The Candidates Test finds the absolute maximum and minimum of a continuous function on a closed interval. You collect the candidates, including critical points and the two endpoints, plug each into the original function, and compare the outputs. For AP Calculus, use the original function values, not derivative values, when choosing absolute extrema.

Candidates Test for AP Calculus

The Candidates Test is the AP Calculus procedure for finding absolute extrema on a closed interval. If a function is continuous on $[a,b]$ , the absolute maximum and absolute minimum can only occur at critical points inside the interval or at the endpoints.

The steps are simple: find where $f'(x)=0$ or $f'(x)$ is undefined, keep only the critical points in the interval, add the endpoints, evaluate the original function $f$ at every candidate, and compare the y-values. The largest y-value is the absolute maximum value, and the smallest y-value is the absolute minimum value.

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Why This Matters for the AP Calculus Exam

On a closed interval, absolute extrema can only happen at critical points or at the endpoints, so you only need to check a short list of locations. This is one of the cleanest, most reliable procedures in AP Calculus, which makes it a common multiple-choice setup and a frequent building block in free-response work where you analyze a function's behavior or solve an applied optimization problem.

The exam rewards clear reasoning, so naming your candidates, evaluating the original function (not the derivative) at each one, and comparing values gives you a complete, checkable argument. Strong communication here also carries over to later topics like optimization in 5.10 and 5.11.

Key Takeaways

Absolute extrema on a closed interval occur only at critical points or at the endpoints.
Critical points are where the first derivative equals zero or is undefined (and the point is in the interval).
Always plug candidates into the original function to get y-values, then compare.
The largest y-value is the absolute maximum; the smallest y-value is the absolute minimum.
Throw out any critical point that falls outside the given interval.
The question asks for either an x-location or a y-value, so read carefully and answer what is asked.

Finding Absolute Extrema

To find the absolute extrema of a function on a closed interval, apply the Candidates Test. Recall from topic 5.2 that absolute extrema are the maximum and minimum values of a function, and that the Extreme Value Theorem guarantees these exist when a function is continuous on a closed interval.

The key idea is that absolute extrema on a closed interval can only occur at critical points or at the endpoints. That gives you a finite list of places to check.

Candidates Test Steps

Find the critical points of the function on the interval. Critical points are where the first derivative equals zero or is undefined.
Evaluate the function at the critical points. This gives the y-values at those points.
Evaluate the function at the endpoints of the interval. This gives the y-values at the endpoints.
Compare the y-values of the critical points and endpoints. The largest y-value is the absolute maximum, and the smallest y-value is the absolute minimum.

Think of yourself as collecting evidence at a short list of suspects, called candidates, then comparing outputs to decide which one wins.

Candidates Test Walkthrough

Use the function:

$h(x)=x^3+3x^2+16$

At what $x$ does the absolute maximum of $h$ over the closed interval $[-4,0]$ occur?

Step 1: Find Critical Points

Take the derivative and set it equal to 0.

$h'(x)=3x^2+6x$

Since $h'(x)$ is defined everywhere, find where $h'(x)=0$ .

$0=3x^2+6x$

$0=3x(x+2)$

So there are critical points at $x=-2,0$ . Both lie in $[-4,0]$ , so both are candidates.

Step 2: Evaluate Critical Points

Evaluate $h(x)$ at $x=-2,0$ .

$h(-2)=20$

$h(0)=16$

Step 3: Evaluate Endpoints

Evaluate $h(x)$ at the endpoints of $[-4,0]$ .

$h(-4)=0$

$h(0)=16$

We already found $h(0)$ above.

Step 4: Compare Outputs

Compare all y-values to find the largest, since the question asks for the absolute maximum. Here $h(-2)=20$ is the largest.

So the absolute maximum of $h$ over $[-4,0]$ occurs at $x=-2$ .

How to Use This on the AP Calculus Exam

Problem Solving

List your candidates explicitly: critical points inside the interval plus both endpoints.
Find critical points by solving $f'(x)=0$ and also checking where $f'(x)$ is undefined.
Drop any critical point that is not inside the given interval.
Plug each candidate into the original function $f$ , not the derivative, to get y-values.
Compare y-values: largest is the absolute max, smallest is the absolute min.

Common Trap

Read the question wording carefully. If it asks "at what x" does the extremum occur, your answer is an x-value. If it asks for the absolute maximum value, your answer is a y-value. Mixing these up is an easy way to lose an otherwise correct problem.

Candidates Test Practice Problems

Candidates Test Question 1

Let $h(x)=-x^4+2x^2$ . At what $x$ does the absolute maximum of $h$ over the closed interval $[0,2]$ occur?

Candidates Test Question 2

Let $h(x)=-x^{5}+5x$ . At what $x$ does the absolute minimum of $h$ over the closed interval $[-2,1]$ occur?

Candidates Test Solution 1

Step 1: Find Critical Points

$h'(x)=-4x^3+4x$

Since $h'(x)$ is defined everywhere, find where $h'(x)=0$ .

$0=-4x^3+4x$

$0=-4x(x^2-1)$

$0=-4x(x+1)(x-1)$

Critical points are at $x=-1,0,1$ . Do all fit in $[0,2]$ ? No. Only $0$ and $1$ fall in the interval, so those are the only critical-point candidates.

Step 2: Evaluate Critical Points

$h(0)=0$

$h(1)= 1$

Step 3: Evaluate Endpoints

We already found $h(0)=0$ .

$h(2)=-8$

Step 4: Compare Outputs

Compare y-values to find the largest. Here $h(1)=1$ is the largest.

So the absolute maximum of $h$ over $[0,2]$ occurs at $x=1$ .

Candidates Test Solution 2

Step 1: Find Critical Points

$h'(x)=-5x^4+5$

Since $h'(x)$ is defined everywhere, find where $h'(x)=0$ .

$0=-5x^4+5$

$0=-5(x^4-1)$

$0=-5(x^2+1)(x+1)(x-1)$

The critical points are at $x=-1,1$ . Both fall in $[-2,1]$ , so both are candidates.

Step 2: Evaluate Critical Points

$h(-1)=-4$

$h(1)= 4$

Step 3: Evaluate Endpoints

$h(-2)=22$

We already found $h(1)=4$ .

Step 4: Compare Outputs

Compare y-values to find the smallest. Here $h(-1)=-4$ is the smallest.

So the absolute minimum of $h$ over $[-2,1]$ occurs at $x=-1$ .

Common Misconceptions

Forgetting the endpoints. Endpoints are candidates too. A critical point is not automatically the absolute extremum; an endpoint can beat it.
Stopping at critical points. Finding critical points only locates where extrema might be. You still must evaluate and compare to decide.
Using critical points outside the interval. Only critical points inside the given closed interval count. Throw out the rest.
Plugging into the derivative instead of the function. Compare values of the original function $f$ , not $f'$ , when deciding which candidate wins.
Confusing critical points are undefined only when f' is zero. A critical point also occurs where $f'$ is undefined, such as a corner or vertical tangent inside the interval. Do not skip those.
Mixing up x and y answers. "Where" the extremum occurs is an x-value; the extremum "value" is a y-value. Answer what the question actually asks.

Vocabulary

The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.

Term	Definition
absolute extrema	The maximum or minimum values of a function over its entire domain or a specified interval.
critical point	A point in the domain of a function where the derivative is zero or undefined, which are candidates for local and absolute extrema.
derivative	The instantaneous rate of change of a function at a specific point, representing the slope of the tangent line to the function at that point.
endpoints	The boundary points of a closed interval where a function's absolute extrema may occur.

Frequently Asked Questions

What is the Candidates Test in AP Calculus?

The Candidates Test is a method for finding absolute maximum and minimum values on a closed interval. You evaluate the original function at critical points and endpoints, then compare the outputs.

When can you use the Candidates Test?

Use it when a function is continuous on a closed interval. The Extreme Value Theorem guarantees absolute extrema exist, and the Candidates Test tells you where to check.

What are candidates for absolute extrema?

Candidates are the endpoints of the interval and any critical points inside the interval. Critical points occur where f prime equals zero or where f prime is undefined.

Do endpoints count in the Candidates Test?

Yes. Endpoints are always candidates on a closed interval, and an endpoint can be the absolute maximum or absolute minimum.

Do you plug candidates into f or f prime?

Plug candidates into the original function f. The derivative helps you find critical points, but the actual extrema are based on the y-values of the function.

What is the biggest AP Calc mistake with the Candidates Test?

The biggest mistake is forgetting endpoints or answering with the wrong type of value. If the question asks where the extremum occurs, give an x-value; if it asks for the extremum value, give a y-value.