AP Calculus BC Quiz: First Derivative Test
20 questions · exam conditions
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First Derivative TestQuestion 1 of 20

A function gg satisfies g′(x)>0g'(x)>0 on (2,5)(2,5) and g′(x)<0g'(x)<0 on (5,9)(5,9); where is a local maximum?

A local minimum at x=5x=5
A local maximum at x=2x=2
A local maximum at x=5x=5
A local minimum at x=9x=9
No local extremum occurs
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AP Calculus BC Quiz

AP Calculus BC Quiz: First Derivative Test

Practice First Derivative Test in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on First Derivative Test, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A function gg satisfies g′(x)>0g'(x)>0 on (2,5)(2,5) and g′(x)<0g'(x)<0 on (5,9)(5,9); where is a local maximum?

  1. A local minimum at x=5x=5
  2. A local maximum at x=2x=2
  3. A local maximum at x=5x=5 (correct answer)
  4. A local minimum at x=9x=9
  5. No local extremum occurs
Explanation: The First Derivative Test determines local extrema by examining derivative sign changes. For function g, g'(x) > 0 on (2, 5) and g'(x) < 0 on (5, 9), showing the derivative changes from positive to negative at x = 5. This positive-to-negative transition indicates the function increases before x = 5 and decreases after x = 5, confirming a local maximum at x = 5. Choice A incorrectly suggests a minimum at x = 5, but the sign pattern (+ to -) specifically characterizes maxima, not minima. When f' transitions from positive to negative at a critical point, that point represents the function's local peak.

Question 2

For uu, u′(x)>0u'(x)>0 on (−1,2)( -1,2), u′(2)=0u'(2)=0, and u′(x)<0u'(x)<0 on (2,8)(2,8); where is a local maximum?

  1. A local minimum at x=2x=2
  2. A local maximum at x=2x=2 (correct answer)
  3. A local maximum at x=−1x=-1
  4. A local minimum at x=8x=8
  5. No local extremum occurs
Explanation: The First Derivative Test determines local extrema by examining sign changes in f'(x). For function u, u'(x) > 0 on (-1, 2), u'(2) = 0, and u'(x) < 0 on (2, 8), showing the derivative transitions from positive to negative at x = 2. This positive-to-negative sign change indicates the function increases before x = 2 and decreases after x = 2, establishing a local maximum at x = 2. Choice A incorrectly identifies a minimum at x = 2, but the sign change pattern (+ to -) definitively characterizes maxima, not minima. When f' changes from positive to negative at a critical point, that point represents the function's local peak.

Question 3

For rr, r′(x)<0r'(x)<0 on (−6,−3)(-6,-3), r′(x)>0r'(x)>0 on (−3,1)(-3,1), and r′(x)<0r'(x)<0 on (1,4)(1,4); where is a local maximum?

  1. A local maximum at x=−3x=-3
  2. A local minimum at x=1x=1
  3. A local maximum at x=1x=1 (correct answer)
  4. A local minimum at x=−3x=-3
  5. No local extremum occurs
Explanation: The First Derivative Test determines extrema by examining sign changes in the derivative. For function r, r'(x) < 0 on (-6, -3), r'(x) > 0 on (-3, 1), and r'(x) < 0 on (1, 4), showing the derivative changes from positive to negative at x = 1. This positive-to-negative transition indicates the function increases before x = 1 and decreases after x = 1, confirming a local maximum at x = 1. Choice D incorrectly identifies a minimum at x = -3, but at x = -3 the sign change is negative-to-positive (indicating a minimum), while at x = 1 it's positive-to-negative (indicating the maximum). Focus on the specific sign change pattern at each critical point to identify the correct extremum type.

Question 4

For tt, t′(x)>0t'(x)>0 on (−9,−5)( -9,-5), t′(x)<0t'(x)<0 on (−5,−2)(-5,-2), and t′(x)>0t'(x)>0 on (−2,0)(-2,0); where is a local minimum?

  1. A local maximum at x=−5x=-5
  2. A local minimum at x=−5x=-5
  3. A local minimum at x=−2x=-2 (correct answer)
  4. A local maximum at x=−2x=-2
  5. No local extremum occurs
Explanation: The First Derivative Test identifies local extrema through derivative sign changes around critical points. For function t, t'(x) > 0 on (-9, -5), t'(x) < 0 on (-5, -2), and t'(x) > 0 on (-2, 0), showing sign changes at both x = -5 and x = -2. At x = -5, the derivative changes from positive to negative (indicating a maximum), while at x = -2, it changes from negative to positive (indicating a minimum). Choice A incorrectly identifies a maximum at x = -5, but the question asks specifically for the local minimum, which occurs at x = -2. Always identify the type of extremum that matches the sign change pattern at each critical point.

Question 5

On (0,2)(0,2), f′(x)<0f'(x)<0; on (2,6)(2,6), f′(x)>0f'(x)>0; and f′(2)=0f'(2)=0. Where is a local minimum?

  1. A local maximum at x=2x=2
  2. A local minimum at x=0x=0
  3. A local minimum at x=2x=2 (correct answer)
  4. A local maximum at x=6x=6
  5. No local extremum occurs
Explanation: The First Derivative Test identifies local extrema through derivative sign changes around critical points. Given f'(x) < 0 on (0, 2), f'(2) = 0, and f'(x) > 0 on (2, 6), the derivative changes from negative to positive at x = 2. This negative-to-positive transition indicates the function decreases before x = 2 and increases afterward, confirming a local minimum at x = 2. Choice A incorrectly identifies a maximum at x = 2, but the sign pattern (- to +) specifically characterizes minima, not maxima. When applying this test, always match the sign change pattern: negative-to-positive signals a minimum, positive-to-negative signals a maximum.

Question 6

For rr, r′(x)<0r'(x)<0 on (−3,2)( -3,2) and r′(x)>0r'(x)>0 on (2,7)(2,7); where is a local minimum?

  1. A local minimum at x=2x=2 (correct answer)
  2. A local maximum at x=2x=2
  3. A local minimum at x=−3x=-3
  4. A local maximum at x=7x=7
  5. No local extremum occurs
Explanation: The First Derivative Test determines local extrema by examining sign changes in the derivative. Given r′(x)<0r'(x) < 0 on (−3,2)(-3, 2) and r′(x)>0r'(x) > 0 on (2,7)(2, 7), the derivative transitions from negative to positive at x=2x = 2. This negative-to-positive sign change means the function decreases before x=2x = 2 and increases afterward, confirming a local minimum at x=2x = 2. Choice B incorrectly suggests a maximum at x=2x = 2, but the sign change pattern (- to +) definitively indicates a minimum, not a maximum. When f′f' changes from negative to positive at a critical point, that point is always a local minimum.

Question 7

A function gg has g′(x)>0g'(x)>0 on (−2,1)(-2,1) and g′(x)<0g'(x)<0 on (1,4)(1,4); where is a local maximum?

  1. A local minimum at x=1x=1
  2. A local maximum at x=−2x=-2
  3. A local maximum at x=1x=1 (correct answer)
  4. A local minimum at x=4x=4
  5. No local extremum occurs
Explanation: The First Derivative Test identifies local extrema through derivative sign changes. Since g'(x) > 0 on (-2, 1) and g'(x) < 0 on (1, 4), the derivative transitions from positive to negative at x = 1. This positive-to-negative sign change means the function increases before x = 1 and decreases after x = 1, establishing a local maximum at x = 1. Choice A incorrectly suggests a minimum at x = 1, but the sign pattern (+ to -) definitively indicates a maximum, not a minimum. To apply this test effectively, remember that a sign change from positive to negative at a critical point always signals a local maximum.

Question 8

For a differentiable function ff, f′(x)<0f'(x)<0 on (−3,−1)(-3,-1) and f′(x)>0f'(x)>0 on (−1,2)(-1,2); where is a local minimum?

  1. A local maximum at x=−1x=-1
  2. A local minimum at x=−1x=-1 (correct answer)
  3. A local minimum at x=−3x=-3
  4. A local maximum at x=2x=2
  5. No local extremum occurs
Explanation: The First Derivative Test determines local extrema by analyzing sign changes in f'(x). Given that f'(x) < 0 on (-3, -1) and f'(x) > 0 on (-1, 2), the derivative changes from negative to positive at x = -1. This sign change from negative to positive indicates the function is decreasing before x = -1 and increasing after x = -1, confirming a local minimum at x = -1. Choice A might seem tempting since x = -1 is a critical point, but the sign change pattern (- to +) specifically indicates a minimum, not a maximum. When f' changes from negative to positive at a critical point, always conclude there's a local minimum.

Question 9

A differentiable vv has v′(x)<0v'(x)<0 on (−3,0)(-3,0) and v′(x)<0v'(x)<0 on (0,5)(0,5) with v′(0)=0v'(0)=0; what occurs at x=0x=0?

  1. A local maximum at x=0x=0
  2. A local minimum at x=0x=0
  3. No local extremum at x=0x=0 (correct answer)
  4. A local maximum at x=5x=5
  5. A local minimum at x=−3x=-3
Explanation: The First Derivative Test requires a sign change in the derivative to establish local extrema. Here, v'(x) < 0 on (-3, 0), v'(0) = 0, and v'(x) < 0 on (0, 5), showing the derivative remains negative on both sides of x = 0. Since there's no sign change in v'(x) around x = 0, the function continues decreasing through this point without forming an extremum. Choice B might seem correct since v'(0) = 0, but a zero derivative alone doesn't create an extremum—the derivative must change sign. When f'(x) maintains the same sign before and after a critical point, no local extremum occurs at that point.

Question 10

Let ff be differentiable with f′(x)<0f'(x)<0 on (−10,−6)( -10,-6) and f′(x)>0f'(x)>0 on (−6,−3)(-6,-3); where is a local minimum?

  1. A local maximum at x=−6x=-6
  2. A local minimum at x=−10x=-10
  3. A local minimum at x=−6x=-6 (correct answer)
  4. A local maximum at x=−3x=-3
  5. No local extremum occurs
Explanation: The First Derivative Test locates extrema by analyzing how f'(x) changes sign around critical points. Given f'(x) < 0 on (-10, -6) and f'(x) > 0 on (-6, -3), the derivative transitions from negative to positive at x = -6. This negative-to-positive sign change means the function decreases before x = -6 and increases afterward, establishing a local minimum at x = -6. Choice A incorrectly identifies a maximum at x = -6, but the sign change pattern (- to +) definitively indicates a minimum. Remember that when f' changes from negative to positive at a critical point, that point is always a local minimum.

Question 11

A function rr has r′(x)>0r'(x)>0 on (0,3)(0,3), r′(3)=0r'(3)=0, and r′(x)>0r'(x)>0 on (3,10)(3,10); what occurs at x=3x=3?

  1. A local maximum at x=3x=3
  2. A local minimum at x=3x=3
  3. No local extremum at x=3x=3 (correct answer)
  4. A local maximum at x=10x=10
  5. A local minimum at x=0x=0
Explanation: The First Derivative Test requires a sign change in f'(x) to establish local extrema. For function r, r'(x) > 0 on (0, 3), r'(3) = 0, and r'(x) > 0 on (3, 10), showing the derivative remains positive on both sides of x = 3. Since there's no sign change in r'(x) around x = 3, the function continues increasing through this point without forming an extremum. Choice B might seem correct since r'(3) = 0, but a zero derivative alone doesn't create an extremum—the derivative must change sign. When f'(x) maintains the same sign before and after a critical point, no local extremum occurs at that point.

Question 12

A differentiable uu has u′(x)<0u'(x)<0 on (−4,2)( -4,2) and u′(x)>0u'(x)>0 on (2,3)(2,3); where is a local minimum?

  1. A local maximum at x=2x=2
  2. A local minimum at x=2x=2 (correct answer)
  3. A local maximum at x=−4x=-4
  4. A local minimum at x=3x=3
  5. No local extremum occurs
Explanation: The First Derivative Test locates extrema by analyzing derivative sign changes around critical points. Given u'(x) < 0 on (-4, 2) and u'(x) > 0 on (2, 3), the derivative transitions from negative to positive at x = 2. This negative-to-positive sign change means the function decreases before x = 2 and increases afterward, confirming a local minimum at x = 2. Choice A incorrectly identifies a maximum at x = 2, but the sign change pattern (- to +) definitively indicates a minimum, not a maximum. When applying this test, remember that a sign change from negative to positive always produces a local minimum.

Question 13

A function tt has t′(x)<0t'(x)<0 on (1,4)(1,4), t′(4)=0t'(4)=0, and t′(x)>0t'(x)>0 on (4,7)(4,7); where is a local minimum?

  1. A local maximum at x=4x=4
  2. A local minimum at x=1x=1
  3. A local minimum at x=4x=4 (correct answer)
  4. A local maximum at x=7x=7
  5. No local extremum occurs
Explanation: The First Derivative Test locates extrema by analyzing derivative sign changes around critical points. Since t'(x) < 0 on (1, 4), t'(4) = 0, and t'(x) > 0 on (4, 7), the derivative changes from negative to positive at x = 4. This negative-to-positive transition means the function decreases before x = 4 and increases afterward, confirming a local minimum at x = 4. Choice A incorrectly suggests a maximum at x = 4, but the sign pattern (- to +) specifically indicates a minimum, not a maximum. When applying this test, remember that a sign change from negative to positive always produces a local minimum.

Question 14

A function uu has u′(x)>0u'(x)>0 on (3,5)(3,5) and u′(x)<0u'(x)<0 on (5,12)(5,12); where is a local maximum?

  1. A local maximum at x=5x=5 (correct answer)
  2. A local minimum at x=5x=5
  3. A local maximum at x=3x=3
  4. A local minimum at x=12x=12
  5. No local extremum occurs
Explanation: The First Derivative Test locates extrema by analyzing derivative sign changes around critical points. Given u'(x) > 0 on (3, 5) and u'(x) < 0 on (5, 12), the derivative changes from positive to negative at x = 5. This positive-to-negative transition indicates the function increases before x = 5 and decreases afterward, establishing a local maximum at x = 5. Choice B incorrectly suggests a minimum at x = 5, but the sign pattern (+ to -) specifically characterizes maxima, not minima. When f' changes from positive to negative at a critical point, that point represents the function's local peak.

Question 15

A function tt has t′(x)<0t'(x)<0 on (−2,0)( -2,0) and t′(x)>0t'(x)>0 on (0,4)(0,4); where is a local minimum?

  1. A local maximum at x=0x=0
  2. A local minimum at x=−2x=-2
  3. A local minimum at x=0x=0 (correct answer)
  4. A local maximum at x=4x=4
  5. No local extremum occurs
Explanation: The First Derivative Test identifies local extrema through derivative sign changes around critical points. Given t'(x) < 0 on (-2, 0) and t'(x) > 0 on (0, 4), the derivative transitions from negative to positive at x = 0. This negative-to-positive sign change means the function decreases before x = 0 and increases afterward, confirming a local minimum at x = 0. Choice A incorrectly suggests a maximum at x = 0, but the sign change pattern (- to +) definitively indicates a minimum, not a maximum. When applying this test, remember that a sign change from negative to positive always produces a local minimum.

Question 16

For ff, f′(x)>0f'(x)>0 on (−1,4)( -1,4), f′(4)=0f'(4)=0, and f′(x)<0f'(x)<0 on (4,6)(4,6); where is a local maximum?

  1. A local minimum at x=4x=4
  2. A local maximum at x=4x=4 (correct answer)
  3. A local maximum at x=−1x=-1
  4. A local minimum at x=6x=6
  5. No local extremum occurs
Explanation: The First Derivative Test locates extrema by analyzing derivative sign changes around critical points. Given f'(x) > 0 on (-1, 4), f'(4) = 0, and f'(x) < 0 on (4, 6), the derivative changes from positive to negative at x = 4. This positive-to-negative transition indicates the function increases before x = 4 and decreases afterward, establishing a local maximum at x = 4. Choice A incorrectly identifies a minimum at x = 4, but the sign pattern (+ to -) specifically indicates a maximum, not a minimum. When f' changes from positive to negative at a critical point, that point represents the function's local peak.

Question 17

A function ww has w′(x)<0w'(x)<0 on (−3,1)( -3,1) and w′(x)>0w'(x)>0 on (1,5)(1,5); where is a local minimum?

  1. A local maximum at x=1x=1
  2. A local minimum at x=−3x=-3
  3. A local minimum at x=1x=1 (correct answer)
  4. A local maximum at x=5x=5
  5. No local extremum occurs
Explanation: The First Derivative Test identifies local extrema through derivative sign changes around critical points. Since w'(x) < 0 on (-3, 1) and w'(x) > 0 on (1, 5), the derivative transitions from negative to positive at x = 1. This negative-to-positive sign change means the function decreases before x = 1 and increases afterward, confirming a local minimum at x = 1. Choice A incorrectly suggests a maximum at x = 1, but the sign change pattern (- to +) definitively indicates a minimum, not a maximum. When applying this test, always remember that negative-to-positive sign changes produce local minima.

Question 18

For ss, s′(x)>0s'(x)>0 on (−2,0)(-2,0) and s′(x)<0s'(x)<0 on (0,3)(0,3) with s′(0)=0s'(0)=0; what occurs at x=0x=0?

  1. A local minimum at x=0x=0
  2. No local extremum at x=0x=0
  3. A local maximum at x=0x=0 (correct answer)
  4. A local minimum at x=−2x=-2
  5. A local maximum at x=3x=3
Explanation: The First Derivative Test identifies local extrema through derivative sign changes. Given s'(x) > 0 on (-2, 0), s'(0) = 0, and s'(x) < 0 on (0, 3), the derivative transitions from positive to negative at x = 0. This positive-to-negative sign change indicates the function increases before x = 0 and decreases afterward, establishing a local maximum at x = 0. Choice B incorrectly concludes no extremum exists, but the clear sign change from positive to negative definitively creates a maximum at x = 0. When f' changes from positive to negative at a critical point, that point is always a local maximum.

Question 19

For hh, h′(x)h'(x) is negative on (−5,−2)(-5,-2), zero at −2-2, and negative on (−2,3)(-2,3); what local extremum occurs?

  1. A local minimum at x=−2x=-2
  2. A local maximum at x=−2x=-2
  3. A local maximum at x=3x=3
  4. No local extremum at x=−2x=-2 (correct answer)
  5. A local minimum at x=−5x=-5
Explanation: The First Derivative Test requires a sign change in f'(x) to establish local extrema. Here, h'(x) is negative on (-5, -2), zero at x = -2, and negative on (-2, 3), showing no sign change around x = -2. Since h'(x) remains negative on both sides of x = -2, the function continues decreasing through this point without forming an extremum. Choice B might appear correct since h'(-2) = 0, but a zero derivative alone doesn't guarantee an extremum—the derivative must change sign. When f'(x) maintains the same sign on both sides of a critical point, no local extremum occurs there.

Question 20

For hh, h′(x)<0h'(x)<0 on (−1,1)(-1,1), h′(1)=0h'(1)=0, and h′(x)<0h'(x)<0 on (1,6)(1,6); what occurs at x=1x=1?

  1. A local maximum at x=1x=1
  2. A local minimum at x=1x=1
  3. No local extremum at x=1x=1 (correct answer)
  4. A local maximum at x=6x=6
  5. A local minimum at x=−1x=-1
Explanation: The First Derivative Test requires a sign change in the derivative to establish local extrema. Here, h'(x) < 0 on (-1, 1), h'(1) = 0, and h'(x) < 0 on (1, 6), indicating the derivative remains negative on both sides of x = 1. Since there's no sign change in h'(x) around x = 1, the function continues decreasing through this point without forming an extremum. Choice B might appear reasonable since h'(1) = 0, but a zero derivative alone doesn't guarantee an extremum—sign changes are essential. When f'(x) maintains the same sign before and after a critical point, no local extremum occurs there.