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.

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.

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?

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 $[a,b]$ ?

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?

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.

2025-01-06 20:01:11

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