AP Calculus AB 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
All Questions (9)
A) (-∞, -2) and (0, 3)
B) (-1, 2)
C) (-2, 0) and (3, ∞)
D) (-∞, -1) and (2, ∞)
Correct Answer: A
A function $f$ is increasing on intervals where its derivative, $f'$, is positive. Based on the provided graph, the value of $f'(x)$ is positive (the graph is above the x-axis) on the intervals (-∞, -2) and (0, 3).
A) x = -1
B) x = 0
C) x = 2
D) x = 3
Correct Answer: D
A function $f$ has a relative maximum at a critical point where its derivative, $f'$, changes from positive to negative. According to the graph, $f'(x)$ is positive just before x = 3 and negative just after x = 3. Therefore, $f$ has a relative maximum at x = 3.
A) (-∞, -1) and (2, ∞)
B) (-1, 2)
C) (-2, 0) and (3, ∞)
D) (-∞, -2) and (0, 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, ∞).
A) x = -2
B) x = 0
C) x = 2
D) x = 3
Correct Answer: C
The graph of a function $f$ has a point of inflection where its concavity changes. This occurs when the second derivative, $f''$, changes sign. Since $f''$ is the derivative of $f'$, we look for points where the slope of $f'$ changes sign. This happens at the local extrema of $f'$. The graph of $f'$ has a local maximum at x = 2, which indicates a point of inflection for $f$.
A) A parabola opening upwards with a vertex at x=1.
B) A line with a positive slope.
C) A line with a negative slope passing through the origin.
D) A parabola opening downwards with a vertex at x=1.
Correct Answer: D
The graph of $f$ is increasing for x < 1 and decreasing for x > 1. Therefore, the graph of $f'$ must be positive for x < 1 and negative for x > 1. The graph of $f$ has a local maximum at x=1, so $f'(1)=0$. The graph of $f$ appears to be concave down everywhere, meaning $f'' < 0$, so the slope of $f'$ must be negative. A parabola opening downwards with its vertex at (1,0) satisfies all these conditions.
A) f has a local maximum at x=2.
B) f has a local minimum at x=2.
C) f has a point of inflection at x=2.
D) The graph of f is decreasing at x=2.
Correct Answer: B
At x=2, we are given that $f'(2) = 0$, which means x=2 is a critical point of $f$. To classify this critical point, we use the Second Derivative Test. Since $f''(2) = 3$, which is positive, the graph of $f$ is concave up at x=2. A critical point where the function is concave up corresponds to a local minimum.
A) A graph that is always concave up.
B) A graph that is concave up on (-∞, 2) and concave down on (2, ∞).
C) A graph that is concave down on (-∞, 2) and concave up on (2, ∞).
D) A graph that is always concave down.
Correct Answer: C
The graph of $f$ is concave down when $f''(x) < 0$ and concave up when $f''(x) > 0$. The provided graph shows that $f''(x)$ is negative for x < 2 and positive for x > 2. Therefore, the graph of $f$ must be concave down on the interval (-∞, 2) and concave up on the interval (2, ∞).
A) The graph of $f$ has a local minimum at $x=1$ because the graph of $f'$ has a local minimum at $x=1$.
B) The graph of $f$ has a point of inflection at $x=1$ because the graph of $f'$ has a local minimum at $x=1$.
C) The graph of $f$ is increasing on the interval (0, 2) because the graph of $f'$ is increasing on the interval (0, 2).
D) The graph of $f$ has a local maximum at $x=0$ because $f'(0)=0$.
Correct Answer: B
A point of inflection on the graph of $f$ occurs where the concavity changes, which means $f''$ changes sign. Since $f''$ is the derivative of $f'$, this corresponds to a point where the slope of $f'$ changes sign. At $x=1$, the graph of $f'$ has a local minimum, meaning its slope ($f''$) changes from negative to positive. Therefore, the graph of $f$ has a point of inflection at $x=1$.
A) Increasing and concave up
B) Increasing and concave down
C) Decreasing and concave up
D) Decreasing and concave down
Correct Answer: C
The behavior of $f$ is determined by the signs of $f'$ and $f''$. On the interval (0, 2), the graph of $f'$ is below the x-axis, so $f'(x) < 0$. This means $f$ is decreasing. On the same interval, the graph of $f'$ is increasing, which means its derivative, $f''$, is positive. This means $f$ is concave up. Therefore, on the interval (0, 2), $f$ is decreasing and concave up.