The Extreme Value Theorem

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We examine a fact about continuous functions.

(a): A function $f$ has a global maximum at $x=a$ , if $f(a)\ge f(x)$ for every $x$ in the domain of the function.
(b): A function $f$ has a global minimum at $x=a$ , if $f(a)\le f(x)$ for every $x$ in the domain of the function.

A global extremum is either a global maximum or a global minimum.

Let $f$ be the function given by the graph below.

Find the $x$ -coordinate of the point corresponding to where the function $f$ has a global maximum. $x= \answer [given]{2}$

Observe that $f(2)\ge f(x)$ for all $x$ in the domain of $f$ . Notice that the function $f$ has also a local maximum at $x=2$ .

Find the $x$ -coordinate of the point where the function $f$ has a global minimum. $x= \answer [given]{5}$ Observe that $f(5)\le f(x)$ for all $x$ in the domain of $f$ . Notice that the function $f$ does not have a local minimum at $x=5$ . Recall, a function cannot not have a local extremum at a boundary point.

Find the $x$ -coordinate(s) of the point(s) where the function $f$ has a local minimum. $x= \answer [given]{1}$ Observe that $f(1)\le f(x)$ for all $x$ in the interval, say, $(0,2)$ . But it is not true that $f(1)\le f(x)$ for all $x$ in the domain of $f$ . For example, $f(4.5)<f(1)$ .

Does every function attain a global extremum on its domain? Select the correct answer.

All functions must attain both global minimum and global maximum on their domain. All functions attain a global minimum on their domain. All functions attain a global maximum on their domain. Some functions have no global extremums on their domain.

Check the following graph.

Notice, the function is not continuous at $x=1$ , and, therefore, $f$ is not continuous on its domain, $(-1,5)$ .
Does the function given by the graph above attain a global extremum on its domain? Select the correct answer.

The function attains both global minimum and global maximum on its domain. The function attains a global minimum, but has no global maximum on its domain. The function attains a global maximum, but has no global minimum on its domain. The function has no global extremum on its domain.

Check the graph below.

Notice, the function is continuous on its domain $(-1,5)$ . Does the function given by the graph above attain a global extremum on its domain? Select the correct answer.

Check the following graph.

Notice, the function is continuous on a closed interval $[-1,5]$ . Does the function given by the graph above attain a global extremum on its domain? Select the correct answer.

Find the x-coordinate(s) of the point(s) corresponding to where the function $f$ has a global minimum. $x= \answer [given]{-1},\hspace {0.2in} x= \answer [given]{1}$

Find the x-coordinate(s) of the point(s) corresponding to where the function $f$ has a global maximum. $x= \answer [given]{5}$

Sometimes it is important to know whether a function attains a global extremum on its domain. The following theorem, which comes as no surprise after the previous three examples, gives a simple answer to that question.

Extreme Value Theorem If $f$ is continuous on the closed interval $[a,b]$ , then there are numbers $c$ and $d$ in $[a,b]$ , such that $f(c)$ is a global maximum of $f$ and $f(d)$ is a global minimum of $f$ on $[a, b]$ .

Below, we see a geometric interpretation of this theorem.

Would this theorem hold if we were working on an open interval?

yes no

Consider $\tan (\theta )$ for $-\pi /2 < \theta < \pi /2$ . Does this function achieve its maximum or minimum?

Would this theorem hold if we were working on a closed interval $[a,b]$ , but a function is not continuous on $[a,b]$ ?

yes no

Consider a function $f$ on a closed interval $[-1,1]$ , defined by $f(x)=\frac {1}{x}$ for $x\neq 0$ and $f(0)=0$ . Does this function achieve its maximum or minimum?

Assume that a function $f$ is continuous on a closed interval $[a,b]$ . By the Extreme Value Theorem, $f$ attains both global extremums on the interval $[a,b]$ . How can we locate these global extrema? We have seen that they can occur at the end points or in the open interval $(a,b)$ . If a global extremum occurs at a number $x$ in the open interval $(a,b)$ , then $f$ has a local extremum at $x$ . That means that $f$ has a critical number at $x$ . So, the global extrema of a function $f$ occur either at the end points, $a$ or $b$ , or at critical numbers. If we want to locate the global extrema, we have to evaluate the function at the end points and at critical numbers, and compare the values.

Let $f(x)=x^2e^{-x}$ , for $-2\le x\le 1$ .

(a): Does the function $f$ satisfy the conditions of the Extreme Value Theorem on its domain?
Yes, because $f$ is continuous on $(-2,1)$ . Yes, because $f$ is continuous on $[-2,1]$ . No, because $f$ is not continuous on $[-2,1]$ . No, because $f$ is not differentiable on $(-2,1)$ .
Now we know that the Extreme Value Theorem guarantees that the function $f$ attains both global extremums on its domain!
(b): Locate the global extremums of $f$ on the closed interval $[-2,1]$ .

The global extremums occur at the end points or at critical points.
Let’s find the critical points of $f$ . First, compute the derivative of $f$ . $f'(x)= xe^{-x}(\answer [given]{2-x})$ In order to find the critical points of $f$ , we have to solve the equation $f'(x)= \answer [given]{0}.$ It follows that the function $f$ has only one critical point $\left (c,f(c)\right )$ . Find $c$ . $c= \answer [given]{0}$ In order to locate the global extremums of $f$ , we have to evaluate $f$ at the end points and at the critical point. $f(-2)= \answer [given]{4e^{2}}$ $f(c)= \answer [given]{0}$
$f(1)= \answer [given]{e^{-1}}$ Order the three values, $f(-2)$ , $f(c)$ , and $f(1)$ , from smallest to largest. You should replace $c$ with its value, when you write $f(c)$ in your answer below. $f(\answer [given]{0})<f(\answer [given]{1})<f(\answer [given]{-2})$
Based on this comparison, the function $f$ has the global minimum at $x=\answer [given]{0}$ , and the global maximum at $x=\answer [given]{-2}$ .

For your convenience, the graph of $f$ on the interval $[-2,1]$ is given below.