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 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 $(0,2)$ . 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 attain a global extremum on its domain? Select the correct answer.

Check the following graph.

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) where the function $f$ has a global minimum. $x= \answer [given]{-1}, x= \answer [given]{1}$

Find the x-coordinate(s) of the point(s) 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 last three examples suggest the following theorem.

Extreme Value Theorem If $f$ is a continuous function for all $x$ in the closed interval $[a,b]$ , then there are points $c$ and $d$ in $[a,b]$ , such that $(c,f(c))$ is a global maximum and $(d,f(d))$ is a global minimum 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 and 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 and 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 point $x$ in the open interval $(a,b)$ , then $f$ has a local extremum at $x$ . That means that $f$ has a critical point at $x$ . So, the global extrema of a function $f$ occur either at the end points, $a$ or $b$ , or at critical points. If we want to locate the global extrema, we have to evaluate the function at the end points and at critical points, and compare the values.

Find the extreme values of $f(x) = x^3 - 3x^2 - 1$ on the interval $[-1,4]$ .

The derivative is $f'(x) = 3x^2 - 6x$ $\begin{align*} f'(x) &= 0\\ 3x^2 - 6x &= 0\\ 3x(x-2) &= 0 \end{align*}$

The critical points are $x=0$ and $x=2$ (both these lie in $(-1,4)$ ), with values $f(0) = -1$ and $f(2) = -5$ .

The endpoints have values $f(-1) = -5$ and $f(4) = 15$ .

That gives us a list of values $\{ -1, -5, 15\}$ .

The global maximum value of $f$ is $15$ (occurring at $x=4$ ) and a global minimum value of $-5$ (occurring at $x=-1$ and $x=2$ ).

Let $f(x)=x^2e^{-x}$ , for $-2\le x\le 1$ . Locate the global extremums of $f$ on the closed interval $[-2,1]$ . Does the function $f$ satisfy the conditions of the Extreme Value Theorem on its domain?

Therefore, the Extreme Value Theorem guarantees that the function $f$ attains both global extremums on its domain. The global extremums occur at the end points or at critical points.

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, find the location of the global minimum and global maximum of $f$ . Circle the correct answer.

Find the extreme values of $g(x) = \sin (x) - \cos (x)$ on $[-\pi , \pi ]$ .

To find critical points: $\begin{align*} g'(x) &= 0\\ \cos(x) + \sin(x) &= 0\\ \cos(x)\left( 1 + \frac{\sin(x)}{\cos(x)} \right) &= 0\\ \cos(x) \left( 1 + \tan(x) \right) &= 0 \end{align*}$

Notice that $x$ -values with $\cos (x) = 0$ have $\sin (x) = \pm 1$ , so $g'(x) = 0 \pm 1 \neq 0$ . These values are not critical points.

In the interval $[-\pi , \pi ]$ , the solutions of $\tan (x) = -1$ are $x = -\frac {\pi }{4}, \frac {3\pi }{4}$ . (The solutions not in $(-\pi , \pi )$ are ignored.)

$\begin{align*} g\left( -\frac{\pi}{4} \right) &= \sin\left( -\frac{\pi}{4} \right) - \cos\left( -\frac{\pi}{4} \right)\\ &= -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\\ &= \answer[given]{-\sqrt{2}}\\ \\ g\left( \frac{3\pi}{4} \right) &= \sin\left( \frac{3\pi}{4} \right) - \cos\left( \frac{3\pi}{4} \right)\\ &= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\\ &= \answer[given]{\sqrt{2}} \end{align*}$

The endpoints have values $g( -\pi ) = g(\pi ) = \answer [given]{1}$

The global maximum value of $g$ is $\answer [given]{\sqrt {2}}$ (occurring at $x=\answer [given]{\frac {3\pi }{4}}$ ) and a global minimum value of $\answer [given]{-\sqrt {2}}$ (occurring at $x=\answer [given]{\frac {\pi }{4} }$ ).

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