Math Core

What's the highest a roller coaster climbs, or the lowest cost a factory can reach? Questions like these ask for the largest or smallest value of a function. In this lesson you'll learn when such values are guaranteed to exist, and a short list of places where they can hide.

Absolute and relative extrema

Definition

Absolute and relative extrema

Let $f$ be defined on an interval containing $c$ .

$f(c)$ is an absolute maximum if $f(c) \ge f(x)$ for every $x$ in the interval. An absolute minimum has $f(c) \le f(x)$ for every $x$ .
$f(c)$ is a relative (local) maximum if $f(c) \ge f(x)$ for all $x$ in some open interval around $c$ . A relative minimum is defined the same way with $\le$ .

"Extremum" (plural "extrema") means a maximum or a minimum. An absolute extremum is the champion over the whole interval. A relative extremum only has to beat its neighbors, like a hilltop that isn't the tallest mountain.

When you report an extremum, keep the two parts straight: the value is the $y$ -coordinate $f(c)$ , and it occurs at $x = c$ . "The absolute maximum is $17$ , at $x = 4$ ."

The extreme value theorem

Not every function has a highest or lowest value. $f(x) = x^2$ on the open interval $(0, 1)$ gets closer and closer to $0$ and $1$ but never reaches either, so it has no absolute max or min there. $f(x) = \dfrac{1}{x}$ on $[-1, 1]$ shoots off to $\pm\infty$ . The extreme value theorem names the conditions that rule out these problems.

The extreme value theorem (EVT)

If $f$ is continuous on a closed interval $[a, b]$ , then $f$ has both an absolute maximum and an absolute minimum on $[a, b]$ .

Both conditions matter. An open interval lets the function sneak toward a value it never reaches. A discontinuity lets it jump over or blow up past a value. Like the mean value theorem, the EVT guarantees existence but doesn't tell you where the extrema are. For that you need critical points.

Critical points

Picture a smooth hilltop. The tangent line at the very top is horizontal, so $f' = 0$ there. At a sharp peak, like the top of $y = -\lvert x \rvert$ , the derivative doesn't exist. These are the only two ways a relative extremum can happen at an interior point.

Definition

Critical point

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

Where relative extrema live

If $f$ has a relative extremum at $x = c$ , then $c$ is a critical point of $f$ . (This is sometimes called Fermat's theorem.)

The converse is false: a critical point doesn't have to be an extremum. For $f(x) = x^3$ , $f'(0) = 0$ , but the graph keeps rising right through the origin. Critical points are candidates, not guarantees. The next lessons show how to tell which candidates really are maxima or minima.

Common mistake

A critical point must be in the domain of $f$ . For $f(x) = \dfrac{1}{x}$ , $f'(x) = -\dfrac{1}{x^2}$ is undefined at $x = 0$ , but $0$ is not a critical point because $f(0)$ doesn't exist. Always check the domain before listing a value where $f'$ is undefined.

Worked example: Finding critical points

Find the critical points of $f(x) = x^{2/3}(x - 5)$ .

Solution. Rewrite as $f(x) = x^{5/3} - 5x^{2/3}$ , whose domain is all real numbers. Then

f'(x) = \frac{5}{3}x^{2/3} - \frac{10}{3}x^{-1/3} = \frac{5}{3}x^{-1/3}(x - 2) = \frac{5(x - 2)}{3\sqrt[3]{x}}.

$f'(x) = 0$ when $x = 2$ . $f'(x)$ is undefined when $x = 0$ , and $0$ is in the domain of $f$ . So the critical points are $x = 0$ and $x = 2$ .

The candidates test

On a closed interval, an absolute extremum happens either at a relative extremum inside the interval (a critical point) or at an endpoint. That gives a simple procedure, sometimes called the closed interval method or candidates test.

Candidates test

To find the absolute extrema of a continuous function $f$ on $[a, b]$ :

Find every critical point of $f$ in $(a, b)$ .
Evaluate $f$ at each critical point and at both endpoints $a$ and $b$ .
The largest of these values is the absolute maximum; the smallest is the absolute minimum.

Worked example: Absolute extrema of a cubic

Find the absolute maximum and minimum values of $f(x) = x^3 - 3x^2 + 1$ on $[-1, 4]$ .

Solution. $f$ is a polynomial, so it's continuous on $[-1, 4]$ and the EVT guarantees both extrema exist. $f'(x) = 3x^2 - 6x = 3x(x - 2)$ , so the critical points are $x = 0$ and $x = 2$ , both in $(-1, 4)$ .

$x$	$-1$	$0$	$2$	$4$
$f(x)$	$-3$	$1$	$-3$	$17$

The absolute maximum is $17$ , at $x = 4$ . The absolute minimum is $-3$ , and it occurs at both $x = -1$ and $x = 2$ .

f(x) = x³ − 3x² + 1 on [−1, 4] (solid) and f′(x) = 3x² − 6x (dashed). The relative extrema sit exactly where f′ crosses zero.Open in grapher →

Notice in the graph that the relative maximum at $(0, 1)$ is not the absolute maximum. The right endpoint wins.

Worked example: A trig function

Find the absolute extrema of $f(x) = x - 2\sin x$ on $[0, \pi]$ .

Solution. $f'(x) = 1 - 2\cos x$ . Set it equal to $0$ : $\cos x = \dfrac{1}{2}$ , so $x = \dfrac{\pi}{3}$ is the only critical point in $(0, \pi)$ . Evaluate:

f(0) = 0, \qquad f\!\left(\frac{\pi}{3}\right) = \frac{\pi}{3} - \sqrt{3} \approx -0.685, \qquad f(\pi) = \pi \approx 3.142.

The absolute maximum is $\pi$ , at $x = \pi$ . The absolute minimum is $\dfrac{\pi}{3} - \sqrt{3}$ , at $x = \dfrac{\pi}{3}$ .

Tip

Organize your work in a table of candidates and their $f$ -values, as in the examples. On the AP exam, a candidates table plus the sentence "the largest value is…" is a complete justification for an absolute extremum.

Practice

Practice 1

Find all critical points of $f(x) = 2x^3 - 3x^2 - 12x + 5$ .

Separate answers with commas, e.g. 2, -5

Practice 2

Find the absolute maximum value of $f(x) = x^2 - 6x + 2$ on $[0, 5]$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Find the absolute minimum value of $f(x) = x^3 - 12x$ on $[-3, 5]$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

For which function does the extreme value theorem guarantee an absolute maximum and an absolute minimum on the given interval?

Practice 5

Find all critical points of $f(x) = x^{1/3}(x + 4)$ .

Separate answers with commas, e.g. 2, -5

Practice 6

Find the absolute maximum value of $f(x) = xe^{-x}$ on $[0, 4]$ . Give the exact value or a decimal rounded to three places.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

A function $g$ is continuous on $[0, 6]$ and differentiable on $(0, 6)$ . Its only critical points are $x = 2$ and $x = 5$ , and

g(0) = 4, \quad g(2) = -1, \quad g(5) = 7, \quad g(6) = 3.

Which statement is true?