Candidates test for absolute extrema

AP Calculus AB· AP Calculus AB CED — Analytical Applications of Differentiation· 14 min read

1. Core Concepts of the Candidates Test★★☆☆☆⏱ 3 min

The Candidates Test (also called the Extreme Value Candidates Test or Closed Interval Method) is the standard algorithm for finding absolute (global) maximum and minimum values of a function on a given interval. Per the AP Calculus AB CED, this topic makes up ~1-2% of your total exam score, appearing in both multiple-choice and free-response, often as part of larger optimization or graph analysis questions.

📘 Definition

Candidate Points

All points where an absolute extremum of a function can occur on an interval. These include: 1) critical points inside the interval, 2) endpoints of the interval, 3) discontinuities where the function is defined.

Example:

For $f (x)$ on $[a, b]$ , candidates include $a$ , $b$ , and all critical points between $a$ and $b$ .

The core idea draws from the Extreme Value Theorem: if a function is continuous on a closed bounded interval $[a, b]$ , it must attain both an absolute maximum and an absolute minimum on that interval, and these can only occur at candidate points. Unlike the First or Second Derivative Tests which only classify local extrema, the Candidates Test directly identifies global extrema, the type most often requested on the AP exam.

2. The Candidates Test on Closed Intervals★★☆☆☆⏱ 4 min

Finding absolute extrema on a closed bounded interval $[a, b]$ is the most common AP exam scenario, and forms the base of the Candidates Test procedure. The Extreme Value Theorem guarantees both absolute extrema exist for continuous functions on closed intervals, so the step-by-step procedure is:

Confirm $f (x)$ is continuous on the entire interval $[a, b]$
Find all critical points of $f (x)$ that lie strictly inside the open interval $(a, b)$ . Critical points are points where $f^{'} (x) = 0$ , or $f^{'} (x)$ is undefined but $f (x)$ is defined.
Add the endpoints $x = a$ and $x = b$ to your list of candidate points
Evaluate $f (x)$ at every candidate point
The largest output is the absolute maximum, the smallest output is the absolute minimum.

📐 Worked Example

Find the absolute maximum and absolute minimum of $f (x) = x^{3} - 3 x^{2} + 1$ on the closed interval $[- 1, 4]$ .

1
$f (x)$ is a polynomial, so it is continuous on all real numbers including $[- 1, 4]$ , so the Extreme Value Theorem applies.
2
Compute the first derivative:
$f'(x) = 3x^2 - 6x = 3x(x-2)$
3
Set $f^{'} (x) = 0$ to find critical points at $x = 0$ and $x = 2$ , both inside $(- 1, 4)$ . There are no points where $f^{'} (x)$ is undefined, so these are our only critical points.
4
Full list of candidates: $x = - 1, 0, 2, 4$ .
5
Evaluate $f (x)$ at each candidate: $f (- 1) = (- 1)^{3} - 3 (- 1)^{2} + 1 = - 3$ $f (0) = 0 - 0 + 1 = 1$ $f (2) = 8 - 12 + 1 = - 3$ $f (4) = 64 - 48 + 1 = 17$
6
The largest value is $17$ , so the absolute maximum of $f (x)$ on $[- 1, 4]$ is $17$ at $x = 4$ . The smallest value is $- 3$ , so the absolute minimum is $- 3$ at $x = - 1$ and $x = 2$ .

3. Candidates Test on Open and Infinite Intervals★★★☆☆⏱ 5 min

The Candidates Test can be adapted to find absolute extrema on open intervals $(a, b)$ or infinite intervals $(- \infty, \infty)$ , though the Extreme Value Theorem does not apply here because endpoints are not included in the interval. The key modification is that we use limit values at open endpoints as candidate values, and we must check whether any extremum is actually attained at a point inside the interval. The procedure is:

Confirm $f (x)$ is continuous on the entire open interval
Find all critical points inside the interval, same as for closed intervals
Evaluate $f (x)$ at each critical point
Calculate the limit of $f (x)$ as $x$ approaches each open endpoint from inside the interval (for infinite endpoints, calculate $lim_{x \to \infty} f (x)$ or $lim_{x \to - \infty} f (x)$ )
If the largest candidate value comes from a critical point inside the interval, that is the absolute maximum. If the largest value is only approached as a limit at an endpoint and never attained inside the interval, there is no absolute maximum. The same logic applies for the absolute minimum.

📐 Worked Example

Find all absolute extrema of $f (x) = x^{2} - 4 x + 3$ on the open interval $(0, 3)$ .

1
$f (x)$ is a polynomial, so it is continuous on $(0, 3)$ , so we proceed.
2
Compute the derivative:
$f'(x) = 2x - 4 = 2(x-2)$
3
The only critical point is $x = 2$ , which is inside $(0, 3)$ , and there are no points where $f^{'} (x)$ is undefined.
4
Evaluate $f (x)$ at the critical point:
$f(2) = 4 - 8 + 3 = -1$
5
Calculate limits at the open endpoints:
$\lim_{x \to 0^+} f(x) = 3, \quad \lim_{x \to 3^-} f(x) = 0$
6
Compare candidate values: $- 1, 3, 0$ . The smallest value is $- 1$ , which is attained at $x = 2$ inside the interval, so this is the absolute minimum. The largest candidate value is $3$ , which is only the limit as $x \to 0^{+}$ and $x = 0$ is not included in the interval, so $f (x)$ never actually attains $3$ on $(0, 3)$ . Thus, there is no absolute maximum.

4. Candidates Test for Discontinuous and Piecewise Functions★★★★☆⏱ 5 min

If a function has a discontinuity on the interval of interest (common for piecewise functions and rational functions), the Candidates Test requires one additional step: add all points of discontinuity where $f (x)$ is defined to your candidate list. The Extreme Value Theorem only guarantees extrema for fully continuous functions on closed intervals, so a discontinuity can create a case where an extremum occurs at the discontinuity that would not be captured by only checking critical points and endpoints.

First identify all points of discontinuity of $f (x)$ on the interval
Add any discontinuity where $f (x)$ is defined to your candidate list, along with critical points and endpoints
Evaluate $f (x)$ at all candidates and compare values, same as before. If the discontinuity is a point where $f (x)$ is undefined, it cannot have an extremum there, so it is not added to the list.

📐 Worked Example

Find the absolute maximum and absolute minimum of $f (x) = {x^{2} + 1, 3 x - 1, - 2 \leq x \leq 0 0 < x \leq 2$ on the closed interval $[- 2, 2]$ .

1
$f (x)$ is made of polynomial pieces, so the only possible discontinuity is at the boundary $x = 0$ . We confirm $f (0) = 1$ , $lim_{x \to 0^{+}} f (x) = - 1$ , so there is a jump discontinuity at $x = 0$ , and $f (0)$ is defined, so we add $x = 0$ to the candidate list.
2
Find critical points inside $(- 2, 2)$ : for $- 2 < x < 0$ , $f^{'} (x) = 2 x$ , which equals zero only at $x = 0$ . For $0 < x < 2$ , $f^{'} (x) = 3$ , which is never zero or undefined. So the only point inside the interval is the discontinuity at $x = 0$ .
3
Full candidate list: endpoints $x = - 2, x = 2$ , plus $x = 0$ .
4
Evaluate $f (x)$ at all candidates:
$f(-2) = 5, \quad f(0) = 1, \quad f(2) = 5$
5
Classification: The largest value is $5$ , attained at both endpoints, so the absolute maximum is $5$ . For the minimum, $lim_{x \to 0^{+}} f (x) = - 1$ , which is smaller than all candidate function values, but $f (x)$ never actually attains $- 1$ on $[- 2, 2]$ , because $x = 0$ is defined as $1$ , and all points $x > 0$ have $f (x) > - 1$ . There is no point in the interval where the minimum value is attained, so this function has no absolute minimum on $[- 2, 2]$ .

5. AP-Style Concept Check★★★☆☆⏱ 4 min

✓ Quick check

Test your understanding with this AP-style multiple choice question:

What is the absolute minimum value of $f (x) = 2 x^{3} - 15 x^{2} + 24 x + 2$ on the closed interval $[0, 5]$ ?
- A. $2$
- B. $- 8$
- C. $11$
- D. $41$
Reveal answer
B. $-8$ —
Correct: $f (x)$ is continuous, critical points at $x = 1$ and $x = 4$ (both inside the interval). Candidates are $x = 0, 1, 4, 5$ . Evaluating gives $f (4) = - 8$ , which is the smallest value. Other options correspond to evaluating at the wrong candidate point.

6. Common Pitfalls

Wrong move:

Forgetting to include endpoints of a closed interval in the list of candidates

Why:

Students confuse critical points with all candidate points, and assume all extrema must be local extrema

Correct move:

Always add the two endpoints of a closed interval to your candidate list before evaluating function values

Wrong move:

Including critical points that lie outside the given interval in the candidate list

Why:

Students find all critical points of the function over its entire domain, then forget to filter them to only those inside the interval of interest

Correct move:

After finding all critical points, cross out any that are not within the given interval, only keep valid candidates

Wrong move:

Claiming an absolute extremum exists on an open interval when its value is only attained as a limit at the endpoint

Why:

Students forget that open intervals do not include endpoints, so the function cannot attain a value there

Correct move:

For any open endpoint, only count the limit as an attained extremum if the endpoint is included in the interval

Wrong move:

Forgetting to add points of discontinuity (where $f (x)$ is defined) to the candidate list

Why:

Students assume all tested functions are continuous everywhere, so they skip checking for discontinuities in piecewise or rational functions

Correct move:

For any non-polynomial function, first find all discontinuities on the interval, add any where $f (x)$ is defined to your candidate list

Wrong move:

Comparing $x$ -values instead of $f (x)$ values to find the absolute maximum/minimum

Why:

Students rush after finding candidates and incorrectly pick the largest or smallest $x$ -coordinate instead of the largest/smallest function output

Correct move:

Always explicitly write the value of $f (x)$ at each candidate, then sort the function outputs to find the extrema

Wrong move:

Ignoring critical points where $f^{'} (x)$ is undefined (but $f (x)$ is defined)

Why:

Students only set $f^{'} (x) = 0$ to find critical points, and forget the second part of the critical point definition

Correct move:

After finding where $f^{'} (x) = 0$ , check for points where $f^{'} (x)$ is undefined but $f (x)$ is defined, add any inside the interval to your candidate list

7. Quick Reference Cheatsheet

Category	Rule/Procedure	Notes
Candidate Point Definition	All points where absolute extrema can occur: 1) Critical points inside interval, 2) Endpoints of interval, 3) Discontinuities where $f (x)$ is defined	Discard any candidates that lie outside the interval of interest
Candidates Test (Closed $[a, b]$ )	Confirm $f$ continuous, 2. Find critical points in $(a, b)$ , 3. Add endpoints to candidates, 4. Evaluate $f$ at all candidates, 5. Largest $f$ = absolute max, smallest $f$ = absolute min	EVT guarantees both absolute max and min exist for continuous $f$
Candidates Test (Open $(a, b)$ )	Confirm $f$ continuous, 2. Find critical points, 3. Evaluate $f$ at critical points, 4. Calculate $lim_{x \to a^{+}} f (x)$ and $lim_{x \to b^{-}} f (x)$ , 5. Compare all candidate values, only count attained extrema	No guarantee extrema exist; always check for attainment
Candidates Test (Infinite $(- \infty, \infty)$ )	Same as open interval, add $lim_{x \to - \infty} f (x)$ and $lim_{x \to \infty} f (x)$ as candidate values	Odd-degree polynomials have no absolute extrema; even-degree have one
Critical Point Definition	$x = c$ is critical if $f (c)$ is defined, and $f^{'} (c) = 0$ or $f^{'} (c)$ is undefined	Always include points where $f^{'}$ is undefined but $f$ is defined
Discontinuous Functions	Add all discontinuities where $f (x)$ is defined to the candidate list	This is required for all piecewise functions at piece boundaries
Absolute Extrema Classification	Absolute maximum = largest $f (x)$ among attained candidates; Absolute minimum = smallest $f (x)$ among attained candidates	Never report an extremum that is not attained at a point in the interval

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · MCQ
Find absolute minimum on closed interval
2023 · FRQ
Justify absolute maximum on open interval

What's Next

Mastering the Candidates Test for absolute extrema is a critical prerequisite for optimization problems, the next major topic in Unit 5: Analytical Applications of Differentiation. Optimization problems ask you to find the maximum or minimum of a real-world function over a given interval, and the Candidates Test is the core method you will use to solve these problems. You will also apply this skill when analyzing graphs of derivatives and justifying extrema in free-response questions, so be sure to practice including all candidates and justifying your conclusions to earn full credit on exam day.