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AP Calculus AB 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 a function that is continuous on the closed interval [a, b] and differentiable on the open interval (a, b). The Mean Value Theorem guarantees the existence of a point c within the open interval (a, b) where which of the following is true?

All Questions (7)

Let f be a function that is continuous on the closed interval [a, b] and differentiable on the open interval (a, b). The Mean Value Theorem guarantees the existence of a point c within the open interval (a, b) where which of the following is true?

A) The instantaneous rate of change of f at c equals the average rate of change of f over [a, b].

B) The function f has a local maximum or minimum at c.

C) The instantaneous rate of change of f at c is zero.

D) The function f equals its average value over the interval [a, b] at c.

Correct Answer: A

The Mean Value Theorem states that if a function f is continuous on [a, b] and differentiable on (a, b), then there is at least one point c in (a, b) such that the instantaneous rate of change, f'(c), is equal to the average rate of change over the interval, (f(b) - f(a))/(b - a). [cite: 2481]

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

A) f(x) = x^3 - x

B) f(x) = 1 / (x^2 + 1)

C) f(x) = |x - 1|

D) f(x) = cos(x)

Correct Answer: C

The Mean Value Theorem requires the function to be continuous on the closed interval and differentiable on the open interval. The function f(x) = |x - 1| is continuous on [-2, 2], but it is not differentiable at x = 1, which is within the open interval (-2, 2). Therefore, the theorem's conditions are not met. [cite: 2481]

Consider the function f(x) = x^2 + 2x on the interval [0, 4]. According to the Mean Value Theorem, what value of c in (0, 4) satisfies the conclusion that the instantaneous rate of change at c equals the average rate of change over the interval?

A) c = 1

B) c = 2

C) c = 3

D) c = 4

Correct Answer: B

First, find the average rate of change: (f(4) - f(0)) / (4 - 0) = ((4^2 + 2*4) - (0^2 + 2*0)) / 4 = (16 + 8) / 4 = 24 / 4 = 6. Next, find the instantaneous rate of change: f'(x) = 2x + 2. The Mean Value Theorem guarantees a point c where f'(c) = 6. So, 2c + 2 = 6, which gives 2c = 4, and c = 2. [cite: 2481]

A car travels 180 miles in a 3-hour period. The car's position is given by a function that is continuous and differentiable. Which conclusion can be justified by applying the Mean Value Theorem?

A) The car's instantaneous velocity was exactly 60 mph at least once during the trip.

B) The car's acceleration must have been zero at some point during the trip.

C) The car traveled at a constant speed of 60 mph for the entire 3 hours.

D) The car's maximum speed was 60 mph.

Correct Answer: A

The average velocity (rate of change) over the 3-hour interval is 180 miles / 3 hours = 60 mph. Since the position function is continuous and differentiable, the Mean Value Theorem guarantees that there must be at least one point in time during the interval where the instantaneous velocity (instantaneous rate of change) was exactly equal to the average velocity of 60 mph. [cite: 2480, 2481]

Let f be a function that is differentiable for all real numbers. If f(1) = 5 and f(5) = 21, which of the following statements must be true?

A) f(x) must be a linear function.

B) There exists a number c in the interval (1, 5) such that f(c) = 13.

C) There exists a number c in the interval (1, 5) such that f'(c) = 4.

D) The derivative f'(x) is positive for all x in the interval (1, 5).

Correct Answer: C

The function f is differentiable, which implies it is also continuous. The conditions for the Mean Value Theorem are met on the interval [1, 5]. The average rate of change is (f(5) - f(1)) / (5 - 1) = (21 - 5) / 4 = 16 / 4 = 4. The Mean Value Theorem justifies the conclusion that there must be a point c in (1, 5) where the instantaneous rate of change, f'(c), equals this average rate of change. [cite: 2480]

Geometrically, the conclusion of the Mean Value Theorem states 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...

A) the tangent line to the graph of f is horizontal.

B) the tangent line to the graph of f is parallel to the secant line connecting the points (a, f(a)) and (b, f(b)).

C) the secant line connecting the points (a, f(a)) and (b, f(b)) intersects the graph of f at a third point.

D) the tangent line to the graph of f is perpendicular to the secant line connecting the points (a, f(a)) and (b, f(b)).

Correct Answer: B

The instantaneous rate of change, f'(c), represents the slope of the tangent line at x=c. The average rate of change, (f(b) - f(a))/(b - a), represents the slope of the secant line connecting the endpoints of the interval. The Mean Value Theorem guarantees these two values are equal at some point c, meaning the tangent line is parallel to the secant line. [cite: 2481]

Let f be a function that is continuous on [2, 7] and differentiable on (2, 7). If f(2) = -4 and the derivative f'(x) ≤ 3 for all x in (2, 7), what is the maximum possible value of f(7)?

A) 11

B) 15

C) 19

D) 21

Correct Answer: A

By the Mean Value Theorem, there exists a c in (2, 7) such that f'(c) = (f(7) - f(2)) / (7 - 2). We can rewrite this as f'(c) = (f(7) - (-4)) / 5. Since we know f'(x) ≤ 3 for all x in the interval, it must be that f'(c) ≤ 3. Therefore, (f(7) + 4) / 5 ≤ 3. Multiplying by 5 gives f(7) + 4 ≤ 15. Subtracting 4 gives f(7) ≤ 11. The maximum possible value for f(7) is 11. [cite: 2480]