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AP Calculus BC Practice Quiz: Sketching Graphs of Functions and Their Derivatives

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 graph of f', the derivative of a function f, is shown below. On which of the following intervals is the function f increasing?

All Questions (9)

The graph of f', the derivative of a function f, is shown below. On which of the following intervals is the function f increasing?

A) (-2, 1)

B) (-3, -1) and (1, 3)

C) (-1, 2)

D) (-3, -2) and (1, 3)

Correct Answer: A

A function f is increasing on intervals where its derivative, f', is positive. Based on the provided graph, f'(x) > 0 for all x in the interval (-2, 1). Therefore, f is increasing on the interval (-2, 1).

The graph of f', the derivative of a function f, is shown below. At which value of x does f have a local maximum?

A) x = -2

B) x = -1

C) x = 1

D) x = 2

Correct Answer: C

According to the First Derivative Test, a function f has a local maximum at a critical point x where f'(x) changes from positive to negative. Looking at the graph of f', the function changes from positive (above the x-axis) to negative (below the x-axis) at x = 1. Therefore, f has a local maximum at x = 1.

The graph of f', the derivative of a function f, is shown below. On which of the following intervals is the graph of f concave down?

A) (-∞, -1) and (2, ∞)

B) (-1, 2)

C) (-2, 1)

D) (-∞, -2) and (1, 3)

Correct Answer: A

The graph of a function f is concave down on intervals where its second derivative, f'', is negative. This corresponds to the intervals where the first derivative, f', is decreasing. Based on the graph, f' is decreasing on the intervals (-∞, -1) and (2, ∞).

Let f be a twice-differentiable function. Selected values of f' and f'' are given in the table below. Which of the following statements must be true? | x | f'(x) | f''(x) | |---|---|---| | 0 | 3 | 5 | | 1 | 0 | -2 | | 2 | -4 | 0 | | 3 | 0 | 4 |

A) f has a local maximum at x = 1.

B) f has a local maximum at x = 3.

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

D) The graph of f is concave down on the interval (2, 3).

Correct Answer: A

To determine if f has a local extremum, we use the Second Derivative Test at the critical points where f'(x) = 0. At x = 1, f'(1) = 0 and f''(1) = -2. Since f''(1) < 0, the graph of f is concave down at x=1, and thus f has a local maximum at x = 1. At x = 3, f'(3) = 0 and f''(3) = 4. Since f''(3) > 0, f has a local minimum at x = 3. A point of inflection requires f'' to change sign, which we cannot confirm at x=2 with the given information.

Let f be a function with a first derivative given by f'(x) = x(x - 2)(x + 3)². On which of the following intervals is f decreasing?

A) (-3, 0)

B) (0, 2)

C) (-∞, -3) and (2, ∞)

D) (-∞, 0) and (2, ∞)

Correct Answer: B

A function f is decreasing when its derivative f' is negative. We need to find the intervals where f'(x) = x(x - 2)(x + 3)² < 0. The critical numbers are x = -3, 0, and 2. The term (x+3)² is always non-negative. Thus, the sign of f'(x) is determined by the sign of x(x - 2). This product is negative when one factor is positive and the other is negative, which occurs on the interval (0, 2). Therefore, f is decreasing on (0, 2).

The graph of f'', the second derivative of the function f, is shown below. If f'(1) = 0, what can be concluded about the function f at x = 1?

A) f has a local maximum at x = 1.

B) f has a local minimum at x = 1.

C) f has a point of inflection at x = 1.

D) The behavior of f at x = 1 cannot be determined from the information given.

Correct Answer: B

We are given that x = 1 is a critical point of f because f'(1) = 0. We can use the Second Derivative Test to classify this critical point. From the graph of f'', we can see that f''(1) is a positive value. Since f'(1) = 0 and f''(1) > 0, the function f has a local minimum at x = 1.

The graphs of three functions, labeled I, II, and III, are shown below. One is the graph of a function f, one is the graph of its derivative f', and one is the graph of its second derivative f''. Which of the following correctly identifies the graphs?

A) f = I, f' = II, f'' = III

B) f = II, f' = III, f'' = I

C) f = II, f' = I, f'' = III

D) f = III, f' = I, f'' = II

Correct Answer: C

We can identify the relationships by looking at extrema and zeros. Graph II has horizontal tangents (extrema) at the x-values where Graph I is zero. This suggests that I is the derivative of II. Furthermore, Graph I has a horizontal tangent (extremum) at the x-value where Graph III is zero. This suggests that III is the derivative of I. Therefore, the correct relationship is f = II, f' = I, and f'' = III.

The table below gives selected values for f', the derivative of a twice-differentiable function f. Based on the table, on which interval must the graph of f be concave up? | x | f'(x) | |---|---| | 0 | -5 | | 1 | -2 | | 2 | 0 | | 3 | 1 | | 4 | 1.5 |

A) (0, 1)

B) (1, 2)

C) (2, 3)

D) (0, 4)

Correct Answer: D

The graph of a function f is concave up on intervals where its second derivative, f'', is positive. This occurs when the first derivative, f', is increasing. Looking at the table, the values of f'(x) are consistently increasing from -5 to 1.5 as x increases from 0 to 4. Therefore, f' is increasing on the interval (0, 4), which implies that f must be concave up on this interval.

The graph of f', the derivative of a continuous function f, is shown for the interval [0, 7]. The graph of f' has horizontal tangent lines at x = 1.5 and x = 5. Which of the following statements about f is true?

A) f is decreasing on the interval (3, 7).

B) f has a local maximum at x = 3.

C) The graph of f has a point of inflection at x = 3.

D) The graph of f has points of inflection at x = 1.5 and x = 5.

Correct Answer: D

Points of inflection on the graph of f occur where f'' changes sign. This corresponds to points where f' has a local extremum. The problem states that f' has horizontal tangent lines at x = 1.5 and x = 5, which means f'(x) has local extrema at these points. At x = 1.5, f' changes from increasing to decreasing, so f'' changes from positive to negative. At x = 5, f' changes from decreasing to increasing, so f'' changes from negative to positive. Therefore, the graph of f has points of inflection at both x = 1.5 and x = 5.