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## Using the candidates test to find absolute (global) extrema

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# Finding absolute extrema on a closed interval

AP Calc: FUN‑4 (EU), FUN‑4.A (LO), FUN‑4.A.3 (EK)

## 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.