### Video Transcript

True or False: The function π of
π₯ is equal to two π₯ divided by one plus π₯ squared has an absolute maximum over
the closed interval from negative three to three equal to one.

In this question, weβre given a
function π of π₯. And we need to determine if this
function has an absolute maximum value over the closed interval from negative three
to three equal to one. To answer this question, letβs
start by determining if this function even has an absolute maximum on this
interval. And to do this, we notice weβre
given a closed interval. So we can try and do this by using
the extreme value theorem.

We recall the extreme value theorem
tells us that real-valued continuous functions on a closed interval will attain a
maximum and a minimum value on this interval. And this is particularly useful
because it helps us find the absolute maximum and minimum values on this
interval. And we can apply this theorem in
this case. First, we can see we have the
closed interval from negative three to three. Next, we can see our function π of
π₯ is a rational function. And rational functions are
continuous everywhere except when their denominator is equal to zero. And the denominator in our function
π of π₯ is never equal to zero because π₯ squared is greater than or equal to
zero. So one plus π₯ squared is greater
than or equal to one.

So π of π₯ is continuous for all
real values of π₯. And in particular, itβs continuous
on the given closed interval. Therefore, the extreme value
theorem tells us that π of π₯ does indeed have an absolute maximum on this
interval. But thatβs not all we need to
do. We also need to check that the
output value of this maximum is equal to one. And to do this, weβll clear some
space and find the absolute maximum of this function over this interval.

First, we recall that the local
extreme of π of π₯ must occur at the critical points of π of π₯. And the critical points are the
points where the derivative of our function is equal to zero or where the derivative
of our function does not exist. Next, we evaluate π of π₯ at these
points to find the values of any possible local extrema. And finally, we evaluate π of π₯
at the endpoints of our interval because these can be possible absolute extrema.

So letβs start by finding the
critical points of π of π₯. To do this, weβre going to need to
differentiate our function. And since π of π₯ is a rational
function, weβll do this by using the quotient rule. We recall this tells us for two
differentiable functions π’ of π₯ and π£ of π₯, the derivative of their quotient is
equal to π’ prime of π₯ times π£ of π₯ minus π£ prime of π₯ times π’ of π₯ all
divided by π£ of π₯ all squared.

So to apply the quotient rule to π
of π₯, we set π’ of π₯ to be the function in our numerator, thatβs two π₯, and π£ of
π₯ to be the function in the denominator, thatβs one plus π₯ squared. We then need to find the derivative
of both of these functions. Letβs start with π’ prime of
π₯. Thatβs the derivative of the linear
function two π₯. And the derivative of any linear
function is the coefficient of π₯, which in this case is two.

Next, we need to find π£ prime of
π₯. We can do this term by term by
using the power rule for differentiation. We need to multiply by the exponent
of π₯ and reduce this exponent by one. Applying this, we get π£ prime of
π₯ is two π₯.

Weβre now ready to find π prime of
π₯ by using the quotient rule. We need to substitute our
expressions for π’, π£, π’ prime, and π£ prime into the quotient rule. We get that π prime of π₯ is equal
to two times one plus π₯ squared minus two π₯ multiplied by two π₯ all divided by
one plus π₯ squared all squared. Distributing over the parentheses
in our numerator, we get two plus two π₯ squared minus four π₯ squared all divided
by one plus π₯ squared all squared. And finally, we can simplify our
numerator. Two π₯ squared minus four π₯
squared is equal to negative two π₯ squared, giving us that π prime of π₯ is equal
to two minus two π₯ squared all divided by one plus π₯ squared all squared.

And now weβre ready to use our
expression for π prime of π₯ to find the critical points of π of π₯. These are the points where π prime
of π₯ is equal to zero or π prime of π₯ does not exist. Letβs start by checking if thereβs
any value of π₯ where π prime of π₯ doesnβt exist.

We found the derivative of π prime
of π₯ by using the quotient rule. And the quotient rule is valid
provided π’ and π£ are differentiable at that value of π₯ and π£ of π₯ is not equal
to zero. However, π£ of π₯ was the
denominator of our function π of π₯. And weβve already checked that this
is never equal to zero for any value of π₯. So the quotient rule is valid for
all real values of π₯. Therefore, π prime of π₯ is
defined for all real values of π₯.

So our only critical points will be
when π prime of π₯ is equal to zero. So letβs determine if there are any
values of π₯ where π prime of π₯ is equal to zero. For the quotient of two numbers to
be equal to zero, the numerator must be equal to zero. And of course we can see the
denominator is never equal to zero. Therefore, the critical points of
π of π₯ occur when this numerator is equal to zero.

We need to solve two minus two π₯
squared equals zero. And we can solve this. We add two π₯ squared to both sides
of the equation, divide through by two, and then take the square root, where we
remember we will get a positive and a negative root. We have π₯ is equal to positive or
negative one. Therefore, we found all of the
critical points of π of π₯. Thereβs two: one when π₯ is equal
to negative one and one when π₯ is equal to one.

So we can move on to the second
part of our question. We need to evaluate π of π₯ at its
critical points. Letβs start with π evaluated at
one. We substitute one into our
expression for π of π₯ to get two times one divided by one plus one squared, which
if we evaluate is equal to one. We can do the same to evaluate π
at its other critical point. We substitute π₯ is equal to
negative one to get two times negative one all divided by one plus negative one all
squared, which if we evaluate we get negative one.

Now that weβve evaluated π at all
of its critical points, we need to move on to our third step. We need to evaluate π at the
endpoints of our interval. Thatβs negative three and
three. Letβs start by evaluating π at
three. We substitute three into our
expression to get two times three divided by one plus three squared, which is equal
to 0.6. And we can do the same to evaluate
π at negative three. We see this is equal to negative
0.6.

And now that weβve evaluated π at
all of its critical points and its endpoints, we know the largest of these values
will be the absolute maximum of our function over the interval and the smallest will
be the absolute minimum of the function over this interval. And the largest of these four
values is one. So π has an absolute maximum value
of one on the closed interval from negative three to three, which is what weβre told
in the question.

Therefore, weβve shown that itβs
true that the function π of π₯ is equal to two π₯ divided by one plus π₯ squared
has an absolute maximum value of one on the closed interval from negative three to
three.