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AP Calculus AB Practice Quiz: Estimating Derivatives of a Function at a Point

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

Test your understanding with short quizzes. This quiz has 9 questions to check your progress.

Question 1 of 9

The table below gives selected values for a differentiable function f(x). | x | 2.9 | 3.0 | 3.1 | |---|---|---|---| | f(x) | 8.41 | 9.00 | 9.61 | Which of the following is the best estimate for f'(3.0)?

All Questions (9)

The table below gives selected values for a differentiable function f(x). | x | 2.9 | 3.0 | 3.1 | |---|---|---|---| | f(x) | 8.41 | 9.00 | 9.61 | Which of the following is the best estimate for f'(3.0)?

A) 0.6

B) 6.0

C) 9.0

D) 18.02

Correct Answer: B

The derivative at a point can be estimated by the average rate of change over a small interval containing that point. Using the interval [2.9, 3.1], the estimate is (f(3.1) - f(2.9)) / (3.1 - 2.9) = (9.61 - 8.41) / 0.2 = 1.2 / 0.2 = 6.0. This is the slope of the secant line through the points closest to x=3.

The graph of a differentiable function g(x) is shown. Which of the following is the best estimate for g'(1)?

A) -2

B) -1/2

C) 1/2

D) 2

Correct Answer: A

The value of g'(1) is the slope of the line tangent to the graph of g(x) at x=1. By sketching a tangent line at x=1, one can see that it passes approximately through the points (1, 3) and (2, 1). The slope of this line is (1 - 3) / (2 - 1) = -2 / 1 = -2. This is the best estimate among the choices.

A graphing calculator is used to estimate the derivative of f(x) = x^3 at x = 2. Which of the following calculations does the calculator most likely perform to find the estimate?

A) It symbolically computes f'(x) = 3x^2 and evaluates it at x = 2.

B) It calculates the slope of a secant line over a very small interval, such as [1.999, 2.001].

C) It finds the area under the curve of f(x) from x=0 to x=2.

D) It determines the y-value of the function at x=2, which is f(2) = 8.

Correct Answer: B

Numerical derivative functions on technology (like graphing calculators) estimate the derivative at a point by calculating the average rate of change, or the slope of a secant line, over a very small interval symmetric around that point. This provides a close approximation of the instantaneous rate of change (the slope of the tangent line).

The amount of water in a tank, W(t), in gallons, is a differentiable function of time t in hours. Selected values of W(t) are given in the table below. | t (hours) | 0 | 3 | 5 | 8 | |---|---|---|---|---| | W(t) (gallons) | 150 | 135 | 125 | 110 | What is the best estimate for W'(5)?

A) -10 gallons/hour

B) -5 gallons/hour

C) 5 gallons/hour

D) 125 gallons/hour

Correct Answer: B

To estimate the derivative at t=5, we use the average rate of change over the smallest available interval containing t=5. Since the data points are not evenly spaced, we use the interval [3, 8]. The estimate is (W(8) - W(3)) / (8 - 3) = (110 - 135) / 5 = -25 / 5 = -5 gallons per hour.

The graph of the function f is shown, which has a sharp corner at x=2. Which of the following best describes the value of f'(2)?

A) f'(2) = 0

B) f'(2) = 3

C) f'(2) is positive.

D) f'(2) is undefined.

Correct Answer: D

A function is not differentiable at a point where its graph has a sharp corner or cusp. At x=2, the slope of the tangent line from the left is different from the slope of the tangent line from the right. Since the left-hand and right-hand derivatives are not equal, the derivative f'(2) does not exist and is therefore undefined.

When estimating a derivative f'(c) from a table of values, the most accurate estimate is typically found by using the values closest to c. If the table provides values at x = c-h and x = c+h for a small h > 0, the estimate is given by (f(c+h) - f(c-h)) / 2h. This is known as the symmetric difference quotient. This method is preferred over a one-sided estimate because it:

A) is the exact value of the derivative.

B) averages the rates of change on either side of c.

C) uses more data points than any other method.

D) is the only way to calculate a derivative from a table.

Correct Answer: B

The symmetric difference quotient effectively averages the slope of the secant line to the left of c and the slope of the secant line to the right of c. This balanced approach often provides a better approximation of the slope of the tangent line at c than a one-sided estimate (forward or backward difference), especially for smooth curves.

The temperature of a substance, in degrees Celsius, is modeled by a differentiable function T(m), where m is the time in minutes. Selected values of T(m) are given in the table. | m (minutes) | 0 | 2 | 4 | 6 | |---|---|---|---|---| | T(m) (°C) | 20 | 32 | 40 | 44 | Based on the data, what is the best estimate for T'(6)?

A) 2 °C/min

B) 4 °C/min

C) 6 °C/min

D) 8 °C/min

Correct Answer: A

Since m=6 is an endpoint of the data interval, we cannot calculate a symmetric difference. The best estimate is the average rate of change over the last available interval, which is [4, 6]. The estimate is (T(6) - T(4)) / (6 - 4) = (44 - 40) / 2 = 4 / 2 = 2 °C/min.

The graph of a differentiable function h(x) is shown. At which of the labeled points is h'(x) = 0?

A) Point A

B) Point B

C) Point C

D) Point D

Correct Answer: B

The derivative of a function at a point is the slope of the tangent line at that point. The derivative is zero where the tangent line is horizontal. On the given graph, the function has a local maximum at Point B, where the tangent line is horizontal, so its slope is 0.

Let v(t) be the velocity, in meters per second, of a particle moving along the x-axis. The table shows values of v(t) for selected times t. | t (sec) | 0 | 5 | 10 | 15 | |---|---|---|---|---| | v(t) (m/s) | 0 | 12 | 16 | 12 | Using the data from the table, what is the best approximation for the acceleration of the particle at t=5 seconds?

A) 0.8 m/s²

B) 1.6 m/s²

C) 2.4 m/s²

D) 12 m/s²

Correct Answer: B

Acceleration is the derivative of velocity, so we need to estimate v'(5). The best estimate uses the average rate of change over the interval symmetric about t=5, which is [0, 10]. The acceleration is approximately (v(10) - v(0)) / (10 - 0) = (16 - 0) / 10 = 1.6 m/s².