# Candidates test for absolute extrema

> AP Calculus AB · Analytical Applications of Differentiation
> Source: https://www.owlsprep.com/study/ap-calculus-ab-u5-candidates-test-for-absolute-extrema/

This guide covers the Candidates Test (Closed Interval Method) for finding absolute extrema of functions on closed, open, and infinite intervals, including adaptations for discontinuous and piecewise functions, aligned to AP Calculus AB exam requirements.

**Prerequisites:** How to compute derivatives using all basic differentiation rules; How to find critical points of a function; Definition of absolute maximum and absolute minimum on an interval

## Learning objectives

- Identify candidate points for absolute extrema on any interval
- Apply the Candidates Test to closed, open, and infinite intervals
- Correctly adapt the Candidates Test for discontinuous and piecewise functions
- Justify whether an absolute extremum exists on an interval

## Core Concepts of the Candidates Test

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.

**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.

## The Candidates Test on Closed Intervals

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:

1. Confirm $f(x)$ is continuous on the entire interval $[a,b]$
2. 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.
3. Add the endpoints $x=a$ and $x=b$ to your list of candidate points
4. Evaluate $f(x)$ at every candidate point
5. The largest output is the absolute maximum, the smallest output is the absolute minimum.

> **tip**
>
> On FRQs, you must explicitly list all candidate points and show their corresponding function values to earn full credit; AP readers will not infer you used the Candidates Test if you only write the final answer.

**Worked example:** Find the absolute maximum and absolute minimum of $f(x) = x^3 - 3x^2 + 1$ 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 $x=2$.

## Candidates Test on Open and Infinite Intervals

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:

1. Confirm $f(x)$ is continuous on the entire open interval
2. Find all critical points inside the interval, same as for closed intervals
3. Evaluate $f(x)$ at each critical point
4. 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 $\lim_{x \to -\infty} f(x)$)
5. 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.

> **tip**
>
> AP exam questions often include a trick where no absolute maximum or minimum exists on an open interval. Always explicitly state whether an extremum exists, do not just give a value if it is not attained.

**Worked example:** Find all absolute extrema of $f(x) = x^2 - 4x + 3$ 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.

## Candidates Test for Discontinuous and Piecewise Functions

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.

1. First identify all points of discontinuity of $f(x)$ on the interval
2. Add any discontinuity where $f(x)$ is defined to your candidate list, along with critical points and endpoints
3. 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.

> **tip**
>
> Always treat the point where the piece definition changes as a candidate point for any piecewise function, regardless of whether it is continuous or not.

**Worked example:** Find the absolute maximum and absolute minimum of $f(x) = \begin{cases} x^2 + 1, & -2 \leq x \leq 0 \\ 3x - 1, & 0 < x \leq 2 \end{cases}$ 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 $-2 < x < 0$, $f'(x) = 2x$, 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 $[-2,2]$.

## AP-Style Concept Check

**Check your understanding**

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

1. What is the absolute minimum value of $f(x) = 2x^3 - 15x^2 + 24x + 2$ on the closed interval $[0, 5]$?

   - A. $2$
   - B. $-8$
   - C. $11$
   - D. $41$

   *Why:* 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.

## Common pitfalls

- **Wrong:** Forgetting to include endpoints of a closed interval in the list of candidates
  - Why it fails: Students confuse critical points with all candidate points, and assume all extrema must be local extrema
  - Correct: Always add the two endpoints of a closed interval to your candidate list before evaluating function values
- **Wrong:** Including critical points that lie outside the given interval in the candidate list
  - Why it fails: 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: After finding all critical points, cross out any that are not within the given interval, only keep valid candidates
- **Wrong:** Claiming an absolute extremum exists on an open interval when its value is only attained as a limit at the endpoint
  - Why it fails: Students forget that open intervals do not include endpoints, so the function cannot attain a value there
  - Correct: For any open endpoint, only count the limit as an attained extremum if the endpoint is included in the interval
- **Wrong:** Forgetting to add points of discontinuity (where $f(x)$ is defined) to the candidate list
  - Why it fails: Students assume all tested functions are continuous everywhere, so they skip checking for discontinuities in piecewise or rational functions
  - Correct: For any non-polynomial function, first find all discontinuities on the interval, add any where $f(x)$ is defined to your candidate list
- **Wrong:** Comparing $x$-values instead of $f(x)$ values to find the absolute maximum/minimum
  - Why it fails: Students rush after finding candidates and incorrectly pick the largest or smallest $x$-coordinate instead of the largest/smallest function output
  - Correct: Always explicitly write the value of $f(x)$ at each candidate, then sort the *function outputs* to find the extrema
- **Wrong:** Ignoring critical points where $f'(x)$ is undefined (but $f(x)$ is defined)
  - Why it fails: Students only set $f'(x)=0$ to find critical points, and forget the second part of the critical point definition
  - Correct: 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

## 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 $[a,b]$) | 1. 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 $(a,b)$) | 1. 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 |

## 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.

- [Unit 5 Overview: Analytical Applications of Differentiation](https://www.owlsprep.com/study/ap-calculus-ab-u5-overview/)
- [Determining Concavity](https://www.owlsprep.com/study/ap-calculus-ab-u5-determining-concavity/)
- [Second Derivative Test](https://www.owlsprep.com/study/ap-calculus-ab-u5-second-derivative-test/)

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