The Candidates Test — AP® Calculus AB Lesson

Part 1 · Hook

Why this matters

A water company must know the lowest level its reservoir will reach this month, not just a dip that looks low. A relative minimum answers "lower than nearby"; the company needs "lowest of all". On a closed interval there is a short, sure way to find it: list a few candidates and compare.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. The Extreme Value Theorem guarantees an absolute maximum when $f$ is

continuous on a closed interval
differentiable on an open interval
increasing
defined at both endpoints

Show the answer

Continuity on $[a, b]$ is the hypothesis.

Correct: continuous on a closed interval:
differentiable on an open interval:
increasing:
defined at both endpoints:

2. $f'$ changes from positive to negative at $x = 2$ . Then $f(2)$ is

a relative maximum
a relative minimum
the absolute maximum
not an extremum

Show the answer

The First Derivative Test gives a relative (local) maximum.

Correct: a relative maximum:
a relative minimum:
the absolute maximum:
not an extremum:

Part 4 · See it

See it first

The graph of f(x) = 2x cubed minus 9x squared plus 12x on the closed interval from 0 to 3, with four candidates: endpoints (0, 0) and (3, 9), critical points (1, 5) and (2, 4). A table lists the values; the absolute maximum is 9 at x = 3 and the absolute minimum is 0 at x = 0. The relative maximum (1, 5) and relative minimum (2, 4) are not absolute. — Four candidates: two endpoints and two critical points. The largest value, at an endpoint, is the absolute maximum; the relative maximum at $x = 1$ is not. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

$f$ is continuous on $[a, b]$ by the Extreme Value Theorem, an absolute maximum and minimum exist
An absolute extremum is either at an endpoint or insideif inside, it is at a critical point, where $f'$ is 0 or undefined
So the endpoints and critical points are the only candidatesevaluate $f$ at each one and compare: the largest value is the absolute maximum, the smallest the absolute minimum
You only showed a sign change of $f'$ that proves a relative extremum, a local fact, not an absolute one

Part 6 · Key ideas

Key ideas

Candidates Test (closed interval method): for $f$ continuous on $[a, b]$ , list the endpoints and the critical points in $(a, b)$ , compute $f$ at each, and compare.
No derivative signs needed: just values. Show the table of candidates and values in your answer.
The absolute extrema can be at endpoints. A relative maximum can lose to an endpoint.
On a calculator problem, find critical points with the graph or solver, store them, and evaluate the model at each candidate.

Part 7 · Misconception

A common mistake

The wrong idea: " $f'$ changes from positive to negative at $x = c$ , so $f(c)$ is the absolute maximum."

What actually happens: A sign change is a local argument: it only compares $f(c)$ with nearby values. For an absolute extremum on a closed interval, compare the values at the endpoints and every critical point. Chief readers see this lost point every year.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Values of $f$ and $f'$

f

is continuous on

[0, 6]

. Its only critical points in

(0, 6)

are

x = 1

,

x = 3

and

x = 5

.

Values of $f(x)$ and $f'(x)$ at selected $x$
$x$	0	1	3	5	6
$f(x)$	4	7	-2	1	5
$f'(x)$	3	0	undefined	0	2

1. What is the absolute maximum value of $f$ on $[0, 6]$ ?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Candidates: endpoints $x = 0, 6$ and critical points $x = 1, 3, 5$ . Values: $4, 7, -2, 1, 5$ . The largest is $f(1) = 7$ .

Answer: 7

2. Where does $f$ have its absolute minimum on $[0, 6]$ ?

At $x = 0$ , the left endpoint
At $x = 3$
At $x = 5$ , where $f'(x) = 0$
Nowhere, because $f'(3)$ is undefined

Show the answer

Of the candidate values $4, 7, -2, 1, 5$ , the smallest is $f(3) = -2$ , so the absolute minimum is at $x = 3$ .

At $x = 0$ , the left endpoint: $f(0) = 4$ is not the smallest.
Correct: At $x = 3$ : $f(3) = -2$ is the smallest of the five candidate values.
At $x = 5$ , where $f'(x) = 0$ : $f(5) = 1$ is larger than $f(3) = -2$ .
Nowhere, because $f'(3)$ is undefined: A critical point where $f'$ is undefined is still a candidate, and here it wins.

3. Which list is the complete set of candidates for the absolute extrema of $f$ on $[0, 6]$ ?

$x = 0, 1, 3, 5, 6$
$x = 1, 3, 5$ , the critical points
$x = 0$ and $x = 6$ , the endpoints
$x = 1$ and $x = 5$ , where $f'(x) = 0$

Show the answer

On a closed interval, the candidates are the endpoints and the critical points (where $f'$ is 0 or undefined): $x = 0, 1, 3, 5, 6$ .

Correct: $x = 0, 1, 3, 5, 6$ : Both endpoints and every critical point, including where $f'$ is undefined.
$x = 1, 3, 5$ , the critical points: The endpoints are missing; an absolute extremum can sit at an end.
$x = 0$ and $x = 6$ , the endpoints: The critical points are missing.
$x = 1$ and $x = 5$ , where $f'(x) = 0$ : $x = 3$ , where $f'$ is undefined, and both endpoints are missing.

4. Which justification earns credit for "the absolute maximum of $f$ on $[0, 6]$ is 7"?

" $f'$ changes from positive to negative at $x = 1$ , so $f(1) = 7$ is the absolute maximum."
" $f'(1) = 0$ , so $f(1) = 7$ is the absolute maximum on $[0, 6]$ ."
" $f(1) = 7$ is greater than both endpoint values, $f(0) = 4$ and $f(6) = 5$ ."
"The candidates $x = 0, 1, 3, 5, 6$ give $f = 4, 7, -2, 1, 5$ . The largest is 7, at $x = 1$ ."

Show the answer

An absolute extremum needs a global argument. The Candidates Test lists every endpoint and critical point with its value and picks the largest.

" $f'$ changes from positive to negative at $x = 1$ , so $f(1) = 7$ is the absolute maximum.": A sign change gives a relative maximum, a local fact. It does not compare $f(1)$ with far-away values.
" $f'(1) = 0$ , so $f(1) = 7$ is the absolute maximum on $[0, 6]$ .": A zero of $f'$ is only a candidate.
" $f(1) = 7$ is greater than both endpoint values, $f(0) = 4$ and $f(6) = 5$ .": Close, but the other critical points must be compared too.
Correct: "The candidates $x = 0, 1, 3, 5, 6$ give $f = 4, 7, -2, 1, 5$ . The largest is 7, at $x = 1$ .": A candidates test: every endpoint and critical point, with values, compared.

5. Find the absolute maximum value of $f(x) = x^3 - 3x$ on $[0, 3]$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$f'(x) = 3x^2 - 3 = 0$ at $x = \pm 1$ ; only $x = 1$ is in $[0, 3]$ . Candidates: $f(0) = 0$ , $f(1) = -2$ , $f(3) = 18$ . Maximum: 18.

Answer: 18

6. Find the absolute maximum value of $f(x) = x - \ln x$ on $\left[\frac{1}{2}, 3\right]$ . Give an exact answer.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$f'(x) = 1 - \frac{1}{x} = 0$ at $x = 1$ . Candidates: $f\left(\frac{1}{2}\right)$ $= \frac{1}{2} + \ln 2$ $\approx 1.193$ , $f(1) = 1$ , $f(3) = 3 - \ln 3 \approx 1.901$ . Maximum: $3 - \ln 3$ .

Answer: $3 - \ln(3)$ ≈ 1.901

Part 9 · Summary

Summary

On a closed interval, a continuous function has absolute extrema, and they sit at endpoints or critical points. The Candidates Test evaluates

f

at each of those and compares. A sign change of

f'

shows only a relative extremum; an absolute one needs the comparison.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

Values of ff and f′f'

Summary

What comes next

Connections

Values of $f$ and $f'$