6.2 Absolute and Relative Extrema

In our world, understanding maximum and minimum values is critical. A businessperson may ask, “What is the optimal production to maximize the company’s profit?” while a rocket scientist may ask, “What is the maximum acceleration of a space shuttle?” While what we are going to learn in this chapter is very basic, derivatives can be used to obtain maximum and minimum values in many situations given a functional form.

In order to understand extrema, we will introduce two types of extrema: absolute extrema and relative extrema. Let $f$ be a function with some domain $D$. The function $f$ has an absolute maximum at the value $a$ if $f(a)\ge f(x)$ for all $x$ in $D$. The value of $f(a)$ is called the absolute maximum value of $f(x)$. The function $f$ has an absolute minimum at the value $a$ if $f(a)\le f(x)$ for all $x$ in $D$. The number $f(a)$ is called the absolute minimum value of $f(x)$. The casual definition would be that the absolute maximum is the highest $f(x)$ at a point $a$ in the entire function, and the absolute minimum is the lowest $f(x)$ at a point $a$ in the entire function.

Given the graph below, can you identify the absolute maximum value and the absolute minimum value of the function?

As Figure 6.1 shows, the graph of a function $f$ has an absolute maximum at $a$ and an absolute minimum at $d.$ As we casually defined, ($a$, $f(a)$) is the highest point on the graph, and ($d$, $f(a)$) is the lowest point on the graph.

How about ($b$, $f(b)$) where a valley is and ($c$, $f(c)$) where a peak is? Consider the value of $x$ near $b$, perhaps in the interval of ($a$, $c$). We can say that $f(b)$ is a relative maximum, or local maximum, since it is the largest of those $f(x)$ near $b$. How about value of $x$ near $c$ in the interval ($b$, $d$)? We can say that $f(c)$ is a relative minimum, or local minimum, since it is the smallest of those $f(x)$ near $c$.

The formal definition of relative extrema is as follows. A function $f$ has a relative maximum at $a$ if $f(a)\ge f(x)$ when $x$ is near $a$. Similarly, a function $f$ has a relative minimum at $a$ if $f(a)\le f(x)$ when $x$ is near $a$.

See the graph below. Identify the absolute or relative extreme values at each point $x=-10,-6,-1,1,$ and $10$.

Let’s start with the absolute maximum and minimum. For this function, the absolute maximum is 8 when $x=-10$, and the absolute minimum is −7 when $x=-6$ since those are the highest point and the lowest point of the entire domain [−10, 10]. Now we want to look at the valleys and the peaks. When $x=-6$ and $1$, the curve forms the bottom of a valley, so the local minimums are −7 at $x=-6$ and $-2$ at $x=1$. At $x=-1$, the function $f$ has a local maximum of 4 since it is the top of a peak. Note that −7 at $x=-6$ is both an absolute minimum and a relative minimum. How about $f(x)=10$? It is neither the absolute extremum nor a relative extremum.

Now, let’s look at the following graph:

As you can see, at $x=-3$, the function $f$ has a local and absolute maximum of 7. However, there is no absolute minimum. There are some functions that have both absolute maximum and minimum extrema, while some may have only one or no extrema (for example, consider a linear function with a domain of all values).

How do we know if there are always both an absolute maximum and absolute minimum in a function? The extreme value theorem states that if a function $f$ is continuous on the domain of a closed interval [$a$, $b$], then there exist $\alpha$ and $\beta$ in the domain such that $f$ has an absolute maximum value $f(\alpha )$ and has an absolute minimum value $f(\beta )$. In order to find the absolute extrema of a function, we need to find the relative extrema of the function. Let’s look at the following graph:

The orange lines in the graph represent the tangent lines at relative extrema. They are all horizontal lines, which means that the slope of the tangent line at a local extrema is 0. Does this ring a bell? If we recall how to find the slope of a tangent line at ($a$, $f(a)$), we differentiate the function, and then plug in $x=a$ to find $m=f^{\prime }(x)$. In other words, the values of $x$ such that they make $m=f^{\prime }(x)$ could be the $x$ values at which $f(x)$ has relative extrema. Fermat’s theorem states that if $f$ has a local maximum or minimum at $a$, and if $f^{\prime }(a)$ exists, then $f^{\prime }(a)=0$.

What we have to be careful with about Fermat’s theorem is that the converse is not true. Even when $f^{\prime }(a)=0$, it does not mean that there is a maximum or minimum at $a$. In addition, there may be an extremum at $a$ even when $f^{\prime }(a)$ does not exist. Now, you might feel like we are going in a circle. What do we do to find the absolute extrema of a function then? The answer is that we need to find a critical number. A critical number of $f$ is a number $a$ in the domain $D$ of the function such that either $f^{\prime }(a)=0$ or $f^{\prime }(a)$ does not exist. While the existence of a critical number $a$ does not guarantee the existence of relative extrema at $x=a$, the existence of relative extrema at $a$ means that $a$ is a critical number of $f$.

In summary, the steps to find the absolute extrema of a continuous function $f$ in a closed interval [$a$, $b$] are as follows:

1. Find all the critical numbers of $f$ in the interval ($a$, $b$) by finding the zeros of $f^{\prime }(x)$.

2. Find the value of $f$ at all the critical numbers found in the previous step.

3. Find the value of $f$ at the endpoints of the interval, $a$ and $b$.

4. Compare the values of $f$ from steps 2 and 3. The largest value found in steps 2 and 3 is the absolute maximum value, and the smallest is the absolute minimum value of $f$ in the interval [$a$, $b$].

Example

Let’s practice with an example. Find the absolute maximum and absolute minimum of the function

$f(x)=2x^3-3x^2-36x+10$

in the closed interval [-6, 4].

Step 1: Find the critical numbers.

The derivative of the function $f$ is

$\begin{array}{rl}{f}^{\prime }(x)& =6x^2-6x-36\\ & =6(x^2-x-6)\\ & =6(x-3)(x+2)\end{array}$

Therefore, the critical numbers are $x=3$ and $x=-2$.

Step 2: Find the value of $f$ at the critical numbers.

Given the critical numbers, we have

$\begin{array}{rl}& f(x=3)=-107\\ & f(x=-2)=54\end{array}$

Step 3: Find the value of $f$ at the endpoints of the interval.

$\begin{array}{rl}& f(x=-6)=-314\\ & f(x=4)=-54\end{array}$

Step 4: Compare the values of $f$ from steps 2 and 3.

Given the values found in steps 2 and 3, the absolute maximum is 54 when $x=-2$, and the absolute minimum is −314 when $x=-6$.

We will discuss how to find relative extreme values in the next section.