Candidates test for absolute extrema — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: Identification of critical points, endpoints, and candidate points; the step-by-step candidate test procedure for closed, open, and infinite intervals; classification of candidates as absolute maxima or minima for continuous and discontinuous functions.

You should already know: How to compute derivatives using all basic differentiation rules. How to find critical points of a function. The definition of absolute maximum and absolute minimum on an interval.

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 AB 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 (also called the Extreme Value Candidates Test or Closed Interval Method) is the standard algorithm for finding the absolute maximum and absolute minimum values of a function on a given interval. Per the AP Calculus AB Course and Exam Description (CED), this topic accounts for approximately 1-2% of total exam score, and appears in both multiple-choice (MCQ) and free-response (FRQ) sections, most often as part of a larger optimization or graph analysis question. 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 extrema can only occur at two types of points: critical points inside the interval, or endpoints of the interval. All such points are called candidate points, because they are the only points we need to test to find absolute extrema. Unlike the First or Second Derivative Tests, which only classify local extrema, the Candidates Test directly identifies global (absolute) extrema, which are the values most frequently requested on the AP exam.

2. The Candidates Test on Closed Intervals

The most common scenario for AP exam problems is finding absolute extrema on a closed bounded interval $[a,b]$, and this case forms the base of the Candidates Test. The Extreme Value Theorem guarantees that any continuous function on a closed interval will have both an absolute maximum and an absolute minimum, so we just need to locate them via the following step-by-step procedure:

1. Confirm the function $f(x)$ 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 defined as points where $f^{\prime }(x)=0$, or $f^{\prime }(x)$ is undefined but $f(x)$ is defined.
3. Add the two endpoints $x=a$ and $x=b$ to your list of candidate points, along with any valid critical points from step 2.
4. Evaluate $f(x)$ at every candidate point on the list.
5. The largest output value is the absolute maximum of $f(x)$ on $[a,b]$, and the smallest output value is the absolute minimum.

The intuition behind this procedure is simple: absolute extrema cannot occur anywhere outside the list of candidates, so we only need to compare the function values at these points to find the global extrema.

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: $f^{\prime }(x)=3x^2-6x=3x(x-2)$. Set $f^{\prime }(x)=0$ to find critical points at $x=0$ and $x=2$, both of which lie strictly inside $(-1,4)$. There are no points where $f^{\prime }(x)$ is undefined, so these are our only critical points.
3. Full list of candidates: $x=-1,0,2,4$.
4. Evaluate $f(x)$ at each candidate:
   • $f(-1)=(-1{)}^3-3(-1{)}^2+1=-3$
   • $f(0)=0-0+1=1$
   • $f(2)=8-12+1=-3$
   • $f(4)=64-48+1=17$
5. 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 $x=2$.

Exam 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.

3. The Candidates Test on Open 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 (since 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 ${\lim }_{x\to -\infty }f(x)$). These limits are candidate values for comparison.
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.

The intuition is that on an open interval, a function can grow without bound or approach an extremum at the endpoint that it never actually reaches, so we cannot assume an absolute extremum exists.

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: $f^{\prime }(x)=2x-4=2(x-2)$. The only critical point is $x=2$, which is inside $(0,3)$, and there are no points where $f^{\prime }(x)$ is undefined.
3. Evaluate $f(x)$ at the critical point: $f(2)=4-8+3=-1$.
4. Calculate limits at the open endpoints: ${\lim }_{x\to 0^{+}}f(x)=3$, ${\lim }_{x\to 3^{-}}f(x)=0$.
5. 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.

Exam 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.

4. Candidates Test for Functions with Discontinuities

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. The procedure is:

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.

Worked Example

Find the absolute maximum and absolute minimum of $f(x)=\{\begin{array}{ll}{x}^2+1,& -2\le x\le 0\\ 3x-1,& 0<x\le 2\end{array}$ 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 $-2<x<0$, $f^{\prime }(x)=2x$, which equals zero only at $x=0$. For $0<x<2$, $f^{\prime }(x)=3$, 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 $f(x)$: $f(-2)=5$, $f(0)=1$, $f(2)=5$.
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 $[-2,2]$.

Exam 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.

5. Common Pitfalls (and how to avoid them)

• 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 $f^{\prime }(x)$ is undefined (but $f(x)$ is defined). Why: Students only set $f^{\prime }(x)=0$ to find critical points, and forget the second part of the critical point definition. Correct move: After finding where $f^{\prime }(x)=0$, check for points where $f^{\prime }(x)$ is undefined but $f(x)$ is defined, add any inside the interval to your candidate list.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

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. $41$

Worked Solution: First, $f(x)$ is a polynomial, so it is continuous on $[0,5]$, and we can apply the Candidates Test for closed intervals. Compute the derivative: $f^{\prime }(x)=6x^2-30x+24=6(x-1)(x-4)$. Critical points at $x=1$ and $x=4$, both inside $(0,5)$. The full list of candidates is $x=0,1,4,5$. Evaluate $f(x)$ at each candidate: $f(0)=2$, $f(1)=13$, $f(4)=-8$, $f(5)=-3$. The smallest function value is $-8$, so this is the absolute minimum. 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 the interval $(0,\infty )$. (b) Use the Candidates Test to determine whether $f(x)$ has an absolute maximum and an absolute minimum on $(0,\infty )$. Justify your answer. (c) Find the absolute maximum value if it exists.

Worked Solution: (a) Use the quotient rule to compute the derivative: $$f^{\prime }(x)=\frac{\left(\frac{1}{x}\right)(x)-(\ln x)(1)}{x^2}=\frac{1-\ln x}{x^2}$$ Set $f^{\prime }(x)=0$: the numerator equals zero when $1-\ln x=0⟹x=e$. $f^{\prime }(x)$ is undefined at $x=0$, but $f(x)$ is also undefined at $x=0$, so it is not a critical point. The only critical point is $x=e$.

(b) We are on the open infinite interval $(0,\infty )$. We have one critical point at $x=e$ with $f(e)=\frac{1}{e}\approx 0.368$. Calculate the limits at the endpoints:

• ${\lim }_{x\to 0^{+}}\frac{\ln x}{x}=-\infty$, since $\ln x\to -\infty$ as $x\to 0^{+}$.
• ${\lim }_{x\to \infty }\frac{\ln x}{x}=0$ by L'Hospital's rule. Since $f(x)$ can take arbitrarily small negative values, there is no absolute minimum. The largest candidate value is $\frac{1}{e}$, attained at $x=e$ inside the interval, so there is an absolute maximum.

(c) The absolute maximum value is $\frac{1}{e}$.

Question 3 (Application / Real-World Style)

A company produces $x$ units of a good per day, with daily profit given by $P(x)=-0.5x^3+12x^2-75x+100$, measured in hundreds of dollars, for $0\le x\le 12$ units. What is the maximum possible daily profit the company can earn, and how many units must be produced to achieve it?

Worked Solution: $P(x)$ is a polynomial, so it is continuous on the closed interval $[0,12]$. Compute the derivative: $P^{\prime }(x)=-1.5x^2+24x-75$. Set $P^{\prime }(x)=0$ and solve for $x$: $x=8\pm \sqrt{14}\approx 4.26$ and $11.74$, both inside $(0,12)$. Candidates are $x=0,4.26,11.74,12$. Evaluate $P(x)$: $P(0)=100$, $P(4.26)\approx -40.4$, $P(11.74)\approx 65.3$, $P(12)=64$. The maximum value is approximately $65.3$ hundred dollars at $x\approx 11.74$ units.

In context: The company can earn a maximum daily profit of approximately $\$6530$, achieved by producing roughly 12 units per day.

7. Quick Reference Cheatsheet

8. 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 only reliable general method to find those absolute extrema is the Candidates Test. Without correctly identifying and testing all candidate points, you will not be able to find the correct optimal value in any optimization problem, which is a common high-weight FRQ topic on the AP exam. This topic also connects to curve sketching, where you need to identify absolute extrema to accurately graph a function and describe its global behavior over an interval. Next you will extend this technique to solve constrained optimization problems, where you must first build the target function before testing for extrema.

• Optimization of functions
• First derivative test for local extrema
• Second derivative test for concavity and extrema
• Curve sketching with derivatives