AP Calculus AB Practice Quiz: Using the First Derivative Test to Determine Relative (Local) 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

Let f be a function with a derivative f'. If f'(x) < 0 for x < 2 and f'(x) > 0 for x > 2, which of the following statements provides a correct conclusion about the function f at x = 2?

A)f has a relative minimum at x = 2.B)f has a relative maximum at x = 2.C)f has a point of inflection at x = 2.D)The behavior of f at x = 2 cannot be determined from the given information.

All Questions (7)

Let f be a function with a derivative f'. If f'(x) < 0 for x < 2 and f'(x) > 0 for x > 2, which of the following statements provides a correct conclusion about the function f at x = 2?

A) f has a relative minimum at x = 2.

B) f has a relative maximum at x = 2.

C) f has a point of inflection at x = 2.

D) The behavior of f at x = 2 cannot be determined from the given information.

Correct Answer: A

According to the First Derivative Test, if the derivative of a function changes from negative to positive at a critical point, the function has a relative minimum at that point. Here, f' changes from negative (indicating f is decreasing) to positive (indicating f is increasing) at x = 2.

Let f be a function such that its derivative is given by f'(x) = x^2 - x - 6. At which value of x does f have a relative maximum?

A) x = -3

B) x = -2

C) x = 2

D) x = 3

Correct Answer: B

First, find the critical points by setting f'(x) = 0. Factoring gives f'(x) = (x-3)(x+2) = 0, so the critical points are x = 3 and x = -2. To determine the type of extremum, we test the sign of f' around these points. For x < -2, f' is positive. For -2 < x < 3, f' is negative. Since f' changes from positive to negative at x = -2, the function f has a relative maximum at x = -2.

The function f has a critical point at x = c. Which of the following provides a justification that f has a relative maximum at x = c?

A) f'(c) = 0.

B) The graph of f is concave down at x = c.

C) f'(x) changes from positive to negative at x = c.

D) f'(x) changes from negative to positive at x = c.

Correct Answer: C

The First Derivative Test is used to determine the location of relative extrema. The test states that a function has a relative maximum at a critical point c if its first derivative, f'(x), changes sign from positive to negative at x = c. This indicates the function itself changes from increasing to decreasing.

The derivative of a function g is given by g'(x) = (x-1)e^x. Which of the following statements is true about the function g?

A) g has a relative maximum at x = 1.

B) g has a relative minimum at x = 1.

C) g has a relative maximum at x = 0.

D) g has no relative extrema.

Correct Answer: B

To find relative extrema, first find critical points by setting g'(x) = 0. Since e^x is never zero, g'(x) = 0 only when x - 1 = 0, so x = 1 is the only critical point. For x < 1, the term (x-1) is negative, so g'(x) is negative. For x > 1, the term (x-1) is positive, so g'(x) is positive. Since g'(x) changes from negative to positive at x = 1, the function g has a relative minimum at x = 1.

The graph of f', the derivative of a function f, is shown. At which of the following x-values does f have a relative minimum?

A) x = -2

B) x = 0

C) x = 3

D) There is no relative minimum shown.

Correct Answer: C

A function f has a relative minimum at a critical point where its derivative f' changes from negative to positive. Based on the graph, f' is below the x-axis (negative) for the interval (-2, 3) and above the x-axis (positive) for the interval (3, ∞). Since f' changes from negative to positive at x = 3, f has a relative minimum at x = 3. At x = -2, f' changes from positive to negative, indicating a relative maximum for f.

Let f be a function with derivative f'(x) = x^3 - 3x. Which of the following statements is true?

A) f has a relative minimum at x = 0.

B) f has a relative maximum at x = -√3.

C) f has a relative minimum at x = √3.

D) f has a relative maximum at x = √3.

Correct Answer: C

First, find the critical points by setting f'(x) = 0: x^3 - 3x = x(x^2 - 3) = 0. The critical points are x = 0, x = √3, and x = -√3. We test the sign of f' on the intervals defined by these points. For the interval (0, √3), let's test x=1: f'(1) = 1-3 = -2. For the interval (√3, ∞), let's test x=2: f'(2) = 8-6 = 2. Since f'(x) changes from negative to positive at x = √3, the function f has a relative minimum at x = √3.

The derivative of the function f is given by f'(x) = (x+1)^2(x-3). At which of the critical points of f does the function have neither a relative maximum nor a relative minimum?

A) x = -1 only

B) x = 3 only

C) x = -1 and x = 3

D) f has relative extrema at all its critical points.

Correct Answer: A

The critical points are where f'(x) = 0, which are x = -1 and x = 3. We must test the sign of f' around these points. The term (x+1)^2 is always non-negative. For x < -1, (x-3) is negative, so f'(x) is negative. For -1 < x < 3, (x+1)^2 is positive and (x-3) is negative, so f'(x) is also negative. Since the sign of f' does not change at x = -1, f has neither a relative maximum nor a relative minimum there. At x = 3, f' changes from negative to positive, indicating a relative minimum.