Candidates test for absolute extrema — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Identifying critical points and endpoints as candidates for absolute extrema, applying the Candidates Test to closed intervals and open/unbounded intervals, justifying absolute extrema for continuous and piecewise functions on any domain.

You should already know: How to find critical points of a differentiable function; The Extreme Value Theorem for continuous functions on closed intervals; Basic differentiation rules including the chain and quotient rules.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Candidates test for absolute extrema?

The Candidates Test for absolute extrema (also called the Candidate Point Method, or the Closed Interval Method when applied to closed bounded intervals) is a systematic, theorem-backed procedure to locate the global (absolute) maximum and minimum values of a function over a specified interval. Per the AP Calculus BC Course and Exam Description (CED), this topic is a core skill within Unit 5: Analytical Applications of Differentiation, which accounts for 15–18% of the total AP exam score. This topic appears in both multiple-choice (MCQ) and free-response (FRQ) sections of the exam; FRQ questions almost always require explicit justification that all candidates were checked to earn full credit. The fundamental intuition behind the test is simple: any absolute extremum of a function on a given interval can only occur at one of two types of points: critical points inside the interval, or endpoints of the interval (if the interval is bounded). The test works by evaluating the function at every candidate point, then comparing output values to find the largest (absolute maximum) and smallest (absolute minimum).

2. Applying the Candidates Test on Closed Bounded Intervals

The most common application of the Candidates Test is for continuous functions on a closed bounded interval $[a,b]$, where the Extreme Value Theorem guarantees that both an absolute maximum and absolute minimum exist. The step-by-step procedure for this case is:

1. Find all critical points of $f(x)$ that lie strictly inside the open interval $(a,b)$. A critical point is any point where $f(x)$ is defined, and $f^{\prime }(x)=0$ or $f^{\prime }(x)$ is undefined.
2. Add the endpoints $x=a$ and $x=b$ to your list of candidates.
3. Evaluate $f(x)$ at every candidate point on your list.
4. The largest value of $f(x)$ from your evaluations is the absolute maximum, and the smallest is the absolute minimum.

Unlike classification of relative extrema, you do not need to use the first or second derivative test here—comparing function values directly is sufficient, which makes this method straightforward and less error-prone.

Worked Example

Find the absolute maximum and absolute minimum of $f(x)=x^3-6x^2+9x+2$ on the interval $[0,4]$.

1. Compute the first derivative: $f^{\prime }(x)=3x^2-12x+9=3(x-1)(x-3)$.
2. $f^{\prime }(x)$ is defined for all $x$ in $(0,4)$, so critical points occur where $f^{\prime }(x)=0$, giving $x=1$ and $x=3$, both inside the interval.
3. List all candidates: endpoints $x=0,x=4$, plus critical points $x=1,x=3$.
4. Evaluate $f$ at each candidate: $f(0)=2$, $f(1)=6$, $f(3)=2$, $f(4)=6$.
5. Conclusion: The absolute maximum value of $f$ on $[0,4]$ is $6$ (attained at $x=1$ and $x=4$), and the absolute minimum value is $2$ (attained at $x=0$ and $x=3$).

Exam tip: On AP FRQ, you do not need to classify critical points as local extrema first—just evaluate $f$ at all candidates and compare; this saves time and avoids unnecessary work that can introduce errors.

3. Applying the Candidates Test on Open or Unbounded Intervals

The Candidates Test extends to open intervals $(a,b)$ or unbounded intervals (like $(a,\infty )$ or $(-\infty ,\infty )$) with one key adjustment: there are no endpoints to evaluate, so instead we calculate the limit of $f(x)$ as $x$ approaches each open boundary (including $\pm \infty$ for unbounded intervals). We then compare function values at critical points to these limit values to determine if an absolute extremum exists.

Unlike closed intervals, the Extreme Value Theorem does not guarantee that an absolute extremum exists, so the test can confirm when no extremum exists. A useful shortcut for this case: if a continuous function has exactly one critical point on an interval, and that critical point is a local maximum, it must be the absolute maximum on the interval (the same logic holds for local minimum = absolute minimum).

Worked Example

Find the absolute maximum and absolute minimum of $f(x)=x{e}^{-x}$ on the interval $(0,\infty )$.

1. $f(x)$ is continuous on $(0,\infty )$, so the test applies.
2. Compute the derivative: $f^{\prime }(x)=e^{-x}(1-x)$. $f^{\prime }(x)$ is defined for all $x>0$, and never equals zero except at $x=1$, which is inside the interval.
3. Candidates: critical point $x=1$, plus limits at the open boundaries: ${\lim }_{x\to 0^{+}}x{e}^{-x}=0$, ${\lim }_{x\to \infty }x{e}^{-x}=0$.
4. Evaluate $f$ at $x=1$: $f(1)=\frac{1}{e}\approx 0.368$.
5. Conclusion: $\frac{1}{e}$ is larger than the limiting value of $0$, so the absolute maximum is $\frac{1}{e}$ at $x=1$. $f(x)$ never actually reaches $0$ on $(0,\infty )$, so no absolute minimum exists.

Exam tip: Always explicitly address whether an extremum value is actually attained by the function on the interval—if the function only approaches a minimum value but never reaches it, you must state that no absolute minimum exists, a common AP exam distracter.

4. Applying the Candidates Test to Piecewise Functions

Piecewise-defined functions are a common AP exam question type, and they require an extra step in the Candidates Test: interior breakpoints (points where the function definition changes) must always be added to the candidate list, even if the function is continuous at the breakpoint. This is because the derivative of a piecewise function is almost always undefined at interior breakpoints (even for continuous piecewise functions, the left and right derivatives rarely match), so all interior breakpoints are automatically critical points that must be checked. If a piecewise function has a discontinuity at an interior breakpoint, that point is still added to the candidate list if the function is defined there.

Worked Example

Find the absolute maximum and minimum of $$f(x)=\{\begin{array}{ll}{x}^2+1& 0\le x\le 2\\ -x+7& 2<x\le 5\end{array}$$ on the closed interval $[0,5]$.

1. Confirm $f$ is continuous on $[0,5]$: ${\lim }_{x\to 2^{-}}f(x)=2^2+1=5$, ${\lim }_{x\to 2^{+}}f(x)=-2+7=5$, so $f$ is continuous at $x=2$, and continuous everywhere else on $[0,5]$.
2. Find critical points on each piece: For $0\le x<2$, $f^{\prime }(x)=2x$, which equals zero only at $x=0$ (an endpoint). For $2<x<5$, $f^{\prime }(x)=-1$, which is never zero.
3. Add the interior breakpoint $x=2$ to the candidate list, since $f^{\prime }(2)$ is undefined.
4. List all candidates: endpoints $x=0,x=5$, breakpoint $x=2$.
5. Evaluate: $f(0)=1$, $f(2)=5$, $f(5)=2$.
6. Conclusion: Absolute maximum is $5$ at $x=2$, absolute minimum is $1$ at $x=0$. If we had forgotten to add $x=2$ as a candidate, we would have incorrectly chosen $2$ as the maximum.

Exam tip: Never forget to add all interior breakpoints of piecewise functions to your candidate list—AP exam writers regularly design problems where the absolute extremum occurs at the breakpoint to test this skill.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Forgetting to include critical points where $f^{\prime }(x)$ is undefined, only including points where $f^{\prime }(x)=0$. Why: Students associate critical points with derivative zero from basic polynomial examples, so they miss points with corners, cusps, or vertical tangents. Correct move: Always explicitly check for points where $f^{\prime }(x)$ is undefined, and confirm $f(x)$ is defined there before adding to the candidate list.
• Wrong move: On open intervals, claiming an absolute minimum exists when $f(x)$ approaches a value but never reaches it. Why: Students confuse the limit of $f(x)$ with an attained value of $f(x)$. Correct move: After calculating endpoint limits for open intervals, always check if the value is actually attained at a point in the interval before claiming an extremum exists.
• Wrong move: For piecewise functions, forgetting to add interior breakpoints to the candidate list. Why: Students only check for critical points on each individual piece and ignore junctions between pieces. Correct move: After finding critical points for each piece, add any breakpoint that lies strictly inside the interval to your candidate list before evaluating.
• Wrong move: Justifying an absolute extremum by only noting it is a local extremum, without comparing to all candidates. Why: Students confuse local (relative) extrema with absolute extrema, and think the first/second derivative test is sufficient. Correct move: Always state that you evaluated $f$ at all critical points and endpoints (or limits for open intervals) and compared values to justify the absolute extremum.
• Wrong move: Adding critical points that lie outside the given interval to the candidate list, then using their $f$ values in the comparison. Why: Students find all critical points of the function over its entire domain and forget to filter to those inside the given interval. Correct move: After finding all critical points of the function, cross out any that are not strictly inside the given interval before adding to candidates.
• Wrong move: On closed intervals, not checking if $f$ is continuous before applying the test. Why: Students assume all functions are continuous, but discontinuous functions on closed intervals may not have extrema that follow the rule. Correct move: Explicitly confirm continuity on the interval at the start of the problem, and add discontinuities inside the interval to the candidate list.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

What is the absolute minimum value of $f(x)=2x^3-3x^2-12x+1$ on the closed interval $[-2,3]$? A) $-21$ B) $-19$ C) $-7$ D) $1$

Worked Solution: $f(x)$ is a polynomial, so it is continuous on $[-2,3]$, so we apply the Candidates Test. First, find the derivative: $f^{\prime }(x)=6x^2-6x-12=6(x-2)(x+1)$. Critical points at $x=2$ and $x=-1$, both inside the interval. Add endpoints $x=-2$ and $x=3$ to the candidate list, then evaluate $f$ at each candidate: $f(-2)=-3$, $f(-1)=8$, $f(2)=-19$, $f(3)=-8$. The smallest value is $-19$, so the correct answer is B.

Question 2 (Free Response)

Let $f(x)=\frac{\ln x}{x}$ for $x>0$. (a) Find all critical points of $f(x)$ on $(0,\infty )$. Justify your answer. (b) Use the Candidates Test to determine whether $f$ has an absolute maximum and absolute minimum on $(0,\infty )$. Justify your answer. (c) What are the absolute maximum and minimum of $f(x)$ on $[1,e^2]$? Justify.

Worked Solution: (a) Use the quotient rule to find the derivative: $$f^{\prime }(x)=\frac{(1/x)(x)-(\ln x)(1)}{x^2}=\frac{1-\ln x}{x^2}$$ $f^{\prime }(x)$ is defined for all $x>0$, and equals zero only when $1-\ln x=0⟹x=e$. The only critical point on $(0,\infty )$ is $x=e$. (b) Candidates are the critical point $x=e$, plus limits at the open boundaries: ${\lim }_{x\to 0^{+}}\frac{\ln x}{x}=-\infty$, ${\lim }_{x\to \infty }\frac{\ln x}{x}=0$. Evaluate $f(e)=\frac{1}{e}\approx 0.368$. Since $f(x)$ approaches $-\infty$ as $x\to 0^{+}$, no absolute minimum exists. $\frac{1}{e}$ is larger than the limiting value of $0$, so the absolute maximum is $\frac{1}{e}$ at $x=e$. (c) On the closed interval $[1,e^2]$, $f$ is continuous. Candidates are endpoints $x=1,x=e^2$, and critical point $x=e$ inside the interval. Evaluate: $f(1)=0$, $f(e)=\frac{1}{e}$, $f(e^2)=\frac{2}{e^2}\approx 0.271$. Comparing values, the absolute maximum is $\frac{1}{e}$ at $x=e$, and the absolute minimum is $0$ at $x=1$.

Question 3 (Application / Real-World Style)

A company produces $x$ units of a good per day, and the daily profit (in thousands of dollars) is modeled by $P(x)=-0.001x^3+0.3x^2+2x-50$, for $0\le x\le 200$, where $x$ can be treated as a continuous variable. The company can produce between 0 and 200 units per day. Use the Candidates Test to find the maximum possible daily profit, and the number of units that produces this maximum.

Worked Solution: $P(x)$ is a polynomial, so it is continuous on $[0,200]$, so we apply the Candidates Test. First, compute the derivative: $P^{\prime }(x)=-0.003x^2+0.6x+2$. Set $P^{\prime }(x)=0$, multiply through by 1000 to get $3x^2-600x-2000=0$. Use the quadratic formula to solve: the only positive root is $x\approx 203.28$, which lies outside the interval $[0,200]$. The only candidates are the endpoints $x=0$ and $x=200$. Evaluate: $P(0)=-50$ (thousand dollars), $P(200)=-8000+12000+400-50=3350$. The maximum possible daily profit is $\$3,350,000$, achieved when the company produces 200 units per day. In context, the profit function is increasing over the entire allowed production interval, so the maximum profit occurs at the maximum allowed production level.

7. Quick Reference Cheatsheet

8. What's Next

Mastering the Candidates Test for absolute extrema is a critical prerequisite for the next core topic in Unit 5: optimization, where you will set up and solve real-world problems that require finding the maximum or minimum of a function over a specified interval. Without the systematic candidate-checking process you learned here, you will often miss extrema at endpoints or critical points with undefined derivatives, leading to incorrect solutions. This topic also supports later topics across the course, including particle motion problems where you find maximum speed or displacement over a time interval, and area/volume optimization problems in Unit 8: Applications of Integration. Next steps for your study: Optimization Mean Value Theorem First Derivative Test for Relative Extrema Particle Motion with Derivatives