PrepGo

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

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

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

Question 1 of 7

For a continuous function f on the closed interval [a, b], the absolute extrema are guaranteed to occur at which of the following locations?

All Questions (7)

For a continuous function f on the closed interval [a, b], the absolute extrema are guaranteed to occur at which of the following locations?

A) Only at the endpoints x=a and x=b.

B) Only at the critical points of f in the open interval (a, b).

C) At the critical points of f in the open interval (a, b) or at the endpoints x=a and x=b.

D) Only at points where the second derivative, f''(x), is equal to zero.

Correct Answer: C

This question directly tests the core concept of the Candidates Test. The provided content states that 'Absolute (global) extrema of a function on a closed interval can only occur at critical points or at endpoints.'

Let f(x) = x³ - 6x² + 9. What are all the candidate x-values for the location of the absolute extrema of f on the closed interval [-1, 5]?

A) x = 0, x = 4

B) x = -1, x = 5

C) x = -1, x = 0, x = 4, x = 5

D) x = -1, x = 2, x = 5

Correct Answer: C

To find the candidates for absolute extrema on a closed interval, we must identify the endpoints and all critical points within the interval. The endpoints are x=-1 and x=5. To find the critical points, we find the derivative: f'(x) = 3x² - 12x. Setting f'(x) = 0 gives 3x(x - 4) = 0, so the critical points are x=0 and x=4. Both critical points are within the interval [-1, 5]. Therefore, the complete set of candidates is {-1, 0, 4, 5}.

When using the Candidates Test for a differentiable function on a closed interval, what is the primary justification for including critical points in the search for absolute extrema?

A) An absolute extremum is always located at a critical point.

B) The behavior of a function's derivative being zero or undefined signals a potential local maximum or minimum, which could also be the absolute extremum.

C) The endpoints of a closed interval are, by definition, also critical points.

D) A function must have a critical point on any closed interval.

Correct Answer: B

This question links the two pieces of provided content. The behavior of the derivative (being zero or undefined) identifies critical points. These points are potential locations for local extrema because the function may change from increasing to decreasing (or vice versa). An absolute extremum on an interval will either be one of these local extrema or occur at an endpoint.

The derivative of a function g is given by g'(x) = x² - x - 6. On the closed interval [-3, 4], what are all the candidate x-values for the location of the absolute extrema of g?

A) x = -2, x = 3

B) x = -3, x = 4

C) x = -3, x = -2, x = 3, x = 4

D) x = -3, x = 0, x = 4

Correct Answer: C

The candidates for absolute extrema are the endpoints and the critical points within the interval. The endpoints are x=-3 and x=4. Critical points occur where g'(x) = 0. We solve x² - x - 6 = 0, which factors to (x-3)(x+2) = 0. The critical points are x=3 and x=-2. Both of these values are within the interval [-3, 4]. Thus, the complete set of candidates is {-3, -2, 3, 4}.

A function h is continuous on the closed interval [0, 10]. The critical points of h are found to be x = -2, x = 3, and x = 12. Which of the following sets contains all the candidates for the location of the absolute maximum of h on [0, 10]?

A) { -2, 0, 3, 10, 12 }

B) { 3 }

C) { 0, 10 }

D) { 0, 3, 10 }

Correct Answer: D

According to the Candidates Test, absolute extrema on a closed interval can only occur at the endpoints or at critical points *within* the interval. The endpoints are x=0 and x=10. Of the given critical points x = -2, 3, and 12, only x=3 lies within the interval [0, 10]. Therefore, the complete and correct set of candidates is {0, 3, 10}.

If a function f is continuous on a closed interval [a, b], which statement provides the most accurate justification for the procedure of testing only critical points and endpoints for absolute extrema?

A) An absolute extremum must occur at a point where the derivative is zero, which is a critical point.

B) The Extreme Value Theorem guarantees an absolute extremum exists, and if it occurs in the interval's interior, it must be at a local extremum, which is a critical point.

C) Every continuous function must have at least one critical point in a closed interval where an extremum can occur.

D) An absolute extremum must occur at an endpoint because the function's domain is bounded there.

Correct Answer: B

This is the most precise theoretical justification. The Extreme Value Theorem guarantees the existence of absolute extrema for a continuous function on a closed interval. If an extremum is not at an endpoint, it must be in the interior (a, b). An interior extremum must also be a local extremum. By Fermat's Theorem, a point of local extremum must be a critical point (where the derivative is zero or undefined). Therefore, the search can be limited to endpoints and critical points. Option A is false (could be an endpoint or where f' is undefined). Option C is false (e.g., f(x)=x on [0,1]). Option D is false (could be a critical point).

The function f is continuous on the closed interval [-1, 5]. The derivative of f, f', has values given in the table below. | x | -1 | 0 | 1 | 2 | 3 | 4 | 5 | |------|----|---|---|---|---|---|---| | f'(x)| -4 | 0 | 2 | 3 | 0 |-2 |-5 | Based on the table, what are all the candidate x-values for the location of the absolute extrema of f on [-1, 5]?

A) x = 0, x = 3

B) x = -1, x = 5

C) x = -1, x = 0, x = 3, x = 5

D) x = -1, x = 1, x = 4, x = 5

Correct Answer: C

The candidates for the location of absolute extrema on a closed interval are the endpoints of the interval and any critical points within the interval. The endpoints are given as x=-1 and x=5. Critical points occur where the derivative is zero or undefined. From the table, we can see that f'(x) = 0 at x=0 and x=3. Both of these critical points are within the interval [-1, 5]. Therefore, the complete set of candidates is {-1, 0, 3, 5}.