Introduction

Welcome to AP Calculus AB/BC's Topic 5.2 - Extreme Value Theorem, Global Versus Local Extrema, and Critical Points! As always, my goal is to make sure this article covers everything you need to know for the AP Calculus Exam by using practice problems and simple explanations. With that being said, let's get started by reviewing something you should already have learned from precalculus.

Identifying Extrema

From precalculus, you should already know how to identify global vs local extrema graphically. If you don't remember, please review how to do it graphically. You can watch this video by Patrick J as a refresher.

To make sure you understand how to identify global and local extrema graphically, answer the following question.

Question

1. Use the graph below to identify all global and local extrema.

Solution

1.
Global Extrema:

Relative Extrema:

Extreme Value Theorem

Yes, I know, another theorem. Don't worry, this is perhaps the easiest of them all.

Technical Definition

According to the extreme value theorem (EVT), if a function $f$ is continuous over the interval $[a,b]$, then $f$ has at least one minimum value and at least one maximum value on $[a,b]$.

Take a look at the graph below.

The EVT tells us that because $f$ is continuous over the interval $[1,7]$, $f$ must have at least one minimum value and at least one maximum value on $[1,7]$. Because we have the graph of $f$, we can clearly find the minimum and maximum values. Notice how the minimum and maximum values are not at the endpoints. Make sure you understand that the EVT does not guarantee that the minimum and maximum values occur the endpoints. 

Critical Points

A critical point is a point that could be a minimum or maximum extrema. We are not 100% of whether or not it is a minimum or maximum extrema without more information. For this lesson, we are trying to find where the candidates for minimum or maximum extrema are.

Let's look at the graph above and identify what is true about $f^{\prime }(x)$ at each of the relative extrema. Well, we notice that at our relative extrema at $x=2,5,6$,  $f^{\prime }(x)=0$. We also notice that at our relative extrema $x=4$, $f^{\prime }(x)$ does not exist. Based on this, what have we determined are conditions for a critical point?

There are two ways to a critical point exists:

1. $f^{\prime }(x)$ does not exist

2. $f^{\prime }(x)=0$

Questions

1. Find all the critical points for $f(x)=4x^2+2x+\sec (\pi )$.

2. Find all the critical points for $g(x)=\frac{4}{\sqrt{9-x^2}}$.

3. Find all the critical points for $h(x)=|x|$.

4. Find all the critical points for $i(x)=x\sqrt{x-a}$ in terms of the constant $a$.

5. Find all the critical points for $j(x)=\frac{1}{x}+\ln x$.

note: problems four and five were taken from John's Solution Set.

Solutions

1.
We can find all the critical points for $f(x)$ by looking at where $f^{\prime }(x)=0$ or does not exist.

$f^{\prime }(x)=8x+2$
$f^{\prime }(x)=8\left(x+\frac{1}{4}\right)$
$f^{\prime }(x)=0$
$8\left(x+\frac{1}{4}\right)=0$
$x=-\frac{1}{4}$ is the location of our only critical point. Note that there are no places where $f^{\prime }(x)$ doesn't exist, so there are no other critical points.

2.
We can find all the critical points for $g(x)$ by looking at where $g^{\prime }(x)=0$ or does not exist. First, it might be helpful to rewrite $g(x)$ to make it easier to find our derivative.

$g(x)=4(9-x^2{)}^{-\frac{1}{2}}$
$g^{\prime }(x)=-2(9-x^2)(2x)$
$g^{\prime }(x)=-4(x)(9-x^2)$
$g^{\prime }(x)=4(x)(x^2-9)$
$g^{\prime }(x)=4(x)(x+3)(x-3)$
$g^{\prime }(x)=0$
$4(x)(x+3)(x-3)=0$
$x=-3,0,3$

But wait! We may be tempted to circle the $x$-values above, but we have to make sure that they are in the domain of $g(x)$ to begin with. We cannot have a critical point if the point does not exist in the domain of $g(x)$. We know that the denominator can never be equal to zero, and that the inside of a square root must be positive. Let's solve for our domain restricits.

$9-x^2>0$ (not we are not greater than or equal to because this is in the denominator, and the denominator can never be equal to zero)

$-(x^2-9)>0$
$-(x+3)(x-3)>0$

Thus, the domain of the function is $(-3,3)$. Therefore, $x=-3$ and $x=3$ are not critical points (because they are outside the domain). However, $x=0$ is a critical point.

3.
We can find all the critical points for $h(x)$ by looking at where $h^{\prime }(x)=0$ or does not exist.
$h(x)$ is not a "traditional" function in the sense we cannot find a "traditional" $h^{\prime }(x)$. However, we can think of the graph of $h(x)=|x|$. We know that at $x=0$, $h^{\prime }(x)$ does not exist (because the left-sided derivative of $-1$ does not equal the right-sided derivative of $1$). Therefore, there exists a critical point at $x=0$ because $h^{\prime }(x)$ does not exist.

Questions 4 and 5:

Check this video for detailed solutions from the problems creator! Make sure you always check your domains before circling your critical points.

4.
$x=a$

5.
$x=1$

Practice Questions