6.4 Extreme values of functions

In many practical problems, we need to find the greatest (maximum) value or the least (minimum) value—there can be more than one of each—of a function.
The maximum and minimum values of a function are called the extreme values or extrema of the function.
Extremum is the singular form of extrema. The plural forms of maximum and minimum are maxima and minima, respectively.
Differentiation can help us locate the extreme values of a function.

In calculus, there are two types of “maximum” and “minimum,” which are distinguished by the two prefixes: absolute and relative.

Table of Contents

Absolute Maxima and Minima

The concepts of absolute maximum and minimum were introduced in Chapter 4. Let’s review the definitions.

Let the function $f$ be defined on a set $E$ . We say $f$ has an absolute maximum on $E$ at a point $p$ if $f (x) \leq f (p) for all x in E,$ and an absolute minimum value on $E$ at $q$ if
$f (x) \geq f (q) for all x in E .$

Absolute maxima and absolute minima (plural forms of maximum and minimum) are also referred to as global maxima and global minima.

Previously, we learned that:

Theorem 1. The Extreme Value Theorem: If $f$ is continuous on a closed interval $[a, b]$ , then $f$ attains both its absolute maximum $M$ and absolute minimum $m$ in $[a, b]$ . That is, there are numbers $p$ and $q$ in $[a, b]$ such that $f (p) = M$ and $f (q) = m$ .

We should emphasize that:

The continuity of the function on an open interval (instead of a closed interval) is not sufficient to guarantee the existence of the absolute maximum and minimum of the function.
If the function fails to be continuous even at one point in the interval $[a, b]$ , the extreme value theorem may fail to be true (although a discontinuous function may have max and min).

For more information, see the Section on the Extreme Value Theorem.

Local (or Relative) Maxima and Minima

Figure 1. Graph of a function

y = f (x)

Figure 1 shows the graph of function

f

with absolute maximum at

x = b

and absolute minimum at

x = p

. The point

(c, f (c))

is higher than all nearby points on the curve, although it is not as high as

(b, f (b))

. That is, if we consider only values of

x

sufficiently close to

c

, then

f (c)

is the largest value of those values of

f

(Figure 2). In this case, we say

f

has a local (or relative) maximum at

x = c

Figure 2. A zoomed in section of the graph

y = f (x)

shown in the previous figure. The function has a local maximum at

x = c

because we can consider an open interval

I = (x_{1}, x_{2})

around

c

such that

f (x) \leq f (c)

for all

x

I

Similarly, we say

f

has a local (or relative) minimum at

x = d

because

f (x) \geq f (d)

for all

x

very close to

d

(Figure 1). Other local maxima and minima are denoted on Figure 3.

Geometrically speaking local maxima and local minima are respectively the “peaks” and “valleys” of the curve.

Figure 3. Local maxima and local minima of a function are the peaks and valleys of its graph. The precise definitions are as follows.

Definition 1. A function $f$ is said to have a local (or relative) maximum at a point $c$ within its domain $D$ if there is some open interval $I$ containing $c$ such that $f (x) \leq f (c) for all x \in I .$ The concept of local (or relative) minimum is similarly defined by reversing the inequality.

Every absolute maximum or minimum that is not an endpoint of an interval is a local maximum or local minimum, respectively. An endpoint is precluded from being a local extremum because we cannot find an open interval around an endpoint that is contained in the domain of the function.

Example 1

The graph of a function

f

is illustrated in Figure 4. Find its absolute and local extrema.

Figure 4.

Solution

The lowest point of the graph is

(x_{3}, f (x_{3}))

, and therefore, the function has an absolute minimum at

x_{3}

. Because

x_{3}

is an interior point of the interval

[a, b]

f (x_{3})

is also a local minimum. The absolute maximum occurs at

b

, but because

b

is an endpoint and the function is not defined on its right side,

f (b)

is not a local maximum. The other local minima occur at

x_{1}

and

x_{5}

Because $x_{2} \notin D o m (f)$ , the function cannot have a maximum there. Moreover, it is evident that $f (x_{4})$ is a local maximum. We claim that $f (x_{6})$ is a local maximum because if we zoom in, we realize that for all $x$ close enough to $x_{6}$ ,
$f (x) \leq f (x_{6}) .$

Figure 5.

All the absolute and local extrema are shown in Figure 6.

Figure 6.

Example 2

The graph of a function

f

is illustrated in the following figure. Specify where its extrema occur.

Figure 7.

Solution

Because there is no greatest and no least value, the function does not have an absolute maximum or absolute minimum. For every

x_{0} \in (a, b)

, there is some neighborhood

I

x_{0}

that is completely contained in

(a, b)

(Figure 8(a)), and for each

x \in I

, we have

f (x) \leq f (x_{0})

and

f (x) \geq f (x_{0})

(because

f (x) = f (x_{0})

).¹ Therefore, the function has at every

x

(a, b)

a local maximum and a local minimum. The function has a local maximum at

x = a

and local minimum at

x = b

(Figure 8(b)).

Figure 8.

It is evident from Figure 9 that at a local extremum, the tangent line is either parallel to the

x

-axis (slope = 0) or has no tangent line. The following theorem helps us locate all the possible values of

c

for which there is a local extremum.

Figure 9. If a function has a local extremum at an interior point

x_{0}

, then either its derivative vanishes there

f^{'} (x_{0}) = 0

or its derivative does not exist (including

f^{'} (x_{0}) = + \infty

- \infty

Theorem 2. (Fermat’s Theorem): Suppose

f

is a function that is defined on an open interval containing the point

c

. If

f (c)

is a local maximum or minimum, then either

f

is not differentiable at

c

(meaning

f^{'} (c)

does not exist) or

f^{'} (c) = 0

Notice that differentiability, or even continuity, of $f$ at other points is not required.
The geometrical interpretation of the above theorem is: At a local max or min, $f$ either has no tangent, or $f$ has a horizontal tangent.

Show the proof

We shall give the proof for the case of a local minimum at $x = c$ . According to the definition, we have $f (c) \leq f (c + h)$ or $0 \leq f (c + h) - f (c)$ for all $h$ sufficiently close to zero (that is, when $c + h$ is near $c$ ). If $f^{'} (c)$ does not exist, there is nothing else to prove. So suppose $f^{'} (c) = lim_{h \to 0} \frac{f (c + h) - f (c)}{h}$ exists as a definite number. We need to show $f^{'} (c) = 0$ . When $h$ is small, we have $\frac{f (c + h) - f (c)}{h} \geq 0 if h > 0$ and $\frac{f (c + h) - f (c)}{h} \leq 0 if h < 0$ because the numerator in both cases is either positive or zero ( $f (c + h) - f (c) \geq 0$ ). If we let $h \to 0^{+}$ , from the first case, we have
$f^{'} (c) \geq 0,$ and if we let $h \to 0^{-}$ , from the second case, we have $f^{'} (c) \leq 0.$ Because we have assumed that $f^{'} (c)$ exists, we must have the same limit in both cases, so $0 \leq f^{'} (c) \leq 0.$ This can happen only when $f^{'} (c) = 0$ . The proof for the case of a local maximum is similar.

The above theorem states a necessary condition for a local extremum. That the condition is not sufficient is evident from a glance at the point $(r, f (r))$ in Figure 9. The graph of $f$ has a horizontal tangent at this point, but $f$ does not have an extreme value at $x = r$ . As another example, consider: $f (x) = x^{3}$
$f (x) = x^{3} \Rightarrow f^{'} (x) = 3 x^{2}$
$f^{'} (0) = 0$ but $x = 0$ does not give either a local maximum or a local minimum of $f$ , as is obvious from the graph of $y = x^{3}$ (Figure 10(a)). If $g (x) = \sqrt[3]{x}$ , then
$g (x) = x^{1 / 3} \Rightarrow g^{'} (x) = \frac{1}{3} x^{1 / 3 - 1} = \frac{1}{3} x^{- 2 / 3} = \frac{1}{3 \sqrt[3]{x^{2}}}$
and $g^{'} (0)$ is not defined (we may say $g^{'} (0) = + \infty$ ), but $g (0) = 0$ is not a local extremum (Figure 10(b)).

Figure 10. Although at

x = 0

(a) the derivative of

y = x^{3}

is zero and (b) the derivative of

y = \sqrt[3]{x}

does not exist,

x = 0

does not give a local maximum or a local minimum.

Critical Points

A number in the domain of the function at which the derivative is zero or the derivative does not exist has a special name. It is called a critical number.

Definition 2. Critical point: A point $c$ in the domain of a function $f$ is called a critical point (or critical number) of $f$ if $f^{'} (c) = 0 or f^{'} (c) does not exist.$
The number $f (c)$ is called a critical value of $f$ .

Recall that if $f^{'} (c) = + \infty$ or $f^{'} (c) = - \infty$ , we say $f^{'} (c)$ does not exist because $+ \infty$ and $- \infty$ are not numbers.

By the above definition, we can reword Fermat’s theorem as:

Fermat’s Theorem: If $f (c)$ is a local maximum or a local minimum, then $x = c$ is a critical number of $f$ .

According to the above theorem, every single local extreme value is a critical value, but not every critical value is necessarily a local extreme value.

We mentioned that every absolute extreme value, with the exception of an absolute extreme value that occurs at an endpoint, is also a local extreme value. Hence:
An absolute maximum or minimum of a function occurs either at a critical point or at an endpoint of its domain.

This provides us a method to find the absolute maximum and the absolute minimum of a differentiable function on a finite closed interval $[a, b]$ .

Strategy for finding absolute extrema of a continuous function $f$ on a finite closed interval $[a, b]$ :

Step 1: Find $f^{'} (x)$
Step 2: Find all critical values: Set $f^{'} (x) = 0$ and solve it for $x$ . Also find every value of $x$ for which $f^{'} (x)$ does not exist. Evaluate $f$ at each of these numbers that lie between $a$ and $b$ .
Step 3: Evaluate $f (a)$ and $f (b)$ .
Step 4: The largest value of $f$ from Steps 2 and 3 is the absolute maximum of $f$ and the least value of $f$ from these steps is the absolute minimum of $f$ on $[a, b]$ .

Example 3

Find the absolute maximum and minimum value of the function

f (x) = \frac{1}{3} x^{3} - 4 x

on the interval

[- 3, 4]

Solution

Step 1: Finding the derivative of $f$

f (x) = \frac{1}{3} x^{3} - 4 x \Rightarrow f^{'} (x) = x^{2} - 4

Step 2: Finding the critical values of $f$ . The function is differentiable everywhere, so all the critical points are obtained by setting

f^{'} (x) = 0

f^{'} (x) = x^{2} - 4 = 0

x^{2} = 4

x = \pm 2

Because both

x = 2

and

x = - 2

lie between

- 3

and

4

, we evaluate

f

at both of these numbers

f (2) = \frac{1}{3} (2^{3}) - 4 (2) = - \frac{16}{3} \approx - 5.333

f (- 2) = \frac{1}{3} (- 2)^{3} - 4 (- 2) = \frac{16}{3} \approx 5.333

Step 3:
Evaluating $f$ at the endpoints

f (- 3) = \frac{1}{3} (- 3)^{3} - 4 (- 3) = 3

f (4) = \frac{1}{3} (4^{3}) - 4 (4) = \frac{16}{3} \approx 5.33

Step 4: Comparing the critical values and the endpoint values.

$\large x$	$- 3$	$- 2$	$2$	$4$
$f (x)$	$3$	$\frac{16}{3}$	$- \frac{16}{3}$	$\frac{16}{3}$
		max	min	max

The absolute maximum of $f$ on $[- 3, 4]$ is $16 / 3$ , which occurs at $x = - 2$ and $x = 4$ , and its absolute minimum on this interval is $- 16 / 3$ , which occurs at $x = - 2$ . The graph of $f$ is shown in Figure 11.

Figure 11. Graph of

y = \frac{1}{3} x^{3} - 4 x

for

- 3 \leq x \leq 4

Example 4

Find the extrema of

f (x) = x - 3 (x - 1)^{2 / 3}

[- 1, 2]

Solution

Step 1: Finding the derivative of

f

f (x) = x - 3 (x - 1)^{2 / 3} \Rightarrow f^{'} (x) = 1 - 3 \cdot \frac{2}{3} (x - 1)^{- 1 / 3}

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

Step 2: Finding the critical values of

f

f^{'} (x) = 0

1 - \frac{2}{\sqrt[3]{x - 1}} = 0

\frac{1}{\sqrt[3]{x - 1}} = \frac{1}{2}

\sqrt[3]{x - 1} = 2

x - 1 = 2^{3} = 8

x = 9

But

x = 9

does not lie between

- 1

and

2

.
We notice that

f^{'} (x)

does not exists when

x = 1

. Therefore,

x = 1

is another critical point of

f

, and

f (1) = 1 - 3 (1 - 1)^{2 / 3} = 1.

Step 3: Evaluating

f

at the endpoints of the given interval

\begin{aligned} f (- 1) & = - 1 - 3 (- 2)^{2 / 3} \\ = - 1 - 3 \sqrt[3]{4} \approx - 5.76 \end{aligned}

\begin{aligned} f (2) & = 2 - 3 (2 - 1)^{2 / 3} \\ = 2 - 3 \\ = - 1 \end{aligned}

Step 4: Comparing the critical values and the endpoint values.

$\large x$	$- 1$	$1$	$2$
$f (x)$	$- 1 - 3 \sqrt[3]{4}$	$1$	$- 1$
	min	max

Thus the absolute max of $f$ on $[- 1, 2]$ is $1$ , which occurs at $x = 1$ and the absolute min of $f$ on that interval is $1 - 3 \sqrt[3]{4}$ , which occurs at $x = - 1$ . The graph of $f$ is shown in Figure 12.

Figure 12. Graph of

y = x - 3 (x - 1)^{2 / 3}

when

- 1 \leq x \leq 2

Example 5

Find the absolute maximum and the absolute minimum of

f (x) = \sin 2 x + 2 \cos x

Solution

The domain of

f

(- \infty, \infty)

, but because

f

is periodic, we can examine it for one period. The fundamental period of

\sin 2 x

\frac{2 π}{2} = π

and the fundamental period of

\cos x

2 π

(Figure 13). Hence the fundamental period of

f

2 π

and we can find the absolute max and min on

[0, 2 π]

(or

[- π, π]

Figure 13. The function $f (x) = \sin 2 x + 2 \cos x$ is periodic. Its fundamental period is $2 π$ .

Step 1: $f (x) = \sin 2 x + 2 \cos x \Rightarrow f^{'} (x) = 2 \cos 2 x - 2 \sin x$ Step 2: $f^{'} (x) = 2 \cos 2 x - 2 \sin x = 0$ We can express $\cos 2 x$ in terms of $\sin x$ : $\cos 2 x = 1 - 2 \sin^{2} x$ (see the section on Trigonometric Identities). Thus $f^{'} (x) = 2 (1 - \sin^{2} x) - 2 \sin x = 0$ $- 4 \sin^{2} x - 2 \sin x + 2 = 0$ This is a quadratic equation in terms of $\sin x$ . Let $u = \sin x$ . Thus
$- 4 u^{2} - 2 u + 2 = 0$
$\Rightarrow u = \frac{2 \pm \sqrt{2^{2} - 4 (- 4) 2}}{2 (- 4)} = \frac{2 \pm \sqrt{36}}{- 8}$
$u = \frac{1}{2}, u = - 1$ We have to solve $\sin x = \frac{1}{2}, \sin x = - 1$ when $0 \leq x \leq 2 π$ .
$\sin x = - \frac{1}{2} \Rightarrow x = \frac{π}{6}, x = π - \frac{π}{6} = \frac{5 π}{6}$
and $\sin x = - 1 \Rightarrow x = \frac{3 π}{2} .$ (See Figure 14(a,b))

Figure 14

Thus the critical points (or critical numbers) are
$x = \frac{π}{6}, x = \frac{5 π}{6}, x = \frac{3 π}{2} .$ Let’s evaluate $f$ at these points.
$\begin{aligned} f (\frac{π}{6}) & = \sin \frac{π}{3} + 2 \cos \frac{π}{6} \\ = \frac{\sqrt{3}}{2} + 2 (\frac{\sqrt{3}}{2}) \\ = \frac{3 \sqrt{3}}{2} . \end{aligned}$
$\begin{aligned} f (\frac{5 π}{6}) & = \sin \frac{5 π}{3} + 2 \cos \frac{5 π}{6} \\ = \sin (2 π - \frac{π}{3}) + 2 \cos (π - \frac{π}{6}) \\ = \sin (- \frac{π}{3}) - 2 \cos \frac{π}{6} \\ = - \sin \frac{π}{3} - 2 \cos \frac{π}{6} \\ = - \frac{\sqrt{3}}{3} - 2 \frac{\sqrt{3}}{2} \\ = - \frac{3 \sqrt{3}}{2} . \end{aligned}$
Here we have used the identities $\sin (2 π + θ) = \sin θ$ and $\cos (π - θ) = - \cos θ$ (see Trigonometric Identities).
$\begin{aligned} f (\frac{3 π}{2}) & = \sin 3 π + 2 \cos \frac{3 π}{2} \\ = 0 + 0 \\ = 0. \end{aligned}$
Step 3: Evaluation of $f$ at the endpoints
$f (0) = \sin (2 \cdot 0) + 2 \cos 0 = 0 + 2 (1) = 2$

$f (2 π) = \sin (4 π) + 2 \cos (2 π) = 0 + 2 (1) = 2.$ Step 4:

$\large x$	$\large 0$	$\large \frac{\pi}{6}$	$\large \frac{5\pi}{6}$	$\large \frac{3\pi}{2}$	$\large 2\pi$
$f (x)$	$2$	$\frac{3 \sqrt{3}}{2} \approx 2.598$	$- \frac{3 \sqrt{3}}{2} \approx - 2.598$	$0$	$2$
		max	min

The above table shows that the maximum value of $f$ is $3 \sqrt{3} / 2$ and its absolute minimum is $- 3 \sqrt{3} / 2$ , which occur at $x = π / 6$ and $x = 5 π / 6$ , respectively.

Figure 15. In one period $f (x) = \sin 2 x + 2 \cos x$ takes on its absolute maximum at $x = π / 6$ and its absolute minimum at $x = 5 π / 6$ .

¹ $x \leq y$ means $x < y$ or $x = y .$ , so we can write, for example, $2 \leq 2.$ Here $f (x) = f (x_{0})$ for all $x$ in $I$ , and therefore we can write $f (x) \leq f (x_{0})$ or $f (x) \geq f (x_{0}) .$

Absolute Maxima and Minima

Local (or Relative) Maxima and Minima

Show the proof

Hide the proof

Critical Points