AP Calculus BC Practice Quiz: Using the Second Derivative Test to Determine Extrema
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
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A) f has a relative maximum at x = c.
B) f has a relative minimum at x = c.
C) f has a point of inflection at x = c.
D) The Second Derivative Test is inconclusive at x = c.
Correct Answer: A
According to the Second Derivative Test, if the first derivative at a critical point c is zero (f'(c) = 0) and the second derivative at that point is negative (f''(c) < 0), the function is concave down at that point, indicating a relative (local) maximum.
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 first find the second derivative of g(x). The first derivative is g'(x) = 3x² - 12x. The second derivative is g''(x) = 6x - 12. We evaluate the second derivative at the critical point x = 4: g''(4) = 6(4) - 12 = 24 - 12 = 12. Since g'(4) = 0 and g''(4) > 0, the function has a relative (local) minimum at x = 4.
A) h has a relative maximum at x = 2.
B) h has a relative minimum at x = 2.
C) h has a point of inflection at x = 2.
D) The Second Derivative Test is inconclusive.
Correct Answer: D
The Second Derivative Test is only conclusive if f''(c) is not equal to zero. When f'(c) = 0 and f''(c) = 0, the test fails. It is not possible to determine whether there is a maximum, minimum, or neither at x = c using this test alone. Another test, such as the First Derivative Test, must be used.
A) f has a relative maximum at x = 3, but not necessarily an absolute maximum.
B) f has an absolute maximum at x = 3.
C) f has a relative minimum at x = 3, but not necessarily an absolute minimum.
D) f has an absolute minimum at x = 3.
Correct Answer: B
First, use the Second Derivative Test. Since f'(3) = 0 (because it's a critical point) and f''(3) = -5 < 0, f has a relative (local) maximum at x = 3. Because the function is continuous and has only one critical point on its entire domain, this relative maximum must also be the absolute (global) maximum of the function.
A) A relative maximum at x = -2 and a relative minimum at x = 1.
B) A relative minimum at x = -2 and a relative maximum at x = 1.
C) Relative maxima at both x = -2 and x = 1.
D) Relative minima at both x = -2 and x = 1.
Correct Answer: B
We apply the Second Derivative Test at each critical point. At x = -2, f'(-2) = 0 and f''(-2) = 4. Since f''(-2) > 0, f has a relative minimum at x = -2. At x = 1, f'(1) = 0 and f''(1) = -3. Since f''(1) < 0, f has a relative maximum at x = 1.
A) The First Derivative Test
B) The Extreme Value Theorem
C) The Second Derivative Test
D) The Mean Value Theorem
Correct Answer: C
The conclusion is based on the conditions that the first derivative is zero at a point and the second derivative is positive at that same point. This is the precise statement of the Second Derivative Test for a local minimum. The First Derivative Test uses the sign change of f', the Extreme Value Theorem guarantees extrema on a closed interval, and the Mean Value Theorem relates average and instantaneous rates of change.
A) Relative maximum at x = 1/e
B) Relative minimum at x = 1/e
C) Relative maximum at x = 1
D) Relative minimum at x = 1
Correct Answer: B
First, find the critical points by setting the first derivative to zero. Using the product rule, f'(x) = (1) * ln(x) + x * (1/x) = ln(x) + 1. Setting f'(x) = 0 gives ln(x) = -1, so x = e⁻¹ = 1/e. Next, find the second derivative: f''(x) = 1/x. Now, evaluate f'' at the critical point: f''(1/e) = 1 / (1/e) = e. Since f''(1/e) = e > 0, the function has a relative minimum at x = 1/e.
A) -6
B) -3/2
C) 3/2
D) 6
Correct Answer: D
For f to have a critical point at x = -4, f'(-4) must be 0. f'(x) = 3x² + 2kx. So, f'(-4) = 3(-4)² + 2k(-4) = 48 - 8k = 0. Solving for k gives 8k = 48, so k = 6. To confirm it's a relative maximum with the Second Derivative Test, we check the sign of f''(x) = 6x + 2k. With k=6, f''(x) = 6x + 12. At x=-4, f''(-4) = 6(-4) + 12 = -24 + 12 = -12. Since f''(-4) < 0, the test confirms a relative maximum.
A) g''(c) = 0
B) g''(c) < 0
C) g''(c) > 0
D) g'(x) changes from positive to negative at x = c
Correct Answer: C
The question asks for a condition that is sufficient to conclude a relative minimum at a critical point x=c using the principles of the Second Derivative Test. The test states that if f'(c)=0 and f''(c) > 0, then the function has a relative minimum at x=c. Option D describes the condition for a relative maximum according to the First Derivative Test.