The Extreme Value Theorem and Critical Points

Absolute and relative extrema

A relative extremum compares

f(c)

only with values nearby, on both sides of

c

. So it occurs at a point inside the domain. An endpoint of a closed interval can't be a relative extremum under this definition, though it can be an absolute one. An absolute extremum at an interior point is also a relative one.

On [a, b], f has its absolute maximum at c, which is also a relative maximum. Its absolute minimum is at the endpoint a. The relative minimum at d isn't absolute, since f(a) is smaller.

Example 1

The graph of

f

on the closed interval

[-3, 4]

is made of a line segment, part of a parabola with vertex

(1, -1)

, and another segment. Find each of the following, and where it occurs.

\begin{aligned} &\text{(a) the absolute extrema of } f \\[6pt] &\text{(b) the relative extrema of } f \end{aligned}

The graph of f on [−3, 4]. Segments join (−3, 0) to (−1, 2) and (3, 2) to (4, 3). Between them is a parabola with vertex (1, −1).

(a) The highest point of the graph is the endpoint

(4, 3)

, and the lowest is the vertex

(1, -1)

. The absolute maximum value is

f(4) = 3

, and the absolute minimum value is

f(1) = -1

(b) At

x = -1

f(-1) = 2

, and

f(x) < 2

for every other

x

(-2, 0)

. So

f

has a relative maximum of

2

x = -1

. The vertex gives a relative minimum of

-1

x = 1

. The endpoint

x = 4

isn't a relative maximum, because no open interval around

4

lies in the domain. Neither is

x = 3

, where

f(3) = 2

f(x) < 2

just left of

3

and

f(x) > 2

just right of it.

The Extreme Value Theorem

The theorem says the extrema exist. It doesn't say where they are, and either one can be at an endpoint. Each hypothesis is needed: drop either one, and the conclusion can fail.

f(x) = (x − 1)² + 1 on the open interval (0, 3). The minimum is f(1) = 1. The values approach 5 near x = 3 but never reach it, so there's no maximum.

g(x) = x for 0 ≤ x < 2 and g(x) = 3 − x for 2 ≤ x ≤ 4. The values approach 2 near x = 2 from the left, but g(2) = 1, so there's no maximum.

On the left,

f

is continuous, but the interval is open. For any

x

(0, 3)

, some point between

x

and

3

gives a larger value of

f

, so no value is the largest. On the right, the interval is closed, but

g

jumps at

x = 2

. Its values get as close to

2

as we like and never equal

2

, so again no value is the largest. Each graph still has an absolute minimum:

f(1) = 1

and

g(4) = -1

Example 2

For each function, determine whether the Extreme Value Theorem guarantees an absolute maximum and an absolute minimum on the interval. If it doesn't, determine whether they exist anyway.

\begin{aligned} &\text{(a)}\ \ f(x) = \tan x \ \text{ on } \left[-\tfrac{\pi}{4}, \tfrac{\pi}{3}\right] \\[6pt] &\text{(b)}\ \ g(x) = \ln x \ \text{ on } (0, e] \\[6pt] &\text{(c)}\ \ h(x) = \begin{cases} x^2, & -2 \le x < 1 \\ x - 2, & 1 \le x \le 3 \end{cases} \end{aligned}

(a)

\tan x = \frac{\sin x}{\cos x}

is continuous wherever

\cos x \ne 0

, and

\cos x = 0

only at odd multiples of

\frac{\pi}{2}

. None of those lies in

\left[-\frac{\pi}{4}, \frac{\pi}{3}\right]

, so

f

is continuous on this closed interval. By the Extreme Value Theorem,

f

has an absolute maximum and an absolute minimum there.

(b) The interval

(0, e]

isn't closed, so the theorem doesn't apply. Since

\ln x \le \ln e = 1

for

0 < x \le e

, the absolute maximum is

g(e) = 1

. But

\ln x \to -\infty

x \to 0^+

, so

g

has no absolute minimum.

(c) As

x \to 1^-

h(x) \to 1

, while

h(1) = -1

. So

h

isn't continuous at

1

, and the theorem doesn't apply. The extrema exist anyway. On

[-2, 1)

the values of

x^2

run from

0

up to

4

, reached at

x = -2

. On

[1, 3]

the values of

x - 2

run from

-1

1

. So the absolute maximum is

h(-2) = 4

and the absolute minimum is

h(1) = -1

The graph of h: a parabola with an open circle at (1, 1), then a segment from (1, −1) to (3, 1). The maximum 4 and the minimum −1 are marked.

Critical points

Suppose

f

has a relative maximum at

c

, so

f(x) \le f(c)

for every

x

near

c

. For

x > c

, the quotient

\frac{f(x) - f(c)}{x - c}

has a numerator

\le 0

and a positive denominator. So the quotient is

\le 0

, and its limit as

x \to c^+

\le 0

. For

x < c

the denominator is negative, the quotient is

\ge 0

, and its limit as

x \to c^-

\ge 0

. Both one-sided limits equal

f'(c)

, so

f'(c) \le 0

and

f'(c) \ge 0

. Thus

f'(c) = 0

. A relative minimum works the same way, with the inequalities reversed.

So at a relative extremum, either

f'(c) = 0

f'(c)

doesn't exist. Either way

c

is a critical point: every relative extremum of

f

occurs at a critical point of

f

A relative maximum where f′(c) exists: the tangent is horizontal, so f′(c) = 0.

A relative minimum at a corner: f′(c) doesn't exist.

y = x³ has a horizontal tangent at 0 and no extremum there.

The converse is false: a critical point needn't be a relative extremum. For

f(x) = x^3

f'(0) = 0

, so

0

is a critical point. But

x^3 < 0

for

x < 0

and

x^3 > 0

for

x > 0

, so

f(0) = 0

is neither the largest nor the smallest value near

0

The graph in Example 1 has three critical points. Its parabola has a horizontal tangent at the vertex, so

f'(1) = 0

. At

x = -1

and

x = 3

the graph has corners, so

f'

doesn't exist there. The critical points

-1

and

1

are relative extrema, and

3

isn't. On a closed interval the endpoints aren't counted as critical points. They're checked separately, as candidates for absolute extrema.

Finding critical points

Compute

f'

. Then solve

f'(x) = 0

, and find where

f'(x)

doesn't exist. Keep only the numbers that lie in the domain of

f

Example 3

Find the critical points of

f(x) = x^3(x - 5)^2

By the product rule, with the common factor

x^2(x - 5)

pulled out,

\begin{aligned} f'(x) &= 3x^2(x - 5)^2 + 2x^3(x - 5) \\ &= x^2(x - 5)\bigl(3(x - 5) + 2x\bigr) \\ &= 5x^2(x - 3)(x - 5). \end{aligned}

y = x³(x − 5)² has a horizontal tangent at each critical point: x = 0, 3, and 5. At 0 the graph keeps rising.

The derivative of a polynomial exists everywhere, so the critical points are the zeros of

f'

x = 0

x = 3

, and

x = 5

. Near

0

, both

x^2

and

(x - 3)(x - 5)

are positive on each side, so

f' > 0

on both sides of

0

. The critical point

0

isn't an extremum.

Example 4

Find the critical points of

f(x) = x\sqrt[3]{x - 4}

By the product rule,

\begin{aligned} f'(x) &= \sqrt[3]{x - 4} + \frac{x}{3(x - 4)^{2/3}} \\[4pt] &= \frac{3(x - 4) + x}{3(x - 4)^{2/3}} \\[4pt] &= \frac{4(x - 3)}{3(x - 4)^{2/3}}. \end{aligned}

y = x∛(x − 4). The tangent is horizontal at (3, −3) and vertical at (4, 0). There's no extremum at 4: f(x) < 0 for 0 < x < 4 and f(x) > 0 for x > 4.

The numerator is

0

x = 3

, so

f'(3) = 0

. The denominator is

0

x = 4

, so

f'(4)

doesn't exist, and

4

is in the domain, since

f(4) = 0

. The critical points are

x = 3

and

x = 4

Example 5

Find the critical points of each function.

\begin{aligned} &\text{(a)}\ \ h(x) = x + \cos 2x \ \text{ on } (0, \pi) \\[6pt] &\text{(b)}\ \ g(x) = \frac{e^x}{x - 2} \end{aligned}

(a) By the chain rule, the derivative of

\cos 2x

-2\sin 2x

, so

h'(x) = 1 - 2\sin 2x.

This exists everywhere, so the critical points are its zeros in

(0, \pi)

. There

\sin 2x = \frac{1}{2}

, and

2x

lies in

(0, 2\pi)

, so

2x = \frac{\pi}{6}

2x = \frac{5\pi}{6}

. The critical points are

\frac{\pi}{12}

and

\frac{5\pi}{12}

y = x + cos 2x on (0, π), with a horizontal tangent at each critical point, x = π/12 and x = 5π/12.

(b) By the quotient rule,

\begin{aligned} g'(x) &= \frac{e^x(x - 2) - e^x}{(x - 2)^2} \\[4pt] &= \frac{e^x(x - 3)}{(x - 2)^2}. \end{aligned}

y = eˣ/(x − 2), with its vertical asymptote x = 2 and a horizontal tangent at (3, e³).

Since

e^x > 0

g'(x) = 0

only at

x = 3

. The derivative doesn't exist at

x = 2

, but

g(2)

isn't defined, so

2

isn't in the domain of

g

. The only critical point is

x = 3