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AP Calculus BC Practice Quiz: Using the Mean Value Theorem

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 the function defined by f(x) = x^2 + 3x on the closed interval [0, 2]. The Mean Value Theorem guarantees the existence of a number c in the open interval (0, 2) such that f'(c) is equal to the average rate of change of f on [0, 2]. What is the value of c?

All Questions (7)

Let f be the function defined by f(x) = x^2 + 3x on the closed interval [0, 2]. The Mean Value Theorem guarantees the existence of a number c in the open interval (0, 2) such that f'(c) is equal to the average rate of change of f on [0, 2]. What is the value of c?

A) 1/2

B) 1

C) 3/2

D) 2

Correct Answer: B

First, find the average rate of change of f on [0, 2]: (f(2) - f(0)) / (2 - 0) = ((2^2 + 3*2) - (0^2 + 3*0)) / 2 = (4 + 6) / 2 = 5. Next, find the derivative: f'(x) = 2x + 3. According to the Mean Value Theorem, there is a c in (0, 2) such that f'(c) = 5. Set 2c + 3 = 5, which gives 2c = 2, so c = 1. This value is within the interval (0, 2).

A function f is continuous on the closed interval [1, 5] and differentiable on the open interval (1, 5). If f(1) = 12 and f(5) = 4, which of the following statements must be true?

A) f(x) must be a decreasing function on [1, 5].

B) There exists a number c in (1, 5) such that f(c) = 0.

C) There exists a number c in (1, 5) such that f'(c) = -2.

D) There exists a number c in (1, 5) such that f'(c) = 8.

Correct Answer: C

The Mean Value Theorem applies because the function is continuous on [1, 5] and differentiable on (1, 5). The theorem guarantees that there is a point c in (1, 5) where the instantaneous rate of change equals the average rate of change. The average rate of change is (f(5) - f(1)) / (5 - 1) = (4 - 12) / 4 = -8 / 4 = -2. Therefore, there must exist a c in (1, 5) such that f'(c) = -2.

For which of the following functions does the Mean Value Theorem NOT apply on the interval [-2, 2]?

A) f(x) = x^4 - x^2

B) f(x) = 1 / (x + 3)

C) f(x) = |x|

D) f(x) = sin(x)

Correct Answer: C

The Mean Value Theorem requires the function to be continuous on the closed interval [-2, 2] and differentiable on the open interval (-2, 2). The function f(x) = |x| is continuous on [-2, 2], but it is not differentiable at x = 0, which is within the open interval (-2, 2). Therefore, the conditions of the Mean Value Theorem are not met for this function.

A particle moves along the x-axis so that its position at any time t ≥ 0 is given by a differentiable function x(t). At time t = 1, the particle is at position x = 3, and at time t = 4, the particle is at position x = 15. Which of the following statements must be true for some time t in the interval (1, 4)?

A) The instantaneous velocity of the particle is 4.

B) The instantaneous velocity of the particle is 12.

C) The acceleration of the particle is 0.

D) The particle is at position x = 9.

Correct Answer: A

The function x(t) is differentiable, which implies it is also continuous. The Mean Value Theorem can be applied to x(t) on the interval [1, 4]. The average velocity (average rate of change of position) over this interval is (x(4) - x(1)) / (4 - 1) = (15 - 3) / 3 = 12 / 3 = 4. The Mean Value Theorem guarantees there is a time t in (1, 4) where the instantaneous velocity, x'(t), is equal to the average velocity. Therefore, for some t in (1, 4), the instantaneous velocity must be 4.

Let f be a differentiable function for all real numbers such that f(0) = 5 and f'(x) ≤ 2 for all x. What is the largest possible value for f(3)?

A) 7

B) 8

C) 10

D) 11

Correct Answer: D

By the Mean Value Theorem on the interval [0, 3], there exists a number c in (0, 3) such that f'(c) = (f(3) - f(0)) / (3 - 0). Substituting the known values, we get f'(c) = (f(3) - 5) / 3. We are given that f'(x) ≤ 2 for all x, so f'(c) ≤ 2. Therefore, (f(3) - 5) / 3 ≤ 2. Multiplying by 3 gives f(3) - 5 ≤ 6. Adding 5 to both sides gives f(3) ≤ 11. The largest possible value for f(3) is 11.

The Mean Value Theorem guarantees that for a function f that is continuous on [a, b] and differentiable on (a, b), there is a point c in (a, b) where the tangent line to the graph of f is parallel to the secant line connecting the endpoints of the interval. For the function f(x) = x^3 on the interval [0, 3], what is the slope of this tangent line?

A) 3

B) 9

C) 18

D) 27

Correct Answer: B

The slope of the secant line connecting the endpoints (0, f(0)) and (3, f(3)) is the average rate of change over the interval. First, calculate the function values: f(0) = 0^3 = 0 and f(3) = 3^3 = 27. The slope of the secant line is (f(3) - f(0)) / (3 - 0) = (27 - 0) / 3 = 9. The Mean Value Theorem guarantees that there is a point c where the slope of the tangent line, f'(c), is equal to this average rate of change. Therefore, the slope of the parallel tangent line is 9.

Which of the following is a required condition for the Mean Value Theorem to apply to a function f on a closed interval [a, b]?

A) f must be differentiable on the closed interval [a, b].

B) f must be continuous on the open interval (a, b).

C) f must be differentiable on the open interval (a, b).

D) f(a) must equal f(b).

Correct Answer: C

The Mean Value Theorem has two primary conditions: 1) The function f must be continuous on the closed interval [a, b], and 2) The function f must be differentiable on the open interval (a, b). Option C correctly states one of these conditions. Option A is too strong; differentiability is only required on the open interval. Option B is too weak; continuity is required on the closed interval. Option D is a condition for Rolle's Theorem, a special case of the Mean Value Theorem.