AP Calculus AB Practice Quiz: Determining Intervals on Which a Function Is Increasing or Decreasing

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 first derivative given by f'(x) = x² - 9. For which of the following intervals is the function f(x) increasing?

A)(-3, 3)B)(-∞, -3) onlyC)(-∞, -3) and (3, ∞)D)(3, ∞) only

All Questions (7)

Let f be a function with a first derivative given by f'(x) = x² - 9. For which of the following intervals is the function f(x) increasing?

A) (-3, 3)

B) (-∞, -3) only

C) (-∞, -3) and (3, ∞)

D) (3, ∞) only

Correct Answer: C

A function f(x) is increasing on intervals where its first derivative, f'(x), is positive. First, find the critical points by setting f'(x) = 0. This gives x² - 9 = 0, so x = -3 and x = 3. These points divide the number line into three intervals: (-∞, -3), (-3, 3), and (3, ∞). Test a value from each interval in f'(x). For x < -3 (e.g., x=-4), f'(-4) = (-4)² - 9 = 16 - 9 = 7 > 0. For -3 < x < 3 (e.g., x=0), f'(0) = 0² - 9 = -9 < 0. For x > 3 (e.g., x=4), f'(4) = 4² - 9 = 16 - 9 = 7 > 0. Therefore, f(x) is increasing on the intervals (-∞, -3) and (3, ∞) because f'(x) > 0 on these intervals.

The derivative of the function g is given by g'(x) = (x + 4)(x - 2). On which of the following intervals is g(x) decreasing?

A) (-4, 2)

B) (-∞, -4) only

C) (2, ∞) only

D) (-∞, -4) and (2, ∞)

Correct Answer: A

A function g(x) is decreasing on intervals where its derivative, g'(x), is negative. The critical points are found by setting g'(x) = 0, which yields x = -4 and x = 2. We test the intervals (-∞, -4), (-4, 2), and (2, ∞). For the interval (-4, 2), let's test x = 0. g'(0) = (0 + 4)(0 - 2) = (4)(-2) = -8. Since g'(x) < 0 on this interval, g(x) is decreasing on (-4, 2). For the other intervals, g'(x) is positive, meaning g(x) is increasing.

Let f be a differentiable function. Which of the following provides the justification that f(x) is increasing on the interval (a, b)?

A) f'(x) > 0 for all x in (a, b).

B) f'(x) < 0 for all x in (a, b).

C) f(x) > 0 for all x in (a, b).

D) The graph of f(x) has a local minimum in (a, b).

Correct Answer: A

The first derivative test provides information about the behavior of a function. The core principle is that a function f(x) is increasing on an interval if its first derivative, f'(x), is positive on that same interval. The other options are incorrect: f'(x) < 0 indicates the function is decreasing, f(x) > 0 indicates the function is above the x-axis, and a local minimum cannot exist on an interval where the function is strictly increasing.

The graph of f'(x), the derivative of a function f(x), is shown above. On which of the following intervals is the function f(x) decreasing?

A) (-∞, -2) and (1, ∞)

B) (-2, 1)

C) (-∞, -0.5)

D) (-0.5, ∞)

Correct Answer: B

A function f(x) is decreasing on intervals where its derivative, f'(x), is negative. By looking at the provided graph of f'(x), we need to identify the x-intervals for which the graph is below the x-axis (i.e., where f'(x) < 0). The graph of f'(x) is below the x-axis between x = -2 and x = 1. Therefore, the function f(x) is decreasing on the interval (-2, 1).

The derivative of a function f is given by f'(x) = (x - 5)e^x. On which interval is the function f increasing?

A) (-∞, 5)

B) (-∞, ∞)

C) (0, 5)

D) (5, ∞)

Correct Answer: D

To find where f is increasing, we must determine where f'(x) > 0. The term e^x is always positive for all real numbers x. Therefore, the sign of f'(x) is determined entirely by the sign of the term (x - 5). The expression (x - 5) is positive when x > 5. Thus, f'(x) > 0 when x > 5, and the function f is increasing on the interval (5, ∞).

Let f be a function with derivative f'(x) = x²(x + 2)(x - 4). On which of the following intervals is f decreasing?

A) (-2, 4)

B) (-2, 0) and (0, 4)

C) (-∞, -2) and (4, ∞)

D) (-∞, -2) only

Correct Answer: A

A function f is decreasing where its derivative f'(x) is negative. The critical points are x = 0, x = -2, and x = 4. The term x² is always non-negative. For x ≠ 0, x² is positive and does not affect the sign of f'(x). Therefore, we only need to test the sign of (x + 2)(x - 4). This is a parabola opening upwards with roots at -2 and 4, so it is negative between the roots. Thus, f'(x) < 0 for x in (-2, 0) and (0, 4). Since f'(0) = 0, the function is momentarily flat but does not change its decreasing behavior across x=0. Therefore, f is decreasing on the entire interval (-2, 4).

Let g be a function with a first derivative g'(x) = cos(x). On the interval (0, 2π), on which of the following intervals is g(x) increasing?

A) (0, π/2) and (3π/2, 2π)

B) (π/2, 3π/2)

C) (0, π)

D) (π, 2π)

Correct Answer: A

The function g(x) is increasing on intervals where its derivative, g'(x) = cos(x), is positive. On the interval (0, 2π), the cosine function is positive in the first and fourth quadrants. The first quadrant corresponds to the interval (0, π/2). The fourth quadrant corresponds to the interval (3π/2, 2π). Therefore, g(x) is increasing on (0, π/2) and (3π/2, 2π).