PrepGo

AP Calculus BC Practice Quiz: Using the Candidates Test to Determine Absolute (Global) Extrema

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 7 questions to check your progress.

Question 1 of 7

For a function f(x) that is continuous on the closed interval [a, b], the Candidates Test is used to find the absolute extrema. Which of the following are the only locations where an absolute extremum can occur?

All Questions (7)

For a function f(x) that is continuous on the closed interval [a, b], the Candidates Test is used to find the absolute extrema. Which of the following are the only locations where an absolute extremum can occur?

A) Points where f'(x) = 0 only.

B) The endpoints x=a and x=b only.

C) Points where f'(x) = 0 or f'(x) is undefined.

D) The endpoints x=a, x=b, and any critical points within (a, b).

Correct Answer: D

Based on the Extreme Value Theorem, the absolute (global) extrema of a continuous function on a closed interval can only occur at the critical points within the interval or at the endpoints of the interval. These locations are the 'candidates' that must be tested.

A student is trying to find the absolute maximum value of a differentiable function g(x) on the closed interval [-5, 5]. They find that g'(-2) = 0 and g''(-2) < 0, concluding that a local maximum occurs at x = -2. Why is this information insufficient to determine if g(-2) is the absolute maximum?

A) The second derivative test does not confirm a local maximum.

B) The function might have other critical points in the interval.

C) The values of the function at the endpoints, g(-5) and g(5), have not been considered.

D) The function may not be continuous on the interval.

Correct Answer: C

Absolute extrema on a closed interval can occur at critical points or at the endpoints. Even if x=-2 is a local maximum, the function's value at one of the endpoints (x=-5 or x=5) could be greater, making it the absolute maximum. The Candidates Test requires checking both critical points and endpoints.

Let f(x) = x³ - 3x² + 5. To find the absolute extrema of f on the closed interval [-1, 3], which of the following x-values must be tested?

A) -1, 0, 2, 3

B) 0, 2

C) -1, 3

D) -1, 1, 3

Correct Answer: A

First, find the critical points by taking the derivative: f'(x) = 3x² - 6x. Set f'(x) = 0: 3x(x - 2) = 0, which gives x=0 and x=2. Both critical points are within the interval [-1, 3]. The candidates for absolute extrema are the critical points and the endpoints. Therefore, the values to be tested are x = -1, 0, 2, 3.

What is the absolute minimum value of the function h(t) = 2t³ + 3t² - 12t on the interval [-3, 2]?

A) -20

B) -7

C) 4

D) 9

Correct Answer: B

First, find the critical points. h'(t) = 6t² + 6t - 12 = 6(t² + t - 2) = 6(t+2)(t-1). The critical points are t=-2 and t=1, both of which are in [-3, 2]. Now, test the candidates (endpoints and critical points) in the original function h(t): h(-3) = 2(-27) + 3(9) - 12(-3) = 9 h(-2) = 2(-8) + 3(4) - 12(-2) = 20 h(1) = 2(1) + 3(1) - 12(1) = -7 h(2) = 2(8) + 3(4) - 12(2) = 4 Comparing the values {9, 20, -7, 4}, the absolute minimum value is -7.

The function f is continuous on the closed interval [0, 8]. The derivative of f, f', has the property that f'(x) > 0 for 0 < x < 3, f'(3) = 0, and f'(x) < 0 for 3 < x < 8. Which of the following statements must be true?

A) The absolute maximum of f occurs at x=3.

B) The absolute minimum of f occurs at x=3.

C) The absolute maximum of f occurs at x=0 or x=3.

D) The absolute minimum of f occurs at x=0 or x=8.

Correct Answer: D

The behavior of the derivative f' indicates that f is increasing on (0, 3) and decreasing on (3, 8). This means f has a local maximum at x=3. For a continuous function on a closed interval, the absolute minimum must occur at either an endpoint or a local minimum. Since the only critical point is a local maximum, the absolute minimum must occur at one of the endpoints, x=0 or x=8. We cannot determine the absolute maximum without comparing f(0) and f(3).

If a function g is continuous on the closed interval [a, b] and g'(x) > 0 for all x in the open interval (a, b), where must the absolute maximum of g occur?

A) At x=a.

B) At x=b.

C) At a critical point c where a < c < b.

D) It is impossible to determine without more information.

Correct Answer: B

The condition g'(x) > 0 on (a, b) means that the function g is strictly increasing across the entire interval. For an increasing function, its value continuously rises from left to right. Therefore, the lowest value must be at the left endpoint (x=a) and the highest value must be at the right endpoint (x=b). The absolute maximum must occur at x=b.

A student is asked to find the absolute maximum of f(x) = 4x - x² on [0, 5]. The student's work is shown below: Step 1: f'(x) = 4 - 2x Step 2: 4 - 2x = 0 ⇒ x = 2 Step 3: f(2) = 4(2) - 2² = 8 - 4 = 4 Conclusion: The absolute maximum value is 4. Which statement best describes the error in the student's reasoning?

A) The derivative was calculated incorrectly.

B) The critical point was found incorrectly.

C) The student found a local extremum but failed to test the endpoints of the interval.

D) The student should have used the second derivative test to confirm it was a maximum.

Correct Answer: C

The student correctly identified the only critical point at x=2. However, to find the absolute maximum on a closed interval, one must compare the function's value at the critical point(s) with its value at the endpoints. The student failed to calculate and consider f(0) = 0 and f(5) = 4(5) - 5² = 20 - 25 = -5. Although f(2)=4 is the correct absolute maximum in this case, the justification is incomplete because the endpoints were not checked, which is a required step in the Candidates Test.