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AP Calculus AB Practice Quiz: Using the Second Derivative Test to Determine Extrema

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

Let `f` be a twice-differentiable function. If `f'(c) = 0` and `f''(c) > 0`, which of the following statements must be true about the function `f` at `x = c`?

All Questions (9)

Let `f` be a twice-differentiable function. If `f'(c) = 0` and `f''(c) > 0`, which of the following statements must be true about the function `f` at `x = c`?

A) `f` has a relative minimum.

B) `f` has a relative maximum.

C) `f` has a point of inflection.

D) The Second Derivative Test is inconclusive.

Correct Answer: A

According to the Second Derivative Test, if the first derivative at a point is zero (`f'(c) = 0`), the point is a critical point. If the second derivative at that same point is positive (`f''(c) > 0`), the function is concave up at that point, which indicates a relative minimum.

Let `f` be a function defined by `f(x) = x^3 - 6x^2 + 5`. At which value of `x` does `f` have a relative maximum?

A) 0

B) 2

C) 4

D) 5

Correct Answer: A

First, find the critical points by setting the first derivative to zero: `f'(x) = 3x^2 - 12x = 3x(x - 4)`. The critical points are `x = 0` and `x = 4`. Next, find the second derivative: `f''(x) = 6x - 12`. Now, apply the Second Derivative Test to each critical point. At `x = 0`, `f''(0) = 6(0) - 12 = -12`. Since `f''(0) < 0`, the function has a relative maximum at `x = 0`. At `x = 4`, `f''(4) = 6(4) - 12 = 12`. Since `f''(4) > 0`, the function has a relative minimum at `x = 4`.

A twice-differentiable function `g` has a critical point at `x = a` where `g'(a) = 0`. If `g''(a) = 0`, what conclusion can be drawn about the behavior of the function `g` at `x = a` using the Second Derivative Test?

A) `g` has a relative maximum at `x = a`.

B) `g` has a relative minimum at `x = a`.

C) `g` has a point of inflection at `x = a`.

D) The Second Derivative Test is inconclusive.

Correct Answer: D

The Second Derivative Test can only be used to classify a critical point as a relative maximum or minimum if the second derivative is non-zero at that point. If `g'(a) = 0` and `g''(a) = 0`, the test provides no information about whether `g` has a relative extremum at `x = a`. The First Derivative Test would be needed to classify the critical point.

Let `y = h(x)` be a twice-differentiable function. If `h'(3) = 0` and `h''(3) = -4`, which of the following statements provides the correct justification for the behavior of `h` at `x = 3`?

A) The function `h` has a relative minimum at `x = 3` because `h'(3) = 0` and `h''(3) < 0`.

B) The function `h` has a relative maximum at `x = 3` because `h'(3) = 0` and `h''(3) < 0`.

C) The function `h` has a relative maximum at `x = 3` because `h'(3) = 0` and `h''(3) > 0`.

D) The function `h` has a point of inflection at `x = 3` because `h'(3) = 0`.

Correct Answer: B

This question tests the direct application and justification using the Second Derivative Test. The conditions for a relative maximum are that the first derivative is zero (`h'(3) = 0`), indicating a critical point, and the second derivative is negative (`h''(3) < 0`), indicating the function is concave down at that point.

The function `f(x) = e^x - ex` is continuous on the interval `(0, ∞)`. The function has only one critical point at `x = 1`. Given that `f''(1) = e`, what can be concluded about the function `f` at `x = 1`?

A) `f` has a relative maximum at `x = 1`, but not an absolute maximum.

B) `f` has a relative minimum at `x = 1`, but not an absolute minimum.

C) `f` has an absolute maximum at `x = 1`.

D) `f` has an absolute minimum at `x = 1`.

Correct Answer: D

First, we use the Second Derivative Test. We are given a critical point at `x = 1`. We are also given `f''(1) = e`. Since `e > 0`, the function has a relative minimum at `x = 1`. According to the provided content, when a continuous function has only one critical point on an interval and that point corresponds to a relative extremum, it is also the absolute extremum on that interval. Therefore, `f` has an absolute minimum at `x = 1`.

For the function `g(x) = sin(x) + (1/2)x^2` on the interval `(0, 2π)`, a critical point occurs at `x = 7π/6`. Use the Second Derivative Test to classify this critical point.

A) Relative maximum

B) Relative minimum

C) Point of inflection

D) The test is inconclusive

Correct Answer: B

To use the Second Derivative Test, we need to find the second derivative of `g(x)`. The first derivative is `g'(x) = cos(x) + x`. The second derivative is `g''(x) = -sin(x) + 1`. Now, we evaluate the second derivative at the critical point `x = 7π/6`. `g''(7π/6) = -sin(7π/6) + 1 = -(-1/2) + 1 = 1/2 + 1 = 3/2`. Since `g''(7π/6) > 0`, the function `g` has a relative minimum at `x = 7π/6`.

Let `f` be a twice-differentiable function. The derivative of `f` is given by `f'(x) = x^2(x-3)`. Which of the following statements is true?

A) `f` has a relative maximum at `x = 0`.

B) `f` has a relative minimum at `x = 3`.

C) `f` has a relative maximum at `x = 3`.

D) `f` has a relative minimum at `x = 0`.

Correct Answer: B

The critical points of `f` are where `f'(x) = 0`, which are `x = 0` and `x = 3`. To use the Second Derivative Test, we find `f''(x)`. Using the product rule on `f'(x) = x^3 - 3x^2`, we get `f''(x) = 3x^2 - 6x`. Now we test the critical points. At `x = 3`, `f''(3) = 3(3)^2 - 6(3) = 27 - 18 = 9`. Since `f''(3) > 0`, `f` has a relative minimum at `x = 3`. At `x = 0`, `f''(0) = 3(0)^2 - 6(0) = 0`, so the Second Derivative Test is inconclusive at `x = 0`.

Selected values for a twice-differentiable function `f` and its derivatives are given in the table below. | `x` | `f(x)` | `f'(x)` | `f''(x)` | |---|---|---|---| | 0 | 5 | 0 | -2 | | 1 | 3 | -4 | 0 | | 2 | 1 | 0 | 3 | | 3 | 4 | 6 | 1 | Based on the table, at which value of `x` does `f` have a relative minimum?

A) 0

B) 1

C) 2

D) 3

Correct Answer: C

To find a relative minimum using the Second Derivative Test, we need to find a point `x` where `f'(x) = 0` and `f''(x) > 0`. Looking at the table, at `x = 0`, `f'(0) = 0` and `f''(0) = -2`, which indicates a relative maximum. At `x = 2`, `f'(2) = 0` and `f''(2) = 3`. Since the first derivative is zero and the second derivative is positive, the function has a relative minimum at `x = 2`.

Let `f` be a function with `f'(c) = 0`. Which of the following provides a sufficient justification using the Second Derivative Test for `f` having a local maximum at `x=c`?

A) `f''(x) < 0` for all `x` in an open interval containing `c`.

B) `f''(c) < 0`.

C) `f'(x)` changes from positive to negative at `x=c`.

D) `f''(c) = 0` and `f'''(c) ≠ 0`.

Correct Answer: B

This question asks for the justification based on the Second Derivative Test. The test states that if `f'(c) = 0` (a critical point) and `f''(c) < 0` (concave down at the critical point), then `f` has a local maximum at `x=c`. Option A is a stronger condition than necessary for the test. Option C is the justification for the First Derivative Test, not the Second. Option D relates to a higher-order derivative test. Therefore, `f''(c) < 0` is the direct justification required by the Second Derivative Test.