The Candidates Test

The test

The reason is short. Since

f

is continuous on a closed interval, the Extreme Value Theorem guarantees that an absolute maximum exists. Suppose it occurs at a point

c

inside

(a, b)

. Then

f(c) \ge f(x)

for every

x

near

c

, so

f

has a relative maximum at

c

. By Fermat's theorem,

c

is a critical point. So the maximum is at a critical point or at an endpoint, and the same argument works for the minimum.

The candidates for f on [a, b]: the endpoints a and b and the critical points c₁ and c₂. The absolute maximum is at b, and the absolute minimum is at c₂ (amber).

The test never asks which critical points are relative extrema. A critical point where

f

has no extremum, or a relative maximum lower than an endpoint, loses the comparison, as

c_1

does above.

Both hypotheses are needed. If

f

isn't continuous on

[a, b]

, or the interval isn't closed, there may be no absolute maximum at all. Then the largest of a few computed values proves nothing. An absolute maximum is a value of

f

, so state the value and the

x

where it occurs.

A polynomial

Example 1

Find the absolute maximum and absolute minimum values of

f(x) = -x^4 - 4x^3 + 2x^2 + 12x

on the interval

[-4, 0]

, and justify your answer.

Since

f

is a polynomial, it's continuous on

[-4, 0]

, so the Candidates Test applies. The derivative is

\begin{aligned} f'(x) &= -4x^3 - 12x^2 + 4x + 12 \\[4pt] &= -4\left(x^2(x + 3) - (x + 3)\right) \\[4pt] &= -4(x + 3)(x + 1)(x - 1). \end{aligned}

f'

is defined everywhere and is

0

x = -3

-1

, and

1

. Only

-3

and

-1

lie in

(-4, 0)

. The critical point

x = 1

is outside the interval, so it isn't a candidate. The values at the candidates are

\begin{aligned} f(-4) &= -256 + 256 + 32 - 48 \\ &= -16, \\[6pt] f(-3) &= -81 + 108 + 18 - 36 \\ &= 9, \\[6pt] f(-1) &= -1 + 4 + 2 - 12 \\ &= -7, \\[6pt] f(0) &= 0. \end{aligned}

f

has a relative minimum at

x = -1

, since

f'

changes from negative to positive there, but the endpoint value

f(-4) = -16

is lower. On

[-4, 0]

, the absolute maximum value is

f(-3) = 9

, and the absolute minimum value is

f(-4) = -16

f(x) = −x⁴ − 4x³ + 2x² + 12x on [−4, 0], with its four candidates. The absolute maximum (−3, 9) and absolute minimum (−4, −16) are amber.

Where f′ doesn't exist

A critical point is a point of the domain where

f' = 0

or where

f'

doesn't exist. Both kinds go on the list.

Example 2

Find the absolute maximum and absolute minimum values of

f(x) = x\left|x - 3\right|

\left[1, \frac{7}{2}\right]

Both

x

and

\left|x - 3\right|

are continuous, so

f

is continuous on

\left[1, \frac{7}{2}\right]

. Without the absolute value,

f(x) = \begin{cases} 3x - x^2, & 1 \le x \le 3, \\[4pt] x^2 - 3x, & 3 < x \le \frac{7}{2}. \end{cases}

f'(x) = 3 - 2x

for

1 < x < 3

, which is

0

x = \frac{3}{2}

. For

3 < x < \frac{7}{2}

f'(x) = 2x - 3

, which is positive there. At

x = 3

the slopes from the two sides are

3 - 6 = -3

and

6 - 3 = 3

, so

f'(3)

doesn't exist. The critical points in

\left(1, \frac{7}{2}\right)

are

\frac{3}{2}

and

3

, and

\begin{aligned} f(1) &= 1 \cdot 2 = 2, \\[6pt] f\left(\tfrac{3}{2}\right) &= \tfrac{3}{2} \cdot \tfrac{3}{2} = \tfrac{9}{4}, \\[6pt] f(3) &= 3 \cdot 0 = 0, \\[6pt] f\left(\tfrac{7}{2}\right) &= \tfrac{7}{2} \cdot \tfrac{1}{2} = \tfrac{7}{4}. \end{aligned}

The absolute maximum value is

\frac{9}{4}

, at

x = \frac{3}{2}

. The absolute minimum value is

0

, at

x = 3

, where

f'

doesn't exist. It was found only because points like

x = 3

are on the list.

f(x) = x·|x − 3| on [1, 7/2]. The absolute maximum is at x = 3/2, where f′ = 0, and the absolute minimum is at the corner x = 3.

A trigonometric function

Example 3

Find the absolute maximum and absolute minimum values of

f(x) = \frac{\cos x}{3 + \sin x}

[-\pi, \pi]

, and where each occurs.

Since

3 + \sin x \ge 2

, the denominator is never

0

, and

f

is continuous everywhere. By the quotient rule, with

\sin^2 x + \cos^2 x = 1

\begin{aligned} f'(x) &= \frac{-3\sin x - \sin^2 x - \cos^2 x}{(3 + \sin x)^2} \\[6pt] &= \frac{-3\sin x - 1}{(3 + \sin x)^2}. \end{aligned}

f'

exists everywhere, and

f'(x) = 0

when

\sin x = -\frac{1}{3}

. In

(-\pi, \pi)

that's

c_1 = -\pi + \arcsin\frac{1}{3} \approx -2.802

and

c_2 = -\arcsin\frac{1}{3} \approx -0.340

. At both,

3 + \sin x = \frac{8}{3}

, and

\cos x = \pm\sqrt{1 - \frac{1}{9}} = \pm\frac{2\sqrt{2}}{3}

: negative at

c_1

and positive at

c_2

. So

\begin{aligned} f(-\pi) &= \tfrac{-1}{3 + 0} = -\tfrac{1}{3}, \\[6pt] f(c_1) &= \tfrac{-2\sqrt{2}/3}{8/3} = -\tfrac{\sqrt{2}}{4}, \\[6pt] f(c_2) &= \tfrac{2\sqrt{2}/3}{8/3} = \tfrac{\sqrt{2}}{4}, \\[6pt] f(\pi) &= \tfrac{-1}{3 + 0} = -\tfrac{1}{3}. \end{aligned}

Since

\frac{\sqrt{2}}{4} \approx 0.354 > \frac{1}{3}

, the absolute maximum value is

\frac{\sqrt{2}}{4}

, at

x = c_2 = -\arcsin\frac{1}{3}

, and the absolute minimum value is

-\frac{\sqrt{2}}{4}

, at

x = c_1 = -\pi + \arcsin\frac{1}{3}

. The two endpoint values tie at

-\frac{1}{3}

, and neither wins.

f(x) = cos x/(3 + sin x) on [−π, π]. The maximum √2/4 is near x = −0.34 and the minimum −√2/4 near x = −2.80. Both endpoints give −1/3.

Critical points outside the interval

Only critical points inside

(a, b)

are candidates. A critical point outside the interval says nothing about the values of

f

[a, b]

. When none lies inside, the extrema are at the endpoints.

Example 4

Let

f(x) = \frac{x + 1}{x^2 + 3}

. Find the absolute maximum and absolute minimum values of

f

on each interval.

\begin{aligned} &\text{(a)}\;\; [0, 5] \\[10pt] &\text{(b)}\;\; [2, 5] \end{aligned}

The denominator is never

0

, so

f

is continuous everywhere. By the quotient rule,

\begin{aligned} f'(x) &= \frac{(x^2 + 3) - (x + 1)(2x)}{(x^2 + 3)^2} \\[6pt] &= \frac{3 - 2x - x^2}{(x^2 + 3)^2} \\[6pt] &= \frac{-(x + 3)(x - 1)}{(x^2 + 3)^2}. \end{aligned}

So the critical points of

f

are

x = -3

and

x = 1

(a) Only

x = 1

lies in

(0, 5)

. The candidates give

f(0) = \frac{1}{3}

f(1) = \frac{2}{4} = \frac{1}{2}

, and

f(5) = \frac{6}{28} = \frac{3}{14}

. On

[0, 5]

the absolute maximum value is

\frac{1}{2}

, at

x = 1

, and the absolute minimum value is

\frac{3}{14}

, at

x = 5

(b) Neither

-3

nor

1

lies in

(2, 5)

, so the only candidates are the endpoints:

f(2) = \frac{3}{7}

and

f(5) = \frac{3}{14}

. On

[2, 5]

the absolute maximum value is

\frac{3}{7}

, at

x = 2

, and the absolute minimum value is

\frac{3}{14}

, at

x = 5

. Indeed, both

x + 3

and

x - 1

are positive on

[2, 5]

, so

f' < 0

there and

f

decreases across the whole interval. The value

f(1) = \frac{1}{2}

from (a) is larger than every value of

f

[2, 5]

(a) On [0, 5], the critical point x = 1 is a candidate, and f(1) = 1/2 is the maximum.

(b) On [2, 5], x = 1 is outside (hollow), and the extrema are at the endpoints.

From a table

The test needs only the values of

f

at the candidates. When those come from a table, sign information about

f'

locates the critical points. It can also show that a missing value isn't an extremum.

Example 5

A hiker's elevation is

E(t)

meters,

t

hours after the start of a hike, for

0 \le t \le 8

. The function

E

is differentiable, and the table gives some values of

E

and the sign of

E'

. Give a reason for each answer.

$t$ (hours)	$E(t)$ (meters)	$E'(t)$
$0$	$760$
$0 < t < 2$		Positive
$2$	$1{,}240$	$0$
$2 < t < 4$		Negative
$4$		$0$
$4 < t < 6$		Negative
$6$	$690$	$0$
$6 < t < 8$		Positive
$8$	$1{,}150$

\begin{aligned} &\text{(a)}\;\; \text{Find the highest elevation.} \\[10pt] &\text{(b)}\;\; \text{Find the lowest elevation.} \end{aligned}

Since

E

is differentiable, it's continuous on

[0, 8]

, and the Candidates Test applies. The critical points in

(0, 8)

are

t = 2

4

, and

6

, where

E'(t) = 0

. The table doesn't give

E(4)

E

is differentiable, so it's continuous on

[2, 4]

and differentiable on

(2, 4)

. By the Mean Value Theorem,

E(4) - E(2) = 2E'(c)

for some

c

(2, 4)

. This difference is negative because

E'(c) < 0

, so

E(4) < E(2)

. The hypotheses hold on

[4, 6]

too, and the same argument gives

E(6) < E(4)

. So

690 < E(4) < 1{,}240

, and

E(4)

is neither the largest nor the smallest candidate value.

(a) The other candidate values are

E(0) = 760

E(2) = 1{,}240

E(6) = 690

, and

E(8) = 1{,}150

. The highest elevation is

1{,}240

meters, at

t = 2

hours.

(b) The lowest elevation is

690

meters, at

t = 6

hours.

One possible graph of E, in meters. The table doesn't give E(4) (hollow); any value between 690 and 1,240 fits, and here it's 1,000.