for the positive x axis. Sign fee is equal

to the y coordinate cosine fee is equal to the x coordinate. And tangent of fee is given

by the ratio of the two, which works out to negative 0.3639 up to four decimal places.

This video gives a method for calculating sine cosine and tangent in terms of the unit

circle. Starting from the positive x axis, you draw the angle theta going counterclockwise.

You look at the coordinates of the point on the circle where that angle ends. And the

cosine of that angle theta is the x coordinate, sine of theta is the y coordinate, and tangent

of theta is the ratio. This video gives three properties of the trig function sine and cosine

that can be deduced from the unit circle definition. Recall that the unit circle definition of

sine and cosine for angle theta is that cosine theta is the x coordinate. And sine of theta

is the y coordinate for the point on the unit circle at angle theta. The first property

is what I call the periodic property. This says that the values of cosine and sine are

periodic with period two pi. And what that means is that if you take cosine of an angle

plus two pi, you get the same thing as just if you took cosine of the angle. So when we

write this down, we're assuming that theta is measured in radians. If we want to measure

theta in degrees, the similar statement is that cosine of theta plus 360 degrees is equal

to cosine of theta, we can make the same statements for sine, sine of an angle plus two pi is

equal to sine of the original angle, here, the angle being measured in radians. If we

want to measure the angle in degrees, the statement is that sine of theta plus 360 is

equal to sine of theta, we can see why this is true from the unit circle definition of

sine and cosine. This is our angle theta, then theta plus two pi, the plus two pi adds

a full turn around the unit circle to our angle, so we end up at the same place, theta

and theta plus two pi are just two different names for the same location on the unit circle.

And since sine and cosine give you the y and x coordinates of that point on the unit circle,

they have to have the same value. Similarly, if we consider an angle theta, and an angle

theta minus two pi, the minus two pi means we go the other direction around the unit

circle clockwise, we still end up in the same place. And therefore, cosine of theta minus

two pi, the x coordinate of that position is the same thing as cosine of theta, sine

of theta minus two pi is the same thing as sine of theta, the same statements hold if

we add or subtract multiples of two pi. For example, cosine of theta plus four pi is still

the same thing as cosine of theta. This time, we've just gone to turns around the unit circle

and still gotten back to the same place. So if we want to find cosine of five Pi, that's

the same thing as cosine of pi plus four pi, which is the same thing as cosine of pi. Thinking

about the unit circle, pi is halfway around the unit circle. So cosine of pi means the

x coordinate of this point right here. Well, that point has coordinates negative one, zero,

so cosine of pi must be negative one. If I want to take sine of negative 420 degrees,

well, that's sine of negative 360 degrees, minus 60 degrees, which is the same thing

as sine of minus 60 degrees. Thinking about the unit circle, minus 60 degrees, means I

start at the positive x axis and go clockwise by 60 degrees that lands me about right here.

And so that's one of the special angles that has an x coordinate of one half a y coordinate

of negative root three over two. And therefore sine of negative 60 is negative root three

over two the y coordinate. The next property I call the even odd property, it says that

cosine is an even function, which means that cosine of negative theta is the same thing

as cosine of theta, while sine is an odd function, which means that sine of negative theta is

the negative of sine of theta. To see why this is true, let's look at an angle theta.

And the angle negative theta. A negative angle means you go in the clockwise instead of counterclockwise

direction from the positive x axis.

The coordinates of this point by definition, r cosine theta sine theta, whereas the coordinates

of this point are cosine negative theta, sine of negative theta. But by symmetry, these

two points have the exact same x coordinate, and therefore cosine of theta must equal cosine

of negative theta, while their y coordinates have the same magnitude, but opposite signs.

This one's positive and this one's negative. Have, therefore, sine of negative theta is

the negative of sine of theta. Let's figure out if tan of theta isn't even or odd function.

Well, we know that tan of negative theta, tangent by definition, is sine over cosine.

Well, we know that sine of negative theta is the negative of sine of theta, whereas

cosine of negative theta is cosine of theta. Therefore, we're getting negative sign theta

over cosine theta, which is negative tan of theta. Since tan of negative theta is the

negative of tan of theta, tan theta is an odd function. The last property on this video

is the Pythagorean property, which says that cosine of theta squared plus sine of theta

squared is equal to one. A lot of times this property is written with this shorthand notation,

cosine squared theta plus sine squared theta equals one. But this notation, cosine squared

theta just means you take cosine of theta and square it. This property is called the

Pythagorean property, because it comes from the Pythagorean Theorem. Let me draw a right

triangle on the unit circle. I'll call this angle theta. So the coordinates of this endpoint

here are cosine theta sine theta. Since this is supposed to be a unit circle, the hypotenuse

of my right triangle has length one, the base of my right triangle is just cosine theta,

same thing as the x coordinate of this point. And the height of my triangle is the y coordinate

of the point sine theta. Now the Pythagorean theorem says that this side length squared

plus that so that squared equals one squared, since one squared is the same thing as one,

that gives me the Pythagorean property. But tigrayan property is handy for computing values

of cosine given values of sine and vice versa. And this problem, we're told that sine of

t is negative two sevenths. And T is an angle that lies in quadrant three. When we say the

angle lies in quadrant three, that means the terminal side of the angle lies here in quadrant

three. One way to find cosine of t is to use the fact that cosine squared t plus sine squared

t is equal to one. That is cosine of t squared plus negative 2/7 squared is equal to one,

I can write this as cosine of t squared plus 4/49 is equal to one and so cosine of t squared

is equal to one minus 4/49, which is 4540 nights. taking the square root of both sides,

that goes for the cosine t is plus or minus the square root of 45 over 49, that's plus

or minus the square root of 45 over seven. Now since we're in the third quadrant, we

know that cosine of t, which represents the x coordinate of this point, must also be negative.

Therefore, cosine of t is going to be negative square to 45 over seven. It's also possible

to solve this problem using the Pythagorean theorem for right triangles directly. If we

look at the fact that sine of t is negative two sevenths and ignore the negative sign

for now, we can think of this information as telling us that we have a right triangle

angle theta, whose opposite side is two, and whose high partners is seven.

We call this side here a them but Tiger in theorem says us a squared plus two squared

is seven squared. So a squared plus four is 49. So a squared is 45. And a is plus or minus

the square root of 45. Since I'm worrying about a triangle, I'm going to use the positive

value. Now, cosine of t is going to be adjacent over hypotenuse. So that's going to be the

square root of 45 over seven. Now I go back to thinking about positive and negative signs.

And I noticed that since I'm in the third quadrant, my co sign is be negative, so I

just stick a negative sign in front. This alternative solution, this is many of the

same ideas as the previous solution, and ultimately gets us the same answer. This video gives

three properties of trig functions, the periodic property, the even odd property, and the Pythagorean

property. This video is about the graphs of sine and cosine. I want to graph the functions

y equals cosine t and y equals sine t, where t is in radians, I'm gonna think of this being

the t axis, and this being the y axis. One way to do this is to plot points. So I'll

fill in this chart, using my knowledge of special angles on the unit circle. These points

will be easier to graph, if I convert them all to decimals. Now plot the points for cosine

and connect the dots to get a graph of y equals cosine t from t equals zero to t equals two

pi. To continue the graph for t values less than zero or bigger than two pi, I could plot

more points. Or I could just use the fact that the cosine values repeat. If I add or

subtract two pi to the my angle T, I'll be at the same place on the unit circle. So my

cosine will be exactly the same. Therefore, my values of cosine, which are represented

by my y values on this graph, repeat themselves. For example, when my T value is two pi plus

pi over six about like here, it's cosine is the same as the cosine of just pi over six.

So I'll take this dot here and repeat it over here. Similarly, the when t is like two pi

plus pi over four, I get the same value of cosine is when it's just pi over four. So

this.is going to repeat. And I can continue repeating all my dots. This one repeats over

here at two pi plus, say pi over three. And so my whole graph will repeat something like

this. It also repeats on this side, something like this. Since subtracting two pi from my

t values will also give me the same value of cosine. We can also plot points to get

a graph for sine and extend it by repetition. Going forward, I'll usually write the function

sine and cosine as y equals cosine of x and y equals sine of x. When I read it this way,

notice that x now refers to an angle, while y refers to a value of cosine, or sine. That's

a different meaning of x and y, compared to when we're talking about the unit circle,

where x refers to the cosine value, and y refers to the sine value. Now let's look at

some properties of the graphs of sine and cosine. The first thing you might notice is

that the graph of cosine and the graph of sine are super similar to each other. In fact,

you can think of the graph of cosine as just being the graph of sine shifted to the left

by pi over two. So we can write cosine of x as the sine function of x plus pi over two,

since adding pi over two on the inside, move the graph horizontally to the left by pi over

two. Or we can think of the graph of sine as being constructed from the graph of cosine

by shifting the cosine graph right by pi over two, that means we can write sine of x as

equal to cosine of x minus pi over two, since subtracting pi over two on the inside, shifts

the cosine graph to the right by pi over two.

Next, let's look at domain and range. The domain of sine and cosine is all real numbers.

All right, that is negative infinity to infinity, but the range is just from negative one to

one. That makes sense, because sine and cosine come from the unit circle. The input values

for the domain come from angles. And you can use any numbers and angle positive or negative

as big as you want, just by wrapping a lot of times around the circle. The output values

for the range, that is the actual values of sine and cosine come from the coordinates

on the unit circle. And those coordinates can't be any bigger than one or any smaller

than negative one. So that gives us a range. As far as even an odd behavior, you can tell

from the graph. Here's cosine, that it's symmetric with respect to the y axis and so it must

be even. Whereas the graph of sine is symmetric with respect to the origin and must be odd.

The absolute maximum value have these two functions is one and the absolute minimum

value is negative one. We can also use the words midline amplitude and period to describe

these two functions. The midline is the horizontal line, halfway in between the maximum and minimum

points. Here, the midline is y equals zero, the amplitude is the vertical distance between

a maximum point and the midline. You can also think of the amplitude as the vertical distance

between a minimum point and the midline, or as half the vertical distance between a midpoint

and a max point. For the cosine function and the sine function, the amplitude is one. A

periodic function is a function that repeats at regular horizontal intervals. The horizontal

length of the smallest repeating unit is called the period for Y equals cosine of x, the period

is two pi. Notice that the period is the horizontal distance between successive peaks, or maximum

points, or between successive troughs, or minimum points. algebraically, we can write

cosine of x plus two pi equals cosine of x and sine of x plus two pi equals sine of x

to indicate that the functions repeat themselves over an interval of two pi and have a period

of two pi. In this video, we graphed y equals cosine of x and y equals sine of x. and observe

that they both have a midline at y equals zero, an amplitude of one and a period of

two pi. sine u sort of functions are functions that are related to sine and cosine by transformations

like stretching and shrinking and shifting. This video is about graphing these functions.

Let's start by graphing the function, y equals three sine of 2x. This function is related

to the function y equals sine x. So I'll graph that first. Now, the three on the outside

stretches this graph vertically by a factor of three, while the two on the inside compresses

that horizontally by a factor of one half. If instead I want to graph y equals three

sine 2x plus one, this plus one on the outside shifts everything up by one unit. Let's compare

the midline amplitude and period of our original y equals sine x are transformed y equals three

sine 2x. And our further transformed y equals three sine 2x plus one, the original sine

has a midline at y equals zero, an amplitude of one and a period of two pi. For the transformed

function, y equals three times sine of 2x. The two on the inside shrinks everything horizontally

by a factor of one half. So it changes the period of two pi into a period of one half

times two pi,

which is pi.

Since the two on the inside only affects x values and horizontal distances, it doesn't

affect the midline, which is a y value, or the amplitude, which is a vertical distance.

But the three on the outside does affect these things. Well, in particular, it affects the

amplitude, since everything is stretched out vertically by a factor of three, the amplitude

of one get stretch to an amplitude of three. In this case, the midline doesn't actually

change, because multiplying a y value of zero by three is still a y value of zero. Now on

the third function, we've taken the second function and added one on the outside, so

that shifts everything up by one. Therefore, instead of having a midline at y equals zero,

we now have a midline at y equals one. The amplitude doesn't change though it's still

three because shifting everything up by one doesn't affect the distance between the mid

mind and the end the maximum point. Also, the period is still pi since the period is

a horizontal measure, and adding one on the outside only affects vertical things. Now

next, let's graph the function y equals three times sine of two times quantity x minus pi

over four. This function is very closely related to the First function we graphed on the previous

page, that was y equals three sine of 2x. In fact, if we give the name f of x to that

function, and maybe we can call g of x, this other function, then we can get g of x by

taking f of x and plugging in x minus pi over four in for x. In other words, g of x is f

of x minus pi over four. This relationship gives me an idea for graphing g of x, the

function we want to graph, we can first graph f of x, we already did that on the previous

page. And then we can shift its graph to the right by pi over four, because that's what

you do when you subtract a number on the inside of a function. So here's the graph of y equals

three sine 2x. Recall that it's just the graph of sine stretched vertically by a factor of

three, and shrunk horizontally by a factor of one half. Now, to graph the function that

I want, I'm going to shift this graph over by pi over four to the right. Notice that

since I had my function written in factored form, I could just read off the horizontal

shift. But if I had written it instead, as y equals three sine 2x minus pi over two,

which is algebraically equivalent, it would be easy to get confused and think that I needed

to shift over by pi over two. So it's best to factor first, before figuring out what

the shift is, we're factoring out the coefficient of x. If instead, we wanted to graph this

function, same as the one we just graphed, it's just with a minus one on the outside,

that minus one would just bring everything down by one. Let's take a moment to look at

midline amplitude, and period for the original parent function, y equals sine of x, and our

final transformed function, y equals three sine of two times quantity x minus pi over

four minus one, our original sine function has midline at y equals zero amplitude of

one and period of two pi. For our transform function, the three on the outside stretches

vertically, so it makes the amplitude three. The minus one on the outside shifts everything

down by one. So it brings the midline, y equals zero, down to Y equals negative one, the two

on the inside, shrinks everything horizontally by a factor of one half. So the period becomes

one half times two pi, which is pi. Finally, there's a horizontal shift going on our transformed

function shifts to the right, by pi over four, this horizontal shift is sometimes called

the phase shift. The function we just analyzed was y equals three sine 2x minus pi over four

minus one, which could also be written as y equals three sine 2x minus pi over two minus

one. This is a function of the form y equals a sine B x minus c plus d, where b is positive.

If we have a function of this form, or the similar function with cosine in it, then we

know that the midline is going to be at y equals D. That's because the original midline

of sine or cosine at y equals zero gets shifted up by D, we know that the amplitude is going

to be a because this A multiplied on the outside stretches everything vertically by a factor

of A.

to be a little more accurate, we should say the amplitude is the absolute value of A in

case a is negative. If a is negative, then that amounts to a vertical reflection or a

reflection over the x axis. We know that the period of the original sine or cosine is two

pi. And we know that this factor of B amounts to a horizontal shrink by a factor of one

over B or I guess it could be a horizontal stretch by a factor of one over b If b is

less than one, so because we're starting with a period of two pi, and we're multiplying

by one over B, our new period is going to be two Pi over B. The trickiest thing is the

horizontal shift. And to get that right, I like to factor out this B for my equation.

So instead of writing y equals a cosine bx minus c plus d, I'm going to write y equals

A cosine B times quantity x minus c over b plus d. Similarly, if it's a sine function,

I write y equals a sine B times x minus quantity c over b plus d, then I can read off the horizontal

shift as C over B. And that'll be a shift to the right, if C over B is positive and

a shift to the left, if C over b is negative, this might seem backwards from what you're

used to, but it's because we have that minus sign there. So if C over B is positive, we're

actually subtracting on the inside. So that shifts right, if C of b over b is negative

minus a negative is actually adding something, and that's why it shifts it to the left. So

as one final example, say I wanted to graph y equals 1/3, cosine of one half x plus three

minus five, that would have a midline at y equals minus five, an amplitude of 1/3, a

period of two pi divided by one half, which is four pi, and a horizontal shift. Better

rewrite this horizontal shift of six units to the left, the horizontal shift is sometimes

called the phase shift. And that's all for graphs of sinusoidal functions. This video

is about graphing the trig functions, tangent, secant, cotangent and cosecant. To gain an

intuition for the graph of y equals tangent of x, I think it's handy to look at the slope

of a line at angle theta on the unit circle. The slope of this line is the rise over the

run. But the rise is given by sine of theta, and the run is given by cosine of theta. So

the slope is given by sine theta over cosine theta, which is simply tan of theta. So if

I want to graph y equals tan of x, I can think of x as being the angle and y as being the

slope of the line at that angle. Notice if the angle is zero, the slope is zero. But

as the angle increases towards pi over two, the slope gets bigger and bigger heading towards

infinity. As the angle goes from zero towards negative pi over two, the slope is getting

negative and heading towards negative infinity at exactly pi over two and negative pi over

two, we have a vertical line. And so the slope is undefined.

Using this information, let's graph a rough sketch of y equals tan x. Remember, we're

thinking of x as the angle and y as the slope, we're going to go between an angle of negative

pi over two and pi over two. So we said that the slope was zero when the angle is zero,

and then it heads up towards positive infinity as we go towards the angle goes towards pi

over two with an undefined value at pi over two, it goes negative heading towards negative

infinity as the angle heads towards negative pi over two, also with an undefined value

at negative pi over two. You can also verify that for angles slightly bigger than pi over

two, we have the same line as for angles that are approaching negative pi over two, and

therefore this picture repeats and it turns out that tangent is periodic With period not

to pi, like sine and cosine, but just pi, the period of pi makes sense because if you

take a line and rotate it by 180 degrees, it's the same line with the same slope, and

therefore has the same value of tangents. In this graph of y equals 10x, notice that

the x intercepts, all right values of x have the form negative two pi, negative pi, zero,

pi, two pi, etc, you can write that as pi times k, where k is an integer, that is a

positive or negative whole number or zero. This makes a lot of sense because tangent

of x is sine of x over cosine of x. And so you're going to get x intercepts, that's where

y is zero, which is where the numerator is zero, and sine x is zero, at values of the

form pi, two pi, and so on. From the graph, you can see the vertical asymptotes are at

values like negative three pi over two, negative pi over two, pi over two, and three pi over

two, these values can be written as pi over two times k, where k is an odd integer. Again,

this makes sense from the definition of tangents since vertical asymptotes, will occur where

the denominator is zero, and cosine x is zero, at numbers, like negative pi over two pi over

two, three pi over two, and so on, the domain of tangent is the x axis for which it's defined.

So that's going to be everything except for the vertical asymptotes, we can write that

as x such that x is not equal to pi over two times k, for K, an odd integer. The range

or the y values go all the way from negative infinity to infinity. And the period, as we

mentioned previously, is pi. Since the smallest repeating unit has a horizontal width of pi,

to graph y equals secant x, I'm going to remember that secant is one over cosine. So if I start

with a graph of cosine, I can take the reciprocal of the y values to get the graph of secant,

the reciprocal of one is one, the reciprocal of zero is undefined, so I'm not going to

have a value at pi over two, negative pi over two, three pi over two, or negative three

pi over two. When I take the reciprocal of numbers, just less than one, I'm going to

get numbers just greater than one, but I would take the reciprocal of positive numbers getting

close to zero, I'm going to get really big positive numbers going up towards infinity.

Similarly, on the other side, over here, I have numbers close to zero, but negative,

so their reciprocals will be negative numbers heading towards negative infinity. The reciprocal

of negative one is negative one. And similarly here, so I'm getting kind of positive and

negative buckets and upside down buckets as the graph of my sequence.

Notice that secant has a period of two pi, which makes sense, since cosine has a period

of two pi, it has a range that goes from negative infinity to negative one inclusive, and from

one to infinity. That makes sense because the range of cosine is between one and negative

one, and we're taking the reciprocal of those values. The domain is everything except for

the vertical asymptotes. Now the vertical asymptotes are where cosine is zero. So that

is at values of the form pi over two, three pi over two, etc. That's values of the form

pi over 2k, where k is an odd integer. So the domain is going to be x values such that

x is not equal to pi over 2k. For K and odd integer. The x intercepts of secant Well,

it doesn't have any, because you can't take one over something and get the numbers zero

for your y value. We've seen the graph of y equals Tana x and y equals secant x. This

is the graph of y equals cotangent x. It looks similar to the graph of tangent x, it's just

a decreasing function instead of an increasing one, and it has its vertical asymptotes. And

it's x intercepts in different places. Finally, this green graph is the graph of y equals

cosecant x. It's related to the graph of sine x, since cosecant is one over sine x, and

in fact, if I draw the graph of sine x in between, you can see how it kind of bounces

off. Because it's the reciprocal. I encourage you to memorize the general shape of these

graphs, you can always figure out the details by thinking how about how they're related

to the graphs of cosine of x, and sine x. This video is an introduction to solving trig

equations. Let's start with the equation two cosine x plus one equals zero, I want to find

all the solutions in the interval from zero to two pi, and then get a general formula

for all solutions, not just those in that interval. Let me start by rewriting this equation

to isolate the tricky part, which is cosine of x. So I'm going to write to cosine x equals

negative one, and then divide both sides by two. Now I'm looking for the angles x between

zero and two pi, whose cosine is negative one half. Since negative one half is one of

the special values on the unit circle, I can use my knowledge of the unit circle, to see

that the angle between zero and two pi must be either two pi over three, or four pi over

three, my answer needs to include both of these values. There are no other spots on

the unit circle whose cosine is negative one half. But there are more angles, because we

can always take one of these angles and add multiples of two pi to it. So if I want to

find all solutions, I can take these two principles solutions, two pi over three, and four pi

over three, and simply add multiples of two pi to them. For example, two pi over three

plus two pi, or two pi over three minus two pi, two pi over three plus four pi, and so

on. A much more efficient way to write this is to write two pi over three plus two pi

times k, any integer that is any positive or negative whole number or zero. Similarly,

I can write four pi over three plus two pi k, to capture all solutions, based on the

principal solution for pi over three by adding and subtracting multiples of two pi.

This is my final solution. Next, let's look at a tricky equation involving tangent. As

usual, I'm going to start out by cleaning things up and isolating the tricky part, which

in this case is tangent. So let me add tangent to both sides. That'll give me three, tan

x equals the square root of three. And so tan x is the square root of three over three.

The square root of three over three looks suspiciously similar to value the value of

square to three over two, which is a special value on my unit circle. So my suspicion is

that my unit circle will again help me find this value of x without a calculator. Recall

that tan x is sine x over cosine x. So I'm looking for angles on the unit circle between

zero and two pi with a ratio of sine over cosine will give me square root of three over

three, I actually only need to look in the first quadrant and the third quadrant, because

those are the quadrants where a tangent is positive. And I really only need to look at

angles whose either sine or cosine has a squared of three in it. So by trial and error, I can

see that tan pi over six, which is sine pi over six over cosine pi over six will give

me one half over root three over two That's the same thing as one half times two over

three, which is one over root three. If I rationalize that, I get root three over three,

so that value works. If I try tan of pi over three, instead, I get root three, which is

not equal to root three over three. So pi over three doesn't work. Similarly, I can

work out some the values in the third quadrant, and see that seven pi over six works. But

four pi over three does not. So my answer to part A includes just the two values, pi

over six, and seven pi over six. Now if I want to find all solutions, not just those

in the interval from zero to two pi, I noticed that I can take one of these principal solutions,

and add multiples of two pi to it, because that'll give me the same angle. So I get pi

over six plus two pi k, and pi over six, sorry, seven pi over six plus two pi, K, any integer.

This is a correct answer. But it's not as simple as it could be. Notice that seven pi

over six over here on the unit circle is exactly pi more than pi over six. So instead of taking

both of these and adding multiples of two pi to them, I could get all the same answers

by just taking one of them and adding multiples of pi to it. So a more efficient answer is

to say that x equals pi over six, plus pi times k, for K any integer. This will still

capture all the same solutions. Because when k is even, I'll get this family solutions.

And when k is odd, I'll get this family. For example, when k is one, pi over six plus one

times pi is just the original seven pi over six. If you think about the fact that tangent

has a period of pi, instead of two pi, it makes a lot of sense that you should be able

to write the solutions in this form. In this video, we solved basic trig equations by first

isolating sine, or tangent, or the same thing would work with cosine. And then using the

unit circle, to find principal solutions. Principal solutions are just solutions between

zero and two pi. And then adding multiples of two pi to these principal solutions to

get all solutions. For tangent, we noticed that it was equivalent to just use one principal

solution and add multiples of pi instead of two pi. In this video, we'll introduce the

idea of the derivative using graphs, secant lines and tangent lines.

So I have a function here, drawn in black, the function is y equals f of x. But actually

here f of x equals x squared. I also have a tangent line to my function drawn in red.

This tangent line is the tangent line at the point 1.5 2.25. By a tangent line, I mean

a line that touches the graph of my function at this one point, and heads off in the same

direction as the function. Well, normally to compute the slope, we need two points.

But for the tangent line, we really only have the exact coordinates of this one point, we

could approximate the slope by guessing the coordinates of some other points on the red

line. But in the long run, we'll end up with a more accurate estimate if we do something

else. So what we're going to do instead is we're going to calculate the slope of a secant

line. A secant line is a line that goes through two points on my graph. So in this case, my

secant line is going through my original point, and this other point at x equals three. So

that's the point three, three squared or three nine. Okay, so here's my secant line. To calculate

the slope of my secant line, I use the fact that slope is rise over run. Or in other words,

it's the change in y over the change in x. And that's going to be written in function

notation. F of three minus f of 1.5 is giving me the change in y and three minus one point

gives me the change in x. Wilson's f of x is x squared, this is the same thing as three

squared minus 1.5 squared over three minus 1.5. So that's just nine minus 2.25, over

1.5, which ends up as 4.5. So 4.5 is the slope of this second line. Well, the idea here is

that the slope of my secant line is an approximation of the slope of my tangent line. But in this

example, really the secant line that I've used, it's, its slope is only a very rough

approximation of the tangent line, not very accurate, all. So how could I get a better

approximation of the slope of my tangent line? Well, one thing I could do is I could use

as my second point, instead of using this point way out here, I could use a point closer

to my first point. So for example, I could use the point two, f of two. In other words,

the point two, four, as my second point for my second line, so let me draw that. And let

me calculate its slope, which after some arithmetic is going to give me an answer of 3.5. Well,

I could continue to pick second points for my secant line, closer and closer to my first

point. And I should end up with more and more accurate approximations of my tangent line,

let me make a little chart for this. For my next secant line, I could take my second point,

something pretty close to 1.5, say 1.6. And after some arithmetic, I get a slope of 3.1.

To take my second point to be 1.51, that's going to give me a slope of 3.01, and so on.

To write this more generally, if I take my second point as x, then the slope of my secant

line is going to be given by again the rise over the run. So that's going to be f of x

minus f of 1.5 divided by x minus 1.5. Change in y over the change in x, that's my slope.

Now, there's no reason I necessarily have to take my second point to be on the right

side of my first point, I could be using stead points on the left side here. Continue with

my chart, letting my second point be one, I can do the same computations to get a slope

of a secant line of 2.5. Here, I could get even closer on the left, say something like

1.4 and get a slope of 2.9, and so on. In general, if I have a point x f of x, and I

use a secant line through that point in our original point, I calculate the slope as change

in y over change in x, which is going to be f of 1.5 minus f of x divided by 1.5 minus

Actually, I can rewrite this a little bit to make it look more like the expression up

here. If I multiply the numerator and denominator by negative one, then I can rewrite this as

f of x minus f of 1.5 divided by x minus 1.5. So that these two expressions look exactly

the same. So the only difference here on my mind is that over here, I was thinking of

x as being a little bit bigger than 1.5. And here I'm thinking of x as being a little smaller

than 1.5. But I get the exact same expression for the slope of the secant line either way.

Now, this process of picking points closer and closer to our original point from the

left, and from the right, should remind you of limits. And indeed, the slope of the tangent

line is the limit as x goes to 1.5 of the slope of my secant lines, which are given

by this expression. This quantity is so important that's given its own name, it's called the

derivative of f of x at x equals 1.5. So in other words, the derivative which is written

as f prime at 1.5, is the limit as x goes to 1.5 of f of x minus f of 1.5 over x minus

1.5. Now based on our numerical tables, for example, here, we can see that that limit

seems to be heading towards three, whether X approaches 1.5, from the right or from the

left. So I'll write down the answer of three. You know, it's pretty strong. If you wanted

to have a really precise argument. We'd actually need to use algebra to compute this limit.

Exactly using the formula for the function itself, f of x equals x squared. And we'll

do examples like that in a future video. But for now, the main point is just that the slope

of the tangent line is the limit of the slope of the secant lines, which is given by this

formula. For now, let's look at an animation that shows how the slope of our secant lines

approach the slope of our tangent line. So this black curve here is the function y equals

x squared. The red line is the tangent line through the point where x equals 1.5. And

the blue line is a secant line that goes through the point with x coordinate 1.5. And a second

point with x coordinate 2.5. The points are shown here on the right. So I'm going to use

this slider here and drag my second point closer to my first point. So notice how as

the x coordinate, my second point gets closer and closer to 1.5. My secant line is getting

closer and closer to my tangent line. So the slope of my tangent line really is the limit

of the slope of my secant lines, as my x coordinate of my second point goes to 1.5. So this is

true, even if I start with my second point on the left instead of the right, as I drag

that second point closer and closer to the first point, the slope of my secant lines

gets closer and closer to the slope of that red tangent line. We saw in our example, that

the slope of our tangent line, or the derivative at 1.5, was given by the limit as x goes to

1.5 of f of x minus f of 1.5 divided by x minus 1.5. Well, in general, the derivative

of a function y equals f of x at an x value a is given by f prime of A equals the limit

as x goes to a of f of x minus F of A over x minus a, the function is said to be differentiable

at A. If this limit exists. In particular, both the limit from the left and a limit from

the right have to exist and be equal for the function to be differentiable

at x equals a.

There's another equivalent version of the definition of derivative that's very common

and very useful. If we're looking at the graph of a function, I'm trying to calculate the

slope of the secant line between the points A, F of A and x f of x, then let's introduce

the letter H to be the quantity x minus a. So h represents the run, when I'm calculating

the slope of this secant line, I can write H equals x minus a, or equivalently x equals

a plus H. And so I can rewrite the definition of derivative in terms of H as f prime of

A equals the limit as x goes to a of f of a plus h minus F of A divided by h, just by

substituting in this expression for x. And h for x minus a. Well as x goes to a, x minus

a is going to zero. In other words, h is going to zero. So this is equivalent to the limit

as h goes to zero of f of a plus h minus f of a over h. One way to think of this is that

we're just relabeling this point right here, as the point A plus h, f of a plus H. And

the slope of the tangent line is still the limit of the rise over the run as the run

goes to zero, this is the definition of derivative that we'll use most frequently going forward

when we actually calculate derivatives based on the definition in future videos. But for

now, let's look at some examples to practice recognizing the derivative rather than computing

it. So each of these following two expressions are supposed to represent the derivative of

some function at some value a. So for each example, we're supposed to find the function

and figure out the value of A. Now remember, we've got two definitions of derivative going

on. They're both equivalent, but they look different. One of them looks like derivative

of f at a is the limit as x goes to a of f of x minus F of A over x minus a. And the

other version is the limit as h goes to zero of f of a plus h minus f of a over h. Now,

you might notice That our first expression looks more like this first definition, because

x is going to some number that's not zero. And we have both x and a number in the denominator

here. Whereas our second expression looks more like the second definition, we've got

the age going to zero, and we've just got the age on the denominator here. Okay, so

let's look at this first one here. First, let's figure out what A is here. It seems

like a has got to be negative one. Since x is approaching negative one. That kind of

makes sense, because now here on the denominator, x plus one could be thought of as x minus

negative one. So that's our x minus a with a is negative one. Okay, great. So we've got

a, now we need to find an F. And we will need the numerator here to look like f of x minus

f of a, well, let's try the simplest thing, we can, let's try f of x equals x plus five

squared, then X plus five squared is our f of x. and f of a is our f of negative one

is going to be negative one plus five squared, which is 16. So that matches up perfectly,

we've got our f of x here, our f of a here, and our x minus a at the bottom. That's exactly

the definition of derivative done the first one. All right, now the second one. Now, again,

we need to figure out what a is. And we need to find figure out what f is. This part of

the expression here is supposed to be f of a plus h. So I'm going to guess, make a guess

here, that f of x should be three to some power, let's just try three to the X and see

how that works. Now we need this nine to be f of a. So nine has to be three to the A.

And the only way that can work is if a is two. So continuing on the top, we need f of

a plus h, that is f of two plus h,

two B three to the two plus H and that actually works perfectly. If our f of x is three to

the x, it's all falling into place. So we've got our f of two plus h, r f of two, and r

h. And it all works where f of x being three to the x and a being two. In this video, we

introduced the idea of derivative as a slope of a tangent line. And we gave two equivalent

definitions of the derivative in terms of limits. We'll continue with some interpretations

of derivatives in the next video. This video is algebraically intensive, and is concerned

with calculating derivatives using the limit definition of derivative. Well, there are

actually two versions of the limit definition of derivative. And I'll mostly use this one,

the limit as h goes to zero of f of a plus h minus f of a over h. If you're interested,

you can try reworking the problems using the alternative definition of derivative. The

limit as x goes to a of f of x minus F of A over x minus a first example, find the derivative

of f of x, which is one over the square root of three minus x at x equals negative one.

Well, in other words, we want to find f prime of negative one. So that's the limit as h

goes to zero of f of negative one plus h minus f of negative one over h. Using our definition

of f, that's the limit of one over the square root of three minus negative one plus h minus

one over the square root of three minus negative one, all over h. Let me clean this up a bit.

So this is one over the square root of, let's see, it's three minus negative one, so that's

three plus one, or four minus h minus one over the square root of four

over h. And I guess I can replace the square root of four with two.

Now, unfortunately, I can't just evaluate this by plugging in H equals zero, because

if I try that, I get one of these zero over zero indeterminate forms. You'll run across

these a lot when calculating derivatives by definition, it kind of makes sense because

remember the context where computing slopes of secant lines as these points get closer

and closer together. So our rise and our runs are both going to zero. So it makes sense,

we'll get these zero or zero indeterminate forms, we have to use our algebraic tricks

that we learned before, for rewriting our expression in a way that we can calculate

the limit. And I see two things going on here, there's square roots lurking, and there's

also fractions. So it's anybody's guess which trick I might want to apply First, the trick

for square roots, which would be multiplying the top and the bottom by the conjugate. And

the trick for fractions, which would be adding together my fraction for the common denominator.

I guess I'll try my fraction trick first. So my common denominator for my two fractions

here, is just the product of these two denominators. So that's the square root of four minus h

times two. Let me rewrite my fractions with this common denominator. Continuing here,

I get the limit of two minus the square root of four minus h over squared, a four minus

h times two, all over h. instead of dividing by H, let me multiply by one over h. And let's

see here, let's see if we can evaluate by plugging in H equals zero at this stage. Unfortunately,

when I try to plug in, I'm still getting the zero of zero indeterminate form. But I'm not

out of tricks, I haven't used the conjugate trick yet. So let's try multiplying the top

and the bottom by the conjugate of the top. Once I multiply out here, I'll get four plus

two square to four minus h minus two square to four minus h, minus the square root of

four minus h squared, that's going to cancel out nicely. And on the bottom, I have two

h squared of four minus h times two plus square to four minus h, I'll leave that factored

for now. That's good thing, I have unlimited space here, I'm kind of need it. Now on the

numerator, I'm going to get four minus four minus h, is carrying the denominator along

for the ride now, oh, I see something good. I see in the numerator, we're getting a four

minus four plus H. Subtracting out those fours to zero, and then cancelling out, my H is

that divide by each other, I think I finally got something that I can evaluate without

getting a zero over zero indeterminate form. All right, so here we've got. So as h goes

to zero, I'm just going to get one over two times the square root of four times two plus

a squared of four, which equals 1/16. So by now, you may have forgotten what the original

problem was, I think I have it, let's go back up here, we were looking for the derivative

of f of x, which was one over a squared of three minus x at x equals negative one, we

set up the limit definition, did a bunch of algebra first adding together fractions, then

using the conjugate trick, and eventually found that that derivative equaled 1/16. The

algebra doesn't get much harder than this problem here. In this next example, we're

asked to find the equation of the tangent line to y equals x cubed minus 3x at x equals

two. So the slope of the tangent line is given by the derivative, f prime of two. So let's

calculate the derivative first. f prime of two is the limit as h goes to zero of f of

two plus h minus f of two, all over h. So that's the limit of two plus h cubed minus

three times two plus h minus

two cubed minus three times two. I'm just plugging in first two plus h for x in the

definition of my function, and then I'm plugging in two for x. Now all of that needs to be

over h. Once again, if I try to plug in zero for H in these expressions, I'm just gonna

get something that all cancels out to zero at the top, and also have zero at the bottom,

one of my classic zero over zero indeterminate forms. So instead, I need to use algebra to

simplify things. And hope I can calculate the limit after that. So a good trick for

simplifying here is to multiply out two plus h cubed multiplies out to two cubed, plus

three times two squared times h, plus three times two times h squared plus h cubed. I'm

getting this from the formula for multiplying out of cubic which I've memorized. But you

can also get it more slowly just by writing out two plus h times itself three times and,

and distributing. Now I need to subtract three times two, and three times h. And finally,

I needed to subtract my two cubed and then add my three times to all this over h. Now

if you have my terms cancel out to zero here, two cubed minus two cubed. And let's see I've

got a minus three times two and a plus three times two. And I notice all the terms that

are left have H's and so I'm going to factor out an H from the top from the remaining terms

here. And that gives me let's say, three times two squared, so that's 12, plus six H, plus

h squared minus three over h. Now, h divided by h is one. So I'm just left with the limit

as h goes to zero of 12 plus six h plus h squared minus three, as h goes to zero, I

can just plug in H zero, and I get 12 minus three, which is nine. So my slope of my tangent

line, my derivative is nine. I'm not quite done, I still need to find the equation of

the tangent line, I just know that its slope is nine. So the equation of the tangent line

equation of any line is something like y equals mx plus b. and here m is nine. So I have y

equals 9x plus b, I just need to find the intercept B. Now, usually, to find the intercept,

I need a plug in a point. what point do I have here to plug in? Well, remember, we're

talking about a tangent line here. So we've got the point of tangency, the point where

x equals two, and the corresponding y value is y equals two cubed minus three times two,

or two. So my tangent line has to go through the point to two, which means if I plug in

this point for x and y, I get two equals nine times two plus b, which means that B has to

equal negative 16. So the equation of my tangent line then becomes y equals 9x minus 16. I

found that by first calculating the derivative to get my slope, and then using the point

of tangency, plugging in the x value to get the y value, and plugging Matt in to get b

to finish off the equation. So in this video, we used our tricks for evaluating limits algebraically

to compute some derivatives, using the definition of derivative. This is pretty labor intensive.

So fortunately, pretty soon, we'll learn some shortcut methods for calculating derivatives

without resorting to the definition. We've seen that the derivative of a function y equals

f of x at a point, x equals A represents the slope of a tangent line through the point

A f of a. But if the function f of x represents some practical quantity, like distance as

a function of time, or fuel efficiency as a function of speed, then the derivative will

also have a practical interpretation. This video is about interpreting the derivative

in different contexts.

One of the most famous contexts for interpreting derivatives is problems involving motion.

So let's say I'm on a bike ride, heading straight north from campus. And let's suppose that

y equals f of x represents my distance from campus. So x is the time and hours and y or

f of x is my distance and miles, the distance away from campus. Distance north of campus.

It's kind of fun to see what this graph here means in terms of my bike ride. In particular,

what's going on up here, where my y values reach their maximum. And what about here,

where my function is constant. Please pause the video for a moment and see if you can

make up a story that fits the graph. So here at the top, my distance is no longer increasing,

it's actually starting to decrease. So I must have turned around and be heading back towards

campus again, over here, where my f of x is constant, I probably stopped at a coffee shop

to take a rest or maybe I'm fixing a flat tire. Now let's get to the questions at hand.

Consider these two points, three, f of three, and four F of four, we want to interpret the

slope of the secant line through those two points. Slope is change in y over change in

x. And y here is distance and x here is time. So change in distance over change in time.

Sounds a lot like speed, or more accurately, velocity. Velocity just means speed in a certain

direction, and is positive. If distance is increasing, and negative if it's decreasing.

Speed is the absolute value of velocity, and there's always positive or zero. So in our

case, the velocity here must be negative, because our distance is decreasing. And we

could estimate it very roughly as about, say four minus 12 over four minus three, so about

negative eight miles per hour. But what is this negative eight miles per hour refer

to.

Since we're looking at the change in distance, over this entire hour long interval, the slope

of my secant line gives my average velocity over this interval. It doesn't give my exact

velocity, and exactly three hours or exactly four hours, only my average velocity. If I

want to know my exact velocity at exactly three hours, I need to look instead at the

slope of the tangent line at x equals three. The velocity at an exact instant of time is

sometimes called the instantaneous velocity to distinguish it from the average velocity

over a time interval. Let's think for a minute about why the velocity at exactly three hours

is given by the slope of the tangent line. We saw in a previous video that the slope

of the tangent line is the limit of the slope of the secant lines More precisely, the limit

as x goes to three of f of x minus f of three over x minus three. Wow, well, each of these

ratios represents an average velocity on the interval from three to x. And so the limit

is the limit of average velocities on tinier and tinier intervals of time, one minute,

one second 100th of a second, in the limit, as the length of the time interval goes to

zero, we're going to get the exact velocity at exactly three hours. So to repeat, in this

example, the slope of the secant line represents the average velocity over time interval, and

the derivative at x equals three, written f prime of three, which is also the slope

of the tangent line, that derivative represents the instantaneous velocity at x equals three.

More generally, if f of x represents any quantity that's changing, then the slope of the secant

line represents an average rate of change. While the slope of the tangent line, f prime

of a represents an instantaneous rate of change. Let's see how that works in a couple of other

examples. Let's suppose that f of x represents the temperature of a cup of coffee and degrees

Fahrenheit as a function of time and minutes since he said it on the counter. So let's

interpret the first equation. f of zero is 140. Well, that just means that at time zero,

the temperature is 140 degrees. What about the equation f of 10 minus f of zero is negative

20. That's saying that the temperature goes down by 20 degrees as x the time goes from

zero to 10 minutes. Now, what about this quotient here being equal to negative two? Well, this

quotient looks a lot like the slope of a secant Why, right, so it must be an average rate

of change. And in this context, we have the temperature is decreasing by an average of

two degrees per minute, as x changes from zero to 10 minutes. Finally, the derivative

of f at 15 is negative point five means that at exactly 15 minutes, the temperature is

decreasing at a rate of point five degrees per minute. negative numbers here always mean

decreasing, and f prime is an instantaneous rate of change. Let's look at another example.

Please pause the video and try this one for yourself. Here g of x represents the fuel

efficiency of a Toyota Prius and mpg as a function of x, the speed in miles per hour

that is traveling g of 45 is 52 means that at 45 miles per hour, the fuel efficiency

is 52 miles per gallon. The second statement is saying that as speed increases from 35

to 45 miles per hour, fuel efficiency goes up by 10. That's 10 miles per gallon. The

third statement says that the average rate of change

of fuel efficiency is two miles per gallon per mile per hour, as speed increases from

35 to 40 miles per hour. So going up from 35 to 40, gives you better fuel efficiency

here. On the other hand, when you're going 60 miles per hour, your fuel efficiency is

decreasing at a rate of two miles per gallon per mile per hour. So I bet the optimal fuel

efficiency here occurs somewhere between 40 and 60 miles per hour. In this video, we've

interpreted the slope of the secant line as the average rate of change, and the slope

of the tangent line with the derivative as an instantaneous rate of change. I hope that

these general principles will help you interpret the derivative in a variety of contexts that

you might encounter throughout your life. In this video, we'll think of the derivative

of a function as being a function in itself, we relate the graph of a function to the graph

of its derivative, and we'll talk about where the derivative does not exist. We've seen

that for a function f of x and a number A, the derivative of f of x at x equals A is

given by this formula. But what if we let a very, if we compute f prime of a lots of

different values of a, we can think of the derivative of f prime as itself being a function,

I'm going to rewrite this definition of derivative with x in the place of a just to make it look

a little more like standard function notation. So f prime as a function of x is the limit

as h goes to zero of f of x plus h minus f of x over h. This isn't anything substantially

different from what we've been doing before, it's just a difference in perspective. So

let's do one more example of computing the derivative by hand using the definition, but

a general number x instead of a specific value, the function we're going to use is f of x

equals one over x. And first, let's just write down the definition of derivative in general.

So f prime of x is given by this formula. And using the fact that f of x is one over

x, I can rewrite this as one over x plus h minus one over x all over h. So as usual,

this is a zero over zero indeterminate form. If I plug in zero for H, I'm just going to

get one over x minus one over x on the numerator, which is zero. and plugging in zero for H

gives me zero on the denominator two. So I'll need to use some algebra to rewrite things

to get an A form that I can evaluate it. Let's add together our fractions in the numerator

here. The common denominator I need to use is x plus h times x. So I multiply this fraction

by x over x and the next fraction by x plus h over x plus h. All that's over h. And now

continuing, I get x minus x plus h over x plus h times X and instead of dividing the

whole thing by h here, multiply by one over h to get the limit have x minus x minus h

over x plus h times x times h. Now I can subtract off my x's here. And after I do that, I can

divide my minus h by my H, to get just a minus one on the numerator here. So that's just

the limit of negative one over x plus h times x. And now I'm in a good position because

I can plug in H equals zero and get something that makes sense. Namely, I'm getting a limit

of negative one over x plus zero times x or negative one over x squared as my derivative.

In this example, we're given the graph of a function that's supposed to represent the

height of an alien spaceship above the Earth's surface, we want to graph the rate of change.

The rate of change means the derivative of our function, but we're not given an equation

to work with. So we'll just have to estimate the derivative based on the shape of the graph

by thinking about slopes of tangent lines. I'll start by drawing a new set of axes where

I can graph my derivative. And I'll consider my original function, which I'll call g of

x, piece by piece.

For x values between zero and two, my original function g of x looks like a line, it has

slope negative one, since the rise is negative two, while the run is two, for any point on

the straight line segment, the tangent line will also be a straight line with slope negative

one, and therefore, the derivative will be negative one. For x values between zero and

two, I'm going to ignore the time being what happens when x is exactly zero or exactly

two, and just look at the interval of X values between two and three. Here, g of x is completely

flat. So tangent lines at any of these points will have slope zero. And I'll draw a derivative

of zero. When x is between two and three, I'll postpone worrying about the derivative

when x is exactly three. And just think about the derivative when x is between three and

five, where g of x is flat again, so it's tangent lines will have slopes of zero. And

I'll draw again, a derivative is zero when x is between three and five. Now things get

a little more interesting. As x increases from five to about seven, g of x is an increasing

function. The slope of tangent lines here are positive, starting at about, say, a slope

of three, and decreasing to a slope of zero, when x is seven, I can draw that down here.

As x increases from seven, the tangent lines now have negative slopes, going to a maximum

negative slope of about negative one here, and then heading towards a slope of zero,

when x is just shy of 10. My estimates of three and negative one for the slopes of my

tangent lines are just rough estimates based on approximating the rise over the run. As

x increases from 10, the tangent line slope is positive, and getting steeper and steeper,

so my derivative is going to be positive and increasing. That's the basic shape of the

derivative. Now let's see what happens at these special points like 235 and zero. To

figure out the derivative at x equals two, let's go back to the definition of derivative

as the limit of the slopes of the secant lines. If I draw a secant line, using a point on

the left, I'll just get this line that lines up with this line and has a slope of negative

one. But if I compute the slope of a secant line, using a point on the right, I'll get

a slope of zero. So the limit from the left and the limit from the right of the slopes

of my secret lines will be different. And so my limit does not exist and my derivative

does not exist. And so I'll just draw this as an open circle at x equals two. Next, let's

look at the derivative when x equals three. Remember that the derivative at three is the

limit as h goes to zero, of g of three plus h minus g of three over eight. Well, if H

is bigger than zero, then g of three plus H is going to be about a half, because three

plus H is to the right of three. On the other hand, if H is less than zero, g of three,

plus H is two, because three plus H is actually a number less than three, g of three itself

is equal to one half, based on the filled in bubble here. so if we calculate the limit,

as h goes to zero from the positive side, we get the limit of one half minus one half

over age, which is just the limit of zeros, so that's zero. On the other hand, if we compute

the limit from the left, we get the limit of two minus a half over eight, which is the

limit of three halves over h. And as h goes to zero, that limit is negative infinity.

So once again, the left limit and write limit are not equal. And so the limit of the slopes

of the secant lines does not exist, and there's no derivative at x equals three. And I'll

draw an open circle there to

add x equals five, again, we have a corner. And by the same sort of argument, we can conclude

that the derivative does not exist. And finally, when x equals zero, we can only have a limit

from the right not the left. And so by that sort of technical reason, we don't have a

derivative at that left endpoint either. So we've drawn a rough graph of the rate of change

of the height of our alien spaceship, as it comes closer to Earth beams down to pick up

Earthlings and then makes us escape up to the mothership. It's interesting to observe

that the domain of the original function g of x is from zero to infinity, but the domain

of g prime is somewhat smaller, and just goes from zero to two, then from two to three,

then from three to five, and finally, from five to infinity, missing some places where

the function originally existed. We saw in the previous example, that the derivative

doesn't necessarily exist at all the x values where the original function exists.

Please pause the video for a moment and try to come up with as many different ways as

you can, that a derivative can fail to exist

at an x value x equals a. Well, one kind of boring way that a function can fail to

have a derivative at x equals A is if f of x itself fails to exist. at x equals a, for

example, if it has a hole, like in this picture, we saw in the previous example, with the alien

spacecraft, that a derivative can fail to exist when the function turns a corner. When

we tried to evaluate the derivative in that example, by taking the limit of the slope

of the secant lines, the limit from the left, and a limit from the right did not agree.

A famous example of a function that turns a corner is the absolute value function. For

the absolute value function, f prime of x is negative one, since the slope here is negative

one, when x is less than zero, and it's positive one when x is greater than zero, but f prime

of zero itself does not exist. A function with a casp also fails to have a derivative

at the cusp. In the alien spaceship example, we also saw that F can fail to have a derivative

at a discontinuity. But there's another way that a derivative can fail to exist, even

when f has no cost per corner discontinuity. Let's look at the function f of x equals x

to the 1/3 graphed here, what's going on at x equals zero. At that instant, the tangent

line is a vertical with a slope that's infinite or undefined. So the limit of the slopes of

the secant lines will fail to exist because it'll be infinite. A function is called differentiable

at x equals a, if the derivative exists at a function is differentiable on an open interval,

if f is differentiable at every point in that interval. So all of the examples on the

previous slides are examples of places where a function is not differentiable. All of these

examples are important. But I'm going to focus on the example involving discontinuity. In

general, if f of x is not continuous at x equals a, then f is not differentiable at

x equals a. This is what we saw in the example involving the jump discontinuity. an equivalent

way of saying the same thing is that f is differentiable at x equals a, then f has to

be continuous at x

previous slides are examples of places where a function is not differentiable. All of these

examples are important. But I'm going to focus on the example involving discontinuity. In

general, if f of x is not continuous at x equals a, then f is not differentiable at

x equals a. This is what we saw in the example involving the jump discontinuity. an equivalent

way of saying the same thing is that f is differentiable at x equals a, then f has to

be continuous at x

equals a.

However, if all we know is that f is continuous at x equals a, then we can't conclude anything

about whether or not is differentiable there, f may or may not be differentiable at x equals

a. Remember the square root example, the square root of x is continuous at x equals zero,

but it's not differentiable there because of the corner. In this video, we related the

graph of a function to the graph of its derivative. By thinking about the slopes of tangent lines,

we also looked at several ways that a derivative can fail to exist at a point and noted that

if a function is differentiable, it has to be continuous. This video gives a proof that

differentiable functions are continuous. What we want to show here is that if a function

is differentiable at a number x equals a, then it's continuous at x equals a. Let me

call the function f of x. And I'm going to start out by writing down what it means for

f of x to be differentiable at x equals a. That means that the limit as x goes to a of

f of x minus F of A over x minus a exists and equals this finite number that we call

f prime