AP Calculus AB Quiz: Second Derivative Test

What this quiz covers

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

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.

Question 1

1. Local maximum at $x=3$.
2. Local minimum at $x=3$. (correct answer)
3. Inflection point at $x=3$.
4. Cannot be determined from the information given.
5. Neither a maximum nor minimum because concavity is irrelevant.

Explanation: The Second Derivative Test utilizes concavity information to determine the type of critical point present. Since $p'(3)=0$ establishes a critical point and the function is concave up at $x=3$ (meaning $p''(3)>0$), the graph curves upward at this location. Upward concavity at a critical point creates a valley-shaped curve, confirming a local minimum. Choice A might attract students who confuse upward curvature with maximum behavior, but concave up specifically indicates the valley formation of minimums. Apply this test when both critical point and concavity direction are definitively known.

Question 2

1. Local minimum at $x=16$. (correct answer)
2. Local maximum at $x=16$.
3. Inflection point at $x=16$.
4. Cannot be determined from the information given.
5. Neither a maximum nor minimum because $G'(16)=0$.

Explanation: The Second Derivative Test determines the nature of critical points through concavity analysis. Since $G'(16)=0$ establishes a critical point and the function is concave up at $x=16$ (meaning $G''(16)>0$), the graph curves upward at this point. Upward concavity at a critical point forms a valley-shaped curve, indicating a local minimum. Choice B might confuse students who think upward curvature suggests maximums, but concave up specifically describes the valley formation of minimums. Apply this test when you can verify both critical point existence and the direction of concavity.

Question 3

1. Inflection point at $x=20$.
2. Local minimum at $x=20$. (correct answer)
3. Local maximum at $x=20$.
4. Cannot be determined from the information given.
5. Neither a maximum nor minimum because $s''(20)>0$.

Explanation: The Second Derivative Test examines concavity at critical points to determine their classification. Given $s'(20)=0$ (critical point) and $s''(20)>0$ (positive second derivative), the function is concave up at $x=20$. Positive concavity at a critical point produces upward curvature resembling a valley, confirming a local minimum. Choice C might tempt students who associate positive values with maximums, but positive second derivative specifically indicates the upward curvature of minimums. Use this test when both critical point status and second derivative sign are clearly determined.

Question 4

1. Local minimum at $x=7$. (correct answer)
2. Local maximum at $x=7$.
3. Inflection point at $x=7$.
4. Cannot be determined from the information given.
5. Neither a maximum nor minimum because $f'(7)=0$.

Explanation: The Second Derivative Test utilizes concavity analysis to classify critical points effectively. Since $f'(7)=0$ establishes a critical point and the graph is concave up at $x=7$ (meaning $f''(7)>0$), the function curves upward at this location. Upward concavity at a critical point forms a valley-shaped curve, indicating a local minimum. Choice B might confuse students who think upward direction suggests maximums, but concave up specifically describes the valley formation of minimums. Apply this test when you can confirm both critical point existence and the direction of concavity.

Question 5

1. Local minimum at $x=13$.
2. Cannot be determined from the information given.
3. Local maximum at $x=13$. (correct answer)
4. Inflection point at $x=13$.
5. Neither a maximum nor minimum because $p''(13)<0$.

Explanation: The Second Derivative Test utilizes concavity information to classify critical points effectively. With $p'(13)=0$ confirming a critical point and $p''(13)<0$ indicating negative concavity, the function exhibits downward curvature at $x=13$. Downward concavity at a critical point creates a hill-like shape, establishing a local maximum. Choice A could mislead students who connect negative signs with minimums, but negative second derivative specifically refers to the hill formation characteristic of maximums. Ensure both conditions are met: critical point verification and determinable concavity sign.

Question 6

1. Local minimum at $x=-13$.
2. Local maximum at $x=-13$. (correct answer)
3. Inflection point at $x=-13$.
4. Cannot be determined from the information given.
5. Neither a maximum nor minimum because concavity is negative.

Explanation: The Second Derivative Test utilizes concavity information to classify critical points effectively. With $H'(-13)=0$ confirming a critical point and concave down behavior at $x=-13$ (meaning $H''(-13)<0$), the function exhibits downward curvature. Downward concavity at a critical point creates a hill-like shape, establishing a local maximum. Choice A could mislead students who connect downward direction with minimums, but concave down specifically refers to the hill formation characteristic of maximums. Ensure both conditions are met: critical point verification and determinable concavity sign.

Question 7

1. Local minimum at $x=-6$.
2. Inflection point at $x=-6$.
3. Cannot be determined from the information given.
4. Local maximum at $x=-6$. (correct answer)
5. Neither a maximum nor minimum because $p'(-6)=0$.

Explanation: The Second Derivative Test determines critical point nature through concavity analysis at those points. Since $p'(-6)=0$ establishes a critical point and $p''(-6)<0$ indicates negative concavity, the function is concave down at $x=-6$. Concave down behavior at a critical point forms a hill-shaped curve, indicating a local maximum. Choice A could confuse students who think negative signs correspond to minimums, but negative second derivative specifically means downward concavity (maximum). Apply this test when you have verified both the critical point condition and the sign of concavity.

Question 8

1. Local minimum at $x=-19$.
2. Local maximum at $x=-19$. (correct answer)
3. Inflection point at $x=-19$.
4. Cannot be determined from the information given.
5. Neither a maximum nor minimum because $g'(-19)=0$.

Explanation: The Second Derivative Test utilizes concavity information to classify critical points effectively. With $g'(-19)=0$ confirming a critical point and concave down behavior at $x=-19$ (meaning $g''(-19)<0$), the function exhibits downward curvature. Downward concavity at a critical point creates a hill-like shape, establishing a local maximum. Choice A could mislead students who connect downward direction with minimums, but concave down specifically refers to the hill formation characteristic of maximums. Ensure both conditions are met: critical point verification and determinable concavity sign.

Question 9

1. Local maximum at $x=-2$.
2. Cannot be determined because $g''(-2)>0$.
3. Local minimum at $x=-2$. (correct answer)
4. Point of inflection at $x=-2$.
5. Neither; $g$ has no critical points at $x=-2$.

Explanation: This question tests understanding of the Second Derivative Test for classifying critical points. We have g'(-2) = 0, confirming a critical point exists at x = -2. The Second Derivative Test tells us that when f'(c) = 0 and f''(c) > 0, the function has a local minimum at x = c. Since g''(-2) = 7 > 0, the function is concave up at x = -2, indicating a local minimum. Choice B incorrectly suggests that having g''(-2) > 0 makes the result indeterminate, when in fact this positive value is exactly what allows us to classify the critical point. Remember: positive second derivative means concave up (local minimum), negative second derivative means concave down (local maximum).

Question 10

1. Cannot be determined because $q''(1)<0$.
2. Local minimum at $x=1$.
3. Local maximum at $x=1$. (correct answer)
4. Point of inflection at $x=1$.
5. Neither; $q'(1)=0$ means $q$ is constant near $1$.

Explanation: This problem tests the Second Derivative Test for classifying critical points. Given q'(1) = 0, we have a critical point at x = 1. The Second Derivative Test tells us that when f'(c) = 0 and f''(c) < 0, the function has a local maximum at x = c. Since q''(1) = -9 < 0, the function is concave down at x = 1, which means we have a local maximum. Choice E incorrectly interprets q'(1) = 0 as meaning the function is constant, when it actually just means the tangent line is horizontal at that point. Remember the mnemonic: negative second derivative means the function "frowns" (concave down), creating a local maximum at the critical point.

Question 11

1. Point of inflection at $x=5$.
2. Local maximum at $x=5$.
3. Neither; $p$ is decreasing at $x=5$.
4. Local minimum at $x=5$. (correct answer)
5. Cannot be determined because $p'(5)=0$.

Explanation: This question requires applying the Second Derivative Test to determine the nature of a critical point. We have p'(5) = 0, confirming a critical point at x = 5. The Second Derivative Test states that when f'(c) = 0 and f''(c) > 0, the function has a local minimum at x = c. Since p''(5) = 12 > 0, the function is concave up at x = 5, indicating a local minimum. Choice A incorrectly identifies this as a point of inflection, which would require f''(c) = 0, not f'(c) = 0. When applying the Second Derivative Test, focus on the sign of f''(c): positive means the parabola opens upward (local minimum), negative means it opens downward (local maximum).

Question 12

1. Local maximum at $x=3$. (correct answer)
2. Local minimum at $x=3$.
3. Point of inflection at $x=3$.
4. Cannot be determined because $f'(3)=0$.
5. Cannot be determined because $f''(3)\neq 0$.

Explanation: This problem requires applying the Second Derivative Test to classify a critical point. Since f'(3) = 0, we have a critical point at x = 3. The Second Derivative Test states that when f'(c) = 0 and f''(c) < 0, the function has a local maximum at x = c. Here, f''(3) = -5 < 0, indicating the function is concave down at x = 3, which means we have a local maximum. Choice D incorrectly suggests that having f'(3) = 0 makes the test inconclusive, but this is actually the necessary condition for applying the test. When solving Second Derivative Test problems, always verify both conditions: f'(c) = 0 (critical point exists) and check the sign of f''(c) to determine the classification.

Question 13

1. Local maximum at $x=\tfrac12$.
2. Local minimum at $x=\tfrac12$. (correct answer)
3. Point of inflection at $x=\tfrac12$.
4. Cannot be determined because $t''(\tfrac12)>0$.
5. Neither; $t'(\tfrac12)=0$ implies a horizontal asymptote.

Explanation: This question applies the Second Derivative Test using inequality notation. Given t'(1/2) = 0, we have a critical point at x = 1/2. The Second Derivative Test tells us that when f'(c) = 0 and f''(c) > 0, the function has a local minimum at x = c. Since t''(1/2) > 0, the function is concave up at x = 1/2, indicating a local minimum. Choice E incorrectly interprets t'(1/2) = 0 as indicating a horizontal asymptote, when it simply means the tangent line is horizontal at that point. For test success, remember that the Second Derivative Test provides a definitive classification when f'(c) = 0 and f''(c) ≠ 0, regardless of whether exact values or inequalities are given.

Question 14

1. Local minimum at $x=5$. (correct answer)
2. Local maximum at $x=5$.
3. Inflection point at $x=5$.
4. Cannot be determined because $g'(5)=0$.
5. No extremum because $g''(5)>0$.

Explanation: The Second Derivative Test determines the nature of critical points by analyzing the concavity at those locations. With $g'(5)=0$ confirming a critical point and $g''(5)>0$ indicating positive concavity, the graph is concave up at $x=5$. Concave up curvature at a critical point forms a valley, establishing a local minimum. Choice B could mislead students who associate positive values with maximums, but positive second derivative specifically indicates upward concavity (minimum). The test is applicable when both the critical point condition and second derivative sign are clearly determined.

Question 15

1. Local maximum at $x=-2$ because $g''(-2)<0$. (correct answer)
2. Local minimum at $x=-2$ because $g''(-2)<0$.
3. Point of inflection at $x=-2$ because $g''(-2)<0$.
4. Neither; $g'(-2)=0$ guarantees only an inflection point.
5. Neither; the Second Derivative Test cannot be used unless $g''(-2)=0$.

Explanation: This question tests the Second Derivative Test for classifying critical points. We have g'(-2) = 0, establishing a critical point at x = -2. Since g''(-2) < 0, the Second Derivative Test tells us that g has a local maximum at x = -2 because the function is concave down there. The negative second derivative means the graph curves downward like an upside-down U, creating a peak at x = -2. Choice B incorrectly associates negative concavity with a minimum, which is a common sign error. For the AP exam, memorize: f''(c) < 0 means concave down and local max; f''(c) > 0 means concave up and local min.

Question 16

1. Local maximum at $x=12$.
2. Local minimum at $x=12$. (correct answer)
3. Inflection point at $x=12$.
4. Cannot be determined from the information given.
5. Neither a maximum nor minimum because $s'(12)=0$.

Explanation: The Second Derivative Test determines the nature of critical points through concavity analysis. Since $s'(12)=0$ establishes a critical point and the function is concave up at $x=12$ (meaning $s''(12)>0$), the graph curves upward at this point. Upward concavity at a critical point forms a valley-shaped curve, indicating a local minimum. Choice A might confuse students who think upward curvature suggests maximums, but concave up specifically describes the valley formation of minimums. Apply this test when you can verify both critical point existence and the direction of concavity.

Question 17

1. Local maximum at $x=22$.
2. Local minimum at $x=22$. (correct answer)
3. Inflection point at $x=22$.
4. Cannot be determined from the information given.
5. Neither a maximum nor minimum because concavity is positive.

Explanation: The Second Derivative Test determines the nature of critical points through concavity analysis. Since $f'(22)=0$ establishes a critical point and the function is concave up at $x=22$ (meaning $f''(22)>0$), the graph curves upward at this point. Upward concavity at a critical point forms a valley-shaped curve, indicating a local minimum. Choice A might confuse students who think upward curvature suggests maximums, but concave up specifically describes the valley formation of minimums. Apply this test when you can verify both critical point existence and the direction of concavity.

Question 18

1. Local minimum at $x=-8$.
2. Local maximum at $x=-8$. (correct answer)
3. Inflection point at $x=-8$.
4. Cannot be determined from the information given.
5. Neither a maximum nor minimum because concavity is negative.

Explanation: The Second Derivative Test classifies critical points based on the concavity present at those locations. With $g'(-8)=0$ confirming a critical point and concave down behavior at $x=-8$ (meaning $g''(-8)<0$), the function exhibits downward curvature. Downward concavity at a critical point creates a hill-like shape, establishing a local maximum. Choice A could mislead students who connect downward direction with minimums, but concave down specifically refers to the hill formation characteristic of maximums. Ensure both conditions are satisfied: critical point verification and determinable concavity direction.

Question 19

1. A local minimum at $x=6$.
2. An inflection point at $x=6$.
3. A local maximum at $x=6$. (correct answer)
4. No conclusion can be drawn because $t''(6)<0$.
5. Neither a maximum nor minimum because $t'(6)=0$.

Explanation: This problem tests the Second Derivative Test with a negative second derivative value. Given t'(6) = 0, we confirm x = 6 is a critical point, and since t''(6) = -1 < 0, the function exhibits downward concavity at this location. The combination of zero slope and negative concavity creates a local maximum—picture the top of a smooth hill or an inverted parabola. The graph rises to reach its peak at x = 6, then falls away on both sides. Choice D incorrectly claims that a negative second derivative prevents us from drawing conclusions, but the Second Derivative Test specifically uses the condition f''(c) < 0 at a critical point to identify local maxima. Test-taking strategy: at any critical point where f''(c) < 0, you have a local maximum—no exceptions.

Question 20

1. A local maximum at $x=-4$. (correct answer)
2. A local minimum at $x=-4$.
3. An inflection point at $x=-4$.
4. No conclusion can be drawn because $r'(-4)=0$.
5. Neither a maximum nor minimum because $r''(-4)<0$.

Explanation: This question requires applying the Second Derivative Test to classify a critical point with negative concavity. Given r'(-4) = 0, we have a critical point at x = -4, and with r''(-4) = -9 < 0, the function is concave down at this location. When a critical point occurs where the function is concave down, it creates a local maximum—visualize an inverted bowl or mountain peak. The strongly negative value -9 indicates pronounced downward curvature, forming a clear peak at x = -4. Choice E incorrectly claims that a negative second derivative prevents classification as a maximum or minimum, but the Second Derivative Test specifically uses f''(c) < 0 to identify local maxima at critical points. Quick test tip: "negative second derivative = frowning graph = local maximum."