4.1: Maximum and Minimum Values

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Throughout this module, if something does not exist, write DNE in the answer box.

Recap Video

Take a look at the following video which recaps the ideas from the section.

Example Video

Below is a video showing a worked example.

Problems

There are a lot of theorems and definitions in this section, so let’s recall all of them:

A critical point of a function $f(x)$ is a point $x=c$ in the domain of $f(x)$ where $f'(c) = 0$ or $f'(c)$ does not exist.

If $f(x)$ has a local maximum or minimum at $x=c$ , then either $f'(c) = 0$ or $f'(c)$ does not exist (i.e. $x=c$ is a critical point of $f(x)$ ).

If $f(x)$ is a continuous function on the closed interval $[a,b]$ , then $f(x)$ has an absolute maximum and an absolute minimum on the interval $[a,b]$ .

Consider the following graph:

How many critical points does this function have? $\answer {3}$ .

List the critical points in increasing order:

$x=\answer {-1}$ , and this point on the graph is a
.
$x=\answer {0}$ , and this point on the graph is a
.
$x=\answer {1}$ , and this point on the graph is a
.

Consider the function $f(x) = 2x^3-15x^2+24x-1$ . How many critical points does this function have? $\answer {2}$ .

In increasing order, the critical points are:

$x = \answer {1},\quad x = \answer {4}.$

Consider $f(x) = 3+(x-2)^{2/3}$ . Which of the following describes the full domain of $f(x)$ ?

How many critical points does $f(x)$ have? $\answer {1}$ .

The critical point is at $x = \answer {2}$ .

Here is the graph of $f(x)$ :

Based on this graph, the point $x=2$ is a local .

To find the absolute maximum and minimum values of a continuous function $f(x)$ on the closed interval $[a,b]$ :

Find the critical points of $f(x)$ on $(a,b)$ (i.e. inside the interval)
Plug in these critical points and endpoints into $f(x)$ to get the highest and lowest $y$ -values.

Find the absolute maximum and minimum values of $f(x) = x-\frac {3}{2}x^{2/3}$ on the interval $[0,8]$ .

First, note that $f(x)$ is continuous on the interval $[0,8]$ . Let’s break this into steps:

Step 1: We can find critical points inside the interval. Taking a derivative, we get $f'(x) = \answer {1-x^{-1/3}}.$ This derivative does not exist at $x=\answer {0}$ , but that is ok in this case because $x=0$ is one of the endpoints (and we only care right now about the interval $(0,8)$ ). Setting $f'(x) = 0$ , we get one critical point at $x= \answer {1}$ , which is inside the interval.
Step 2: We found $x=1$ is a critical point in the previous part. We also have $x=0$ and $x=8$ as endpoints. We plug these three points into $f(x)$ : $f(0) = \answer {0},\quad f(1) = \answer {-1/2},\quad f(8) = \answer {2}.$ This tells us that the maximum value is $\answer {2}$ at $x=\answer {8}$ , and the minimum value is $-1/2$ at $x=\answer {1}$ .

Find the absolute maximum and minimum values of the function $f(x) = 2x^3-15x^2+24x-10$ on the interval $[0,2]$ .

$\text {The maximum value is } \answer {1} \text { at } x = \answer {1}.$

$\text {The minimum value is } \answer {-10} \text { at } x = \answer {0}.$

Find the absolute maximum and minimum values of the function $f(x) = \sin ^2(x)+\cos (x)$ on the interval $[0,2\pi ]$ .

$\text {The maximum value is } \answer {5/4}.$

$\text {The minimum value is } \answer {-1}.$

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