Extreme Values of Functions

The maximum and minimum values of a function are collectively called its extreme values. Fermat's Theorem tells us where to look: every local extreme value occurs at a critical point. On a closed interval, we also check the endpoints.

Concept	Meaning
Absolute maximum at $p$	$f (x) \leq f (p)$ for all $x$ in the domain
Absolute minimum at $q$	$f (x) \geq f (q)$ for all $x$ in the domain
Local maximum at $c$	$f (x) \leq f (c)$ for all $x$ near $c$
Local minimum at $c$	$f (x) \geq f (c)$ for all $x$ near $c$
Critical point	f'(c)=0 or f'(c) does not exist

Overview

In many practical problems, we need to find the greatest (maximum) value or the least (minimum) value of a function. The maximum and minimum values of a function are collectively called the extreme values or extrema of the function. (Extremum is the singular form; maxima and minima are the plural forms of maximum and minimum.)

There are two types of extrema, distinguished by the prefixes absolute and local (or relative).

Absolute Maxima and Minima

Definition 1.

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 are also referred to as global maxima and global minima.

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 $f (q) = m$ .

Important remarks:

Continuity on an open interval is not sufficient to guarantee the existence of absolute extrema.
If the function fails to be continuous even at one point in $[a, b]$ , the Extreme Value Theorem may fail (although a discontinuous function may still have extrema).

Local (Relative) Maxima and Minima

Illustration for Extreme Values of Functions — **Figure 1** Graph of a function $y = f (x)$ showing absolute maximum at $x = b$ and absolute minimum at an interior point.

A point may be higher than all nearby points without being the global maximum. Such a point is called a local maximum.

Definition 2.

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.

Geometrically speaking, local maxima and local minima are respectively the peaks and the bottoms of trenches of the curve.

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

Example 1.

The graph of a function $f$ is illustrated in the following figure. Find its absolute and local extrema.

Graph of f showing various labeled x-values where extrema occur

Solution

The lowest point of the graph is

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

, so

f

has an absolute minimum at

x_{3}

. Because

x_{3}

is an interior point,

f (x_{3})

is also a local minimum. The absolute maximum occurs at

b

, but because

b

is an endpoint and

f

is not defined on its right side,

f (b)

is not a local maximum. Other local minima occur at

x_{1}

and

x_{5}

. Because

x_{2} \notin Dom (f)

, the function cannot have a maximum there. It is evident that

f (x_{4})

is a local maximum. We claim

f (x_{6})

is a local maximum because for all

x

close enough to

x_{6}

f (x) \leq f (x_{6})

Zoomed view near x_6 confirming local maximum

Graph of f with all absolute and local extrema labeled — All absolute and local extrema labeled on the graph of $f$ .

Example 2.

The graph of a function $f$ is illustrated in the following figure. Specify where its extrema occur.

Graph of a constant-valued function on (a,b)

Solution

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

x_{0} \in (a, b)

, there is some neighborhood

I

completely contained in

(a, b)

, 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})

on that interval). Therefore, the function has a local maximum and a local minimum at every

x

(a, b)

. The function has a local maximum at

x = a

and a local minimum at

x = b

Two panels showing that the constant function has local maxima at a and local minima at b

Fermat's Theorem and Critical Points

At a local extremum, the tangent line is either horizontal (slope $= 0$ ) or the function has no tangent line at that point.

Theorem 2

Fermat's Theorem. Suppose $f$ is a function defined on an open interval containing the point $c$ . If $f (c)$ is a local maximum or minimum, then either f'(c) does not exist or f'(c)=0.

Proof

We prove the case of a local minimum at

x = c

. By definition:

f (c) \leq f (c + h)

for all

h

sufficiently close to zero. If f'(c) does not exist, there is nothing more to prove. So suppose f'(c)=\lim_{h\to0}\frac{f(c+h)-f(c)}{h} exists as a definite number. We need to show f'(c)=0. When

h

is small,

f (c + h) - f (c) \geq 0

. Therefore:

\frac{f (c + h) - f (c)}{h} \geq 0 if h > 0

\frac{f (c + h) - f (c)}{h} \leq 0 if h < 0.

Letting

h \to 0^{+}

gives f'(c)\geq0, and letting

h \to 0^{-}

gives f'(c)\leq0. Since the limit must be the same from both sides, 0\leq f'(c)\leq0, so f'(c)=0. The proof for a local maximum is similar.

Equivalently: if f'(c) exists and f'(c)\neq0, then $f (c)$ is not a local maximum or minimum.

If f'(c) exists and f'(c)\neq 0, then $f (c)$ is not a local maximum or minimum.

Fermat's Theorem states a necessary condition for a local extremum, it is not sufficient. For example:

$f (x) = x^{3}$ has f'(0)=0, but $x = 0$ is neither a local max nor a local min.
$g (x) = \sqrt[3]{x}$ has g'(0) undefined, but $g (0) = 0$ is not a local extremum.

Critical Points

Definition 3.

A point $c$ in the domain of a function $f$ is called a critical point (or critical number) of $f$ if

f'(c)=0\quad\text{or}\quad f'(c)\text{ does not exist.}

The number $f (c)$ is called a critical value of $f$ .

Note: if f'(c)=+\infty or f'(c)=-\infty, we say f'(c) does not exist because $\pm \infty$ are not numbers.

Theorem 3

Fermat's Theorem (reworded). If $f (c)$ is a local maximum or a local minimum, then $x = c$ is a critical point of $f$ .

Every local extreme value is a critical value, but not every critical value is a local extreme value.

Since every absolute extreme value (except at endpoints) is also a local extreme value:

An absolute maximum or minimum of a function occurs either at a critical point or at an endpoint of its domain.

Strategy for Finding Absolute Extrema

Strategy for finding the 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 for $x$ . Also find every $x$ for which f'(x) does not exist. Evaluate $f$ at each of these numbers that lie in $[a, b]$ .
Step 3: Evaluate $f (a)$ and $f (b)$ .
Step 4: The largest value from Steps 2 and 3 is the absolute maximum of $f$ ; the smallest is the absolute minimum.

Examples

Example 3.

Find the absolute maximum and minimum of $f (x) = \frac{1}{3} x^{3} - 4 x$ on $[- 3, 4]$ .

Solution

**Step 1:** Finding the derivative. f(x)=\frac{1}{3}x^{3}-4x\Rightarrow f'(x)=x^{2}-4. **Step 2:** Finding the critical values. The function is differentiable everywhere, so critical points come from f'(x)=0:

x^{2} - 4 = 0 \Rightarrow x = \pm 2.

Both

x = 2

and

x = - 2

lie in

[- 3, 4]

f (2) = \frac{1}{3} (8) - 8 = - \frac{16}{3} \approx - 5.333

f (- 2) = \frac{1}{3} (- 8) + 8 = \frac{16}{3} \approx 5.333 .

**Step 3:** Evaluating at the endpoints:

f (- 3) = \frac{1}{3} (- 27) + 12 = 3

f (4) = \frac{1}{3} (64) - 16 = \frac{16}{3} \approx 5.33 .

**Step 4:** Comparing values: |

x

- 3

- 2

| $2$ | $4$ | |-----|------|------|-----|-----| |

f (x)

| $3$ |

\frac{16}{3}

- \frac{16}{3}

\frac{16}{3}

| | | | max | min | max | The absolute maximum is

16 / 3

, occurring at

x = - 2

and

x = 4

. The absolute minimum is

- 16 / 3

, occurring at

x = 2

Graph of y = (1/3)x^3 - 4x on [-3, 4] showing the absolute extrema — Graph of $y = \frac{1}{3} x^{3} - 4 x$ for $- 3 \leq x \leq 4$ .

Example 4.

Find the extrema of $f (x) = x - 3 (x - 1)^{2 / 3}$ on $[- 1, 2]$ .

Solution

**Step 1:** Finding the derivative: f'(x)=1-3\cdot\frac{2}{3}(x-1)^{-1/3}=1-\frac{2}{\sqrt[3]{x-1}}. **Step 2:** Finding critical values. Setting f'(x)=0:

1 - \frac{2}{\sqrt[3]{x - 1}} = 0 \Rightarrow \sqrt[3]{x - 1} = 2 \Rightarrow x - 1 = 8 \Rightarrow x = 9.

But

x = 9

does not lie in

[- 1, 2]

. Also, f'(x) does not exist when

x = 1

. So

x = 1

is a critical point with

f (1) = 1 - 0 = 1

. **Step 3:** Evaluating at the endpoints:

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

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

**Step 4:** Comparing values: |

x

- 1

| $1$ | $2$ | |-----|------|-----|-----| |

f (x)

- 1 - 3 \sqrt[3]{4}

| $1$ |

- 1

| | | min | max | | The absolute maximum is $1$ at

x = 1

, and the absolute minimum is

- 1 - 3 \sqrt[3]{4}

x = - 1

Graph of y = x - 3(x-1)^(2/3) on [-1, 2] — Graph of $y = x - 3 (x - 1)^{2 / 3}$ for $- 1 \leq x \leq 2$ .

Example 5.

Find the absolute maximum and minimum of $f (x) = \sin 2 x + 2 \cos x$ .

Solution

The domain of

f

(- \infty, \infty)

. Because

f

is periodic with fundamental period

2 π

, we find the extrema on

[0, 2 π]

Graph of f(x) = sin(2x) + 2cos(x) showing its periodicity with period 2pi — The function $f (x) = \sin 2 x + 2 \cos x$ is periodic with fundamental period $2 π$ .

**Step 1:** f'(x)=2\cos2x-2\sin x. **Step 2:** Setting f'(x)=0:

2 \cos 2 x - 2 \sin x = 0.

Using

\cos 2 x = 1 - 2 \sin^{2} x

2 (1 - 2 \sin^{2} x) - 2 \sin x = 0

- 4 \sin^{2} x - 2 \sin x + 2 = 0.

This is a quadratic in

u = \sin x

- 4 u^{2} - 2 u + 2 = 0 \Rightarrow u = \frac{1}{2} or u = - 1.

Solving in

[0, 2 π]

\sin x = \frac{1}{2} \Rightarrow x = \frac{π}{6}, x = \frac{5 π}{6}

\sin x = - 1 \Rightarrow x = \frac{3 π}{2} .

Unit circle diagram showing solutions to sin x = 1/2 and sin x = -1

Critical points:

x = \frac{π}{6}, x = \frac{5 π}{6}, x = \frac{3 π}{2}

. Evaluating

f

f (\frac{π}{6}) = \sin \frac{π}{3} + 2 \cos \frac{π}{6} = \frac{\sqrt{3}}{2} + \sqrt{3} = \frac{3 \sqrt{3}}{2}

f (\frac{5 π}{6}) = - \frac{3 \sqrt{3}}{2}

f (\frac{3 π}{2}) = 0.

**Step 3:** At the endpoints:

f (0) = 0 + 2 = 2, f (2 π) = 0 + 2 = 2.

**Step 4:** |

x

| $0$ |

\frac{π}{6}

\frac{5 π}{6}

\frac{3 π}{2}

2 π

| |-----|-----|---------|---------|----------|------| |

f (x)

| $2$ |

\frac{3 \sqrt{3}}{2} \approx 2.598

- \frac{3 \sqrt{3}}{2} \approx - 2.598

| $0$ | $2$ | | | | max | min | | | The absolute maximum is

\frac{3 \sqrt{3}}{2}

x = \frac{π}{6}

, and the absolute minimum is

- \frac{3 \sqrt{3}}{2}

x = \frac{5 π}{6}

Graph of f(x) = sin(2x) + 2cos(x) in one period showing absolute max at pi/6 and min at 5pi/6 — In one period, $f (x) = \sin 2 x + 2 \cos x$ takes its absolute maximum at $x = π / 6$ and its absolute minimum at $x = 5 π / 6$ .

Frequently Asked Questions

What is the difference between an absolute and a local extremum?

A local (relative) maximum at

c

means

f (c)

is the largest value of

f

in some neighborhood of

c

. An absolute (global) maximum means

f (c) \geq f (x)

for every

x

in the domain. Every absolute extreme value that is not at an endpoint is also a local extreme value, but not every local extreme value is absolute.

Must every critical point be a local extremum?

No. Fermat's Theorem says a local extremum must be at a critical point, but a critical point need not be a local extremum. For example,

f (x) = x^{3}

has f'(0)=0, but the graph passes through $(0,0)$ without turning around, it is neither a local max nor a local min.

What happens when the second derivative is zero at a critical point?

If f'(c)=0 and f''(c)=0, the Second Derivative Test is inconclusive. You must use the First Derivative Test (checking the sign of f' on either side of

c

) to classify the critical point.

Overview

Absolute Maxima and Minima

Local (Relative) Maxima and Minima

Fermat's Theorem and Critical Points

Critical Points

Strategy for Finding Absolute Extrema

Examples

Frequently Asked Questions

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