AP Calculus BC Practice Quiz: Using the First Derivative Test to Determine Relative (Local) Extrema
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
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Question 1 of 7
All Questions (7)
A) The function's points of inflection.
B) The location of the function's relative (local) extrema.
C) The function's concavity.
D) The function's absolute maximum value on a closed interval.
Correct Answer: B
Based on the provided content, the first derivative of a function is used to determine the location of its relative (local) extrema by analyzing where the derivative changes sign.
A) x = -2
B) x = 0
C) x = 1
D) x = 2
Correct Answer: D
The critical points of f are where f'(x) = 0, which are x=0 and x=2. For x < 0, f'(x) is positive. For 0 < x < 2, f'(x) is negative. For x > 2, f'(x) is positive. A relative minimum occurs where f'(x) changes from negative to positive. This change occurs at x = 2.
A) g has a relative minimum at x = 3 because g'(x) changes from positive to negative.
B) g has a relative maximum at x = 3 because g'(x) changes from positive to negative.
C) g has a relative minimum at x = 3 because g'(x) changes from negative to positive.
D) g has a relative maximum at x = 3 because g'(x) changes from negative to positive.
Correct Answer: B
According to the First Derivative Test, a function has a relative maximum at a point where its first derivative changes sign from positive to negative. This indicates the function changes from increasing to decreasing.
A) f has a relative maximum at x = -1 and a relative minimum at x = 4.
B) f has a relative minimum at x = 4 and no relative extremum at x = -1.
C) f has a relative maximum at x = 4 and a relative minimum at x = -1.
D) f has a relative minimum at x = -1 and no relative extremum at x = 4.
Correct Answer: B
The critical points are x = -1 and x = 4. For x < -1, f'(x) is negative. For -1 < x < 4, f'(x) is also negative. Since f'(x) does not change sign at x = -1, there is no relative extremum there. For x > 4, f'(x) is positive. Since f'(x) changes from negative to positive at x = 4, f has a relative minimum at x = 4.
A) (-3, 3)
B) (-∞, -3) and (3, ∞)
C) (-∞, 9)
D) (0, ∞)
Correct Answer: B
A function is increasing where its first derivative is positive. Since e^x is always positive, the sign of h'(x) is determined by the sign of (x^2 - 9). The expression x^2 - 9 is positive when x < -3 or x > 3. Therefore, h is increasing on the intervals (-∞, -3) and (3, ∞).
A) f has a relative maximum at x = -1.
B) f has a relative minimum at x = -1.
C) f has neither a relative maximum nor a relative minimum at x = -1.
D) The First Derivative Test cannot be used to draw a conclusion at x = -1.
Correct Answer: C
For a relative extremum to exist at a critical point x = c, the first derivative f'(x) must change sign at x = c. In this case, f'(x) is positive on both sides of x = -1. Since the sign of f'(x) does not change, f does not have a relative extremum at x = -1.
A) x = 1 only
B) x = -3 only
C) x = 1 and x = -3
D) g has no relative maximum.
Correct Answer: D
Critical points occur where g'(x) = 0 or is undefined. g'(x) = 0 at x=1. g'(x) is undefined at x=-3. For x < -3, g'(x) is positive. For -3 < x < 1, g'(x) is negative. For x > 1, g'(x) is positive. A relative maximum occurs where g' changes from positive to negative, which happens at x = -3. However, since g' is undefined at x=-3, g is not differentiable there and may have a cusp or vertical asymptote, but the question asks for a relative maximum based on the derivative test. The change from negative to positive at x=1 indicates a relative minimum. Therefore, there is no relative maximum.