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## AP®︎/College Calculus AB

### Unit 5: Lesson 5

Using the candidates test to find absolute (global) extrema# Finding absolute extrema on a closed interval

AP.CALC:

FUN‑4 (EU)

, FUN‑4.A (LO)

, FUN‑4.A.3 (EK)

Sal finds the absolute maximum value of f(x)=8ln(x)-x² over the interval [1,4]. Created by Sal Khan.

## Want to join the conversation?

- Is there any method to identify that critical point
`x = 2`

is maxima or minima, instead of replacing 1, 2 and 4 into`f(x)`

?(23 votes)- You can see whether
`x=2`

is a**local**maximum or minimum by using either the First Derivative Test (testing whether`f'(x)`

changes sign at`x=2`

) or the Second Derivative Test (determining whether`f"(2)`

is positive or negative). However, neither of these will tell you whether`f(2)`

is an**absolute**maximum or minimum on the closed interval [1, 4], which is what the Extreme Value Theorem is talking about here. You need to actually compare the values of the function at the critical numbers and at the endpoints to determine which is the highest and lowest on the interval. Hope that helps!(40 votes)

- Did we really have to check the values of the function at the endpoints to see if they are greater than the local maximum f(2)? I mean, since we didn't find any other x value in the interval [1,4] where f(x) is either 0 or undefined and since the function is continuous, isn't that enough to know for sure that neither of the endpoints has a greater value than f(2)? Because if at least one of them did, shouldn't the function have a local minimum in order to increase to a greater value than f(2)? And if it did, we would have already found that local minimum, but we didn't.(11 votes)
- Once we prove that f(2) is the local maximum by taking derivatives of intervals before and after it, and that there are no other critical points, then you are right, I don't see any other information needed to prove that f(2) is also the absolute maximum over the domain. However, he didn't prove it was the local maximum before proving it was the absolute maximum by comparison to the endpoints.

Hopefully this helps someone 4 years later.(9 votes)

- Is there any proof of the extreme value theorem(2 votes)
- So could we say that the absolute(global) minimum is f(4)? I re-watched the video right at the end but i don't think he said it.(3 votes)
- Yes, the global minimum is at x=4.

So, altogether the extrema are:

local min at x=1

global max at x= 2

global min at x=4(10 votes)

- My question is if any question has a critical value is "infinity". now, we ignore the critical value or put that value in original function to find max. or mini.??(4 votes)
- A critical point cannot have a value of infinity. There may be a critical point because the first derivative diverges toward infinity, but in such a case the first derivative fails to exist at that point.

If the original function is defined at a point and its first derivative fails to exist at that point, then you would proceed to see whether it is an extremum in the usual way -- seeing if the first derivative changes signs by comparing the first derivative to just before vs. just after to see if there is a sign change OR by plugging in the critical point into the original function and then comparing that to points arbitrarily close to it on either side.(4 votes)

- I'm not sure if someone else asked this or not, but wouldn't -1^2 = 1 not -1?(1 vote)
- No. it is (-1)² = 1 but -1² means -(1²) which equals -1(10 votes)

- Sal you said at5:37that ln(2) (.693147...) is going to be some fraction. And that fraction, if greater than 1/2 will result in 8*ln(2) being more than 4. While ln(2) * 8 is greater than 4 (5.54518), I wanted to follow your reasoning and got lost at this point.

Being more than have will eventually give you a negative number when added with -4.(3 votes)- If p > 1/2, then 8*p > 8*1/2 = 4. Multiplying both sides by the same positive number leaves the direction of the inequality unchanged. (Or is this not what you were asking?)(2 votes)

- How to determine minima or maxima (absolute) when there are asymptotes?(2 votes)
- If there are vertical or oblique asymptotes, the chances are that you don't have an absolute max and/or min. Specifically, if there is anywhere the function heads toward positive infinity, you don't have an absolute max. If there is anywhere the function heads toward negative infinity, you don't have an absolute min.

If you have a specific problem in mind, please post that so we can examine the specifics.(3 votes)

- Let's say I'm looking at a graph with two maximums and two minimums (four points). So the absolute max/min would be the highest/lowest value out of the four points on a graph, and the relative or local max/min would be the other two points on the graph?(1 vote)
- No, that is not correct.

First the plural of "maximum" is "maxima" and the plural of "minimum" is "minima".

There is not always an absolute max/min, but if there is it will be one of the local max/min. To find out here is what to do:

If you have a closed interval, then the endpoints are automatically local max/min. If you have an open interval the endpoints are never max/min (because they are not in the domain).

Every max/min is a local max/min.

Identify the largest local max and the least local min. If and only if there is no other point in the domain greater than the greatest local max, then that local max is also an absolute max. Likewise, if and only if there is no other point in the domain that is less than the least local min, then that min is also an absolute min.

If there is any point in the domain greater than the greatest local max, then there is no absolute max. Likewise, if there is any point in the domain less than the least local min, then there is no absolute min.

Note, if you have two or more max that with equal y values and which are greater than all other points in the domain, then both (or however many there are) are absolute max. Similar for there being more than one absolute min. Consider the case of the sine function. Every local min is -1 and every local max is +1 AND there are no points where sine is greater than +1 nor less than -1. Thus, with the sine function all of the local maxima are absolute maxima and all of the local minima are absolute minima. So, sine has infinitely many absolute maxima and infinitely many absolute minima.(5 votes)

- At about4:40, Sal is testing the end points and the critical value. He uses Natural Log in lieu of dividing. Is this another way to use ln?(2 votes)
- There Sal was plugging in the endpoints x=1 and x=4 as well as the critical point x=2, and instead of just going to the calculator (like I think he does at the end of the video) he tries to estimate these values of the natural logs.

The first ln(1) he doesn't need to estimate as we know this is 0. Think what power to we raise e to to get 1 as a result: 0.

[ 1 is e^0 so ln(1) is 0. ]

The next ln(4), we know is some positive number but less than 2. Think what power we raise e to to get 4. Remember e is around 2.72, and if we raise that to more than 2 we'd get something more than 4. So we must raise e to something between 1 and 2 to get 4. So ln(4) is between 1 and 2.

[ 4 is between e^1 and e^2, so ln(4) is between 1 and 2. ]

Finally he reasons similarly that ln(2) is between 0.5 and 1. Think what power we'd raise 2.7 to to get 2. It'd be a power less than 1 but not so small that we get less than 2, so more than 1/2. So 1/2 < ln(2) < 1.

[ 2 is between e^1/2 and e^1 so ln(2) is between 1/2 and 1. ]

He does all these approximations just to figure out that the first two values f(1) and f(4) are both negative, and the last f(2) is positive.

To summarize:

ln(1) = 0 so f(1) = 8*ln(1)-1^2 = -1.

1< ln(4) < 2 so -8 < 8ln(4)-4^2 < 0, so f(4) is negative.

1/2 < ln(2) < 1 so 0 < 8ln(2)-2^2 < 4.

f(2) is positive and hence bigger than both f(1) and f(4), and so f(2) is the absolute maximum.(2 votes)

## Video transcript

Let's say that we've got the function f of x is equal to eight times the
natural log of x minus x squared and it is defined over the closed
interval between one and four, so it's a closed interval. it also includes one and it includes four. You can view this as the
domain of our function as we have defined it. So given this, given this
information, this function definition, what I would like you
to do is come up with the absolute, absolute
maximum value, value of f, of f as defined where f is
defined right over here, where f is defined on
this closed interval. And I encourage you to pause this video and think about it on your own. So the extreme value theorem tells us, look, we've got some closed
interval - I'm going to speak in generalities here -
so let's say that's our X axis and let's say we have some
function that's defined on a closed interval. We have a couple of
different scenarios for what that function might look
like on that closed interval. So, we might hit a maximum
point, we might hit a maximum point, at the
beginning of the interval, something like that. We might hit an absolute
maximum point at the end of the interval, so it might
look something like this, so that's at the end of the interval. Or, we might hit an absolute
maximum point someplace in between and that could look something like this, it could look like this, and at this maximum
point, the slope of the tangent line is zero, so
here the derivative is zero, or we could have a maximum
point someplace in between that looks like this. And if it looks like this,
then here the derivative would be undefined. There's a lot of different tangent lines that you could place, that you
could place right over there. So, what we need to do is, let's test. Let's test the different endpoints. Let's test the function at the beginning, let's test the function at
the end of the interval, and then let's see if there's any points where the derivative is either zero or the derivative is undefined. And these points where
the derivative is either zero it is undefined,
we've seen them before, we call these, of
course, critical numbers. So this would be either,
in either case actually, if we assume that that's
happening to the same number, we would call that a critical number. Critical - a critical number. So those are the different candidates. Now you could have a critical
number in between that, where, say, the slope is
zero, say something like this, but it is at the maximum or minimum. But what we can do is, if we can
find all the critical numbers, we can then test the affect,
the function of the value of the critical numbers and the function of the value at the
endpoints and we can see which of those are the largest. All of those are the
possible candidates for where f hits a maximum value. So, first we could think about
- well actually, let's just, let's find the critical numbers
first, since we have to do it. So, let's take the derivative of f. f prime of x is going to be equal to the derivative of the natural
log of x is one over x. so it's going to be eight over x minus 2x and let's set that equal to zero. So if we focus on this part
right over here, we could add 2x to both sides and we would
get eight over x is equal to 2x. Multiply both sides by x, we get eight is equal to 2x squared. Divide both sides by two, you
get four is equal to x squared. And if we were just purely
solving this equation, we would get x is equal
to plus or minus two. Now, we are saying that the
function is only defined over this interval, so negative
two is a part of its domain, so we are only going to
focus on x is equal to two. This right over here is
definitely a critical number. Now, have we found all
of the critical numbers? Critical numbers. Well, this is the only number
other than negative two, the only number in the
interval that will make f prime of x equal to zero. What about where it's undefined? Well, f prime of x would be
undefined, the only place where it would be undefined is if you
stuck a zero right over here in the denominator, but zero is
not in the interval, so the only critical number in the
interval is x equals two. So, now we just have to test f
at the different endpoints and at the critical number and see
which of those is the highest. So, we're going to test
f of one, f of one, which is equal to eight
times the natural log of one minus one squared. We'll test f of four,
which is equal to eight times the natural log of
four minus four squared, which is, of course, 16, which is 16. And we're going to test f of 2. So these are the endpoints
and this is a critical number. Eight times the natural log
of two minus two squared. Now, which of these is
going to be the largest? And it might be tempting
to get a calculator out, but actually let's see if we
can get a little intuition here. So this is, the natural log of
one is zero, e to the zero power is equal to one. So eight times zero is zero, so
this evaluates to negative one. Now, let's see, what
does this evaluate to? The natural log of four, e is two
point seven, on and on and on, so this number is going
to be between one and two so it's going to be between one and two, and it's actually going to
be, well, between one and two, you multiply that times
eight and you're going to be between eight and 16, and then you subtract
16, so that means you're going to be between
zero and negative eight. So, okay, so that, it's not
clear, at least not without using a calculator, or this very rough
way, which of these is larger. Both of these are
negative numbers, though. Now, what about this? The natural log of two. The natural log of two is
going to be some fraction. It's going to be more than half,
it's going to be more than half, And since it's more than
half, this whole thing is going to be more than four, which means this whole thing is going to be positive. So this is negative, this is
negative, this is positive. And these are only critical
numbers, these are only candidates for our maximum value, so I would go with this one. Our maximum, our maximum value happens when x is equal to two,
and that maximum value is eight natural log of two minus four. That is the absolute maximum
value, absolute max value over the interval, or I guess we could say over the domain that this
function has defined. If we want to verify it with a
calculator we, of course, could. So, we already figured out this
one, but let's see, f of four, eight natural log of four minus
16 is equal to negative five. So that's, that's, this is definitely not, this one is definitely
not the maximum value. And then f of two is eight
natural log of two minus four which, as we said, is
indeed a positive number. So feel pretty good about what we did.