AP Calculus BC 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 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.

Question 1

1. A relative minimum at $x=-2$ (correct answer)
2. A relative maximum at $x=-2$
3. A point of inflection at $x=-2$
4. Cannot be determined because $s''(-2)>0$
5. Neither a maximum nor minimum at $x=-2$ because it is a critical point

Explanation: This problem requires the Second Derivative Test to classify critical points of a function. The test identifies a relative minimum when s'(c) = 0 and s''(c) > 0, showing concave up. Here, x = -2 is a critical point, and s''(-2) = 9 > 0 indicates upward concavity. This positive concavity forms a local trough where the graph bends up. A tempting distractor is choice B, a relative maximum, but positive second derivatives denote minima, not maxima. Ensure the point is indeed critical and the second derivative non-zero to qualify for the test's conclusive application.

Question 2

1. Local minimum at $x=-6$
2. Neither; $M''(-6)<0$ implies concave up
3. Local maximum at $x=-6$ (correct answer)
4. Neither; the test is inconclusive because $M''(-6)<0$
5. Cannot be determined without $M(-6)$

Explanation: The Second Derivative Test is a method to classify critical points of a function by examining the sign of the second derivative at those points. At x=-6, M'(-6)=0 marks a critical point, and M''(-6)=-1, negative, indicates concave down. This suggests a local maximum, as the graph peaks there. The negative value ensures decreasing function away from the point. A tempting distractor is choice B, which claims M''(-6)<0 implies concave up, but it means concave down. To apply the Second Derivative Test effectively, always ensure the first derivative is zero and check if the second derivative is non-zero; if it is zero, switch to the First Derivative Test for classification.

Question 3

1. A relative maximum at $x=\pi$
2. A relative minimum at $x=\pi$ (correct answer)
3. A point of inflection at $x=\pi$
4. Cannot be determined because $\pi$ is irrational
5. Neither a maximum nor minimum at $x=\pi$ because $F''(\pi)>0$

Explanation: This problem requires the Second Derivative Test to classify critical points of a function. The test states that F'(c) = 0 and F''(c) > 0 imply a relative minimum due to concave up. For F, F'(π) = 0 marks the critical point, and F''(π) = 2 > 0 shows upward concavity. This positive value means the graph forms a local minimum at x = π. A tempting distractor is choice A, a relative maximum, but this would require a negative second derivative instead. A transferable strategy is to first locate critical points where the first derivative is zero, then check the second derivative's sign for classification.

Question 4

1. Local maximum at $x=-\pi$
2. Neither; $G''(-\pi)>0$ implies a local maximum
3. Cannot be determined without $G(-\pi)$
4. Local minimum at $x=-\pi$ (correct answer)
5. Neither; the test is inconclusive because $G''(-\pi)>0$

Explanation: The Second Derivative Test is a method to classify critical points of a function by examining the sign of the second derivative at those points. At x=-π, G'(-π)=0 indicates a critical point, and G''(-π)=5, positive, shows concave up. This upward curve suggests a local minimum. The function forms a valley there, increasing away. A tempting distractor is choice B, which incorrectly states G''(-π)>0 implies a local maximum, but positive means minimum. To apply the Second Derivative Test effectively, always ensure the first derivative is zero and check if the second derivative is non-zero; if it is zero, switch to the First Derivative Test for classification.

Question 5

1. Local minimum at $x=\tfrac{1}{2}$ (correct answer)
2. Local maximum at $x=\tfrac{1}{2}$
3. Neither; the test is inconclusive because $k''(\tfrac{1}{2})\neq 0$
4. Neither; it must be an inflection point
5. Cannot be determined without checking $k'(x)$ on both sides

Explanation: The Second Derivative Test is a method to classify critical points of a function by examining the sign of the second derivative at those points. At x=1/2, k'(1/2)=0 marks a critical point, and k''(1/2)=3, positive, indicates concave up. Concave up implies a minimum point, as the graph curves upward like a bowl. Thus, there is a local minimum at x=1/2, with function values rising away from it. A tempting distractor is choice D, which assumes it must be an inflection point, but positive f'' confirms a minimum, not a concavity change. To apply the Second Derivative Test effectively, always ensure the first derivative is zero and check if the second derivative is non-zero; if it is zero, switch to the First Derivative Test for classification.

Question 6

1. Neither; $b''(3)<0$ implies concave up
2. Local minimum at $x=3$
3. Local maximum at $x=3$ (correct answer)
4. Neither; the test is inconclusive because $b'(3)=0$
5. Cannot be determined without checking $b''(x)$ near $3$

Explanation: The Second Derivative Test classifies critical points by checking the second derivative's sign at locations where the first derivative is zero. If negative, like b''(3) = -1/5 < 0, the function is concave down, indicating a peak. This means the point is higher than its neighbors, marking a local maximum. Therefore, x = 3 is a local maximum. Some might think negative implies concave up, but that's incorrect; negative denotes concave down. Ensure the critical point is isolated and the function smooth to apply the test across various functions.

Question 7

1. Local maximum at $x=-1$
2. The test is inconclusive at $x=-1$ (correct answer)
3. Local minimum at $x=-1$
4. Neither; it must be an inflection point
5. Neither; $c''(-1)=0$ implies concave down

Explanation: The Second Derivative Test is applied to critical points to determine their nature through the second derivative. When the second derivative is zero, as with c''(-1) = 0, concavity doesn't clearly indicate a max or min. It could be either or neither, like an inflection point or flat spot. Thus, the test is inconclusive here. A distractor suggesting it's necessarily an inflection point fails because zero second derivative alone doesn't confirm a concavity change; further checks are needed. Always confirm if higher derivatives or other tests like the first derivative test can resolve inconclusive cases.

Question 8

1. Local maximum at $x=-\tfrac{5}{4}$
2. Local minimum at $x=-\tfrac{5}{4}$ (correct answer)
3. Neither; it must be an inflection point
4. Neither; the test is inconclusive because $r''(-\tfrac{5}{4})\neq 0$
5. Cannot be determined without checking $r'(x)$

Explanation: Employing the Second Derivative Test, positive derivatives indicate minima. r''(-5/4) = 16/3 > 0 means concave up, local minimum. Graph bottoms locally. So, x = -5/4 is a local minimum. Needing f value is unnecessary; test uses derivatives only. Confirm differentiability for application strategy.

Question 9

1. Neither; $g''(-9)<0$ implies concave up
2. Local minimum at $x=-9$
3. Local maximum at $x=-9$ (correct answer)
4. Neither; the test is inconclusive because $g''(-9)\neq 0$
5. Cannot be determined without checking $g'(x)$

Explanation: The Second Derivative Test assesses concavity for critical point classification. A negative second derivative, g''(-9) = -11/3 < 0, shows concave down, indicating a local maximum. Values decrease from this peak. Therefore, x = -9 is a local maximum. The idea that negative implies concave up is false; it's the opposite. Always check if the first derivative changes sign or use alternatives if inconclusive.

Question 10

1. Local minimum at $x=0$ because $r''(0)<0$.
2. The Second Derivative Test is inconclusive at $x=0$.
3. Local maximum at $x=0$ because $r''(0)<0$. (correct answer)
4. Point of inflection at $x=0$ because $r'(0)=0$.
5. Cannot be determined without checking $r'(x)$ sign changes.

Explanation: This problem applies the Second Derivative Test with a negative second derivative. We have r'(0) = 0 (critical point) and r''(0) = -9 < 0. According to the Second Derivative Test, when f'(c) = 0 and f''(c) < 0, the function has a local maximum at x = c. The negative second derivative indicates concave down behavior, creating a hill shape at x = 0. Choice A reverses the relationship, incorrectly pairing negative concavity with a minimum. For test success, remember the mnemonic: "negative second derivative = frown shape = maximum" and "positive second derivative = smile shape = minimum."

Question 11

1. Point of inflection at $x=5$ because $t''(5)>0$.
2. Local maximum at $x=5$ because $t''(5)>0$.
3. Local minimum at $x=5$ because $t''(5)>0$. (correct answer)
4. The Second Derivative Test is inconclusive at $x=5$.
5. Cannot be determined without knowing $t(5)$.

Explanation: This problem applies the Second Derivative Test with a positive fractional second derivative. Given t'(5) = 0 and t''(5) = 3/4 > 0, we can classify the critical point at x = 5. Since t''(5) > 0, the Second Derivative Test tells us that t has a local minimum at x = 5. The positive second derivative indicates upward concavity, creating a valley shape. Choice A incorrectly suggests an inflection point despite the non-zero second derivative; inflection points require f'' = 0 or undefined, not just any critical point. For the AP exam, any positive f''(c) value—whether integer, fraction, or decimal—indicates a local minimum at a critical point.

Question 12

1. Local minimum at $x=0$ (correct answer)
2. Neither; it must be an inflection point
3. Local maximum at $x=0$
4. Neither; the Second Derivative Test is inconclusive because $a''(0)>0$
5. Cannot be determined without $a(0)$

Explanation: Utilizing the Second Derivative Test, we classify critical points based on the concavity indicated by the second derivative at points where the first derivative vanishes. Positive concavity, shown by a''(0) = 4 > 0, means the graph curves upward, fostering a local minimum. At such a point, the function dips to its lowest nearby value. Thus, x = 0 is a local minimum. A common error is assuming it's an inflection point, but inflection points require a change in concavity, not just a zero first derivative. Remember, the test requires the point to be a critical point with a non-zero second derivative for definitive classification.

Question 13

1. A point of inflection at $x=-1$
2. A relative minimum at $x=-1$ (correct answer)
3. A relative maximum at $x=-1$
4. Neither a maximum nor minimum at $x=-1$ because $g''(-1)>0$
5. Cannot be determined because $g'(-1)=0$

Explanation: This problem requires the Second Derivative Test to classify critical points of a function. The Second Derivative Test indicates that if g'(c) = 0 and g''(c) > 0, then there is a relative minimum at x = c due to the graph being concave up. For g, g'(-1) = 0 marks a critical point, and g''(-1) = 7, which is positive, showing upward concavity. This upward concavity implies the function curves toward the tangent line from below, forming a local valley. A tempting distractor is choice C, a relative maximum, but this is incorrect as a negative second derivative, not positive, signals a maximum. To use the test reliably, ensure the function is twice differentiable and the first derivative vanishes at the point in question.

Question 14

1. Local maximum at $x=-\tfrac{1}{2}$
2. Neither; it must be an inflection point
3. Local minimum at $x=-\tfrac{1}{2}$ (correct answer)
4. Neither; the test is inconclusive because $S''(-\tfrac{1}{2})\neq 0$
5. Cannot be determined without $S(-\tfrac{1}{2})$

Explanation: The Second Derivative Test is used to classify critical points of a function by examining the sign of the second derivative at those points. When the first derivative is zero at a point, indicating a critical point, the second derivative reveals the concavity: a positive value means the graph is concave up, like a U-shape. This concavity suggests that the function values are higher on either side, confirming a local minimum. In this case, since $S''(-\tfrac{1}{2}) = \tfrac{1}{9} > 0$, the point is a local minimum. A tempting distractor might claim it's inconclusive because the second derivative isn't zero, but the test is only inconclusive when it is zero, not when it's positive or negative. Always ensure the first derivative is zero before applying the test to confirm it's a critical point eligible for classification.

Question 15

1. A local maximum at $x=8$
2. An inflection point at $x=8$
3. The Second Derivative Test is inconclusive at $x=8$
4. A local minimum at $x=8$ (correct answer)
5. Neither a local maximum nor a local minimum at $x=8$

Explanation: This problem assesses your understanding of the Second Derivative Test for classifying critical points. When v'(c) = 0 and v''(c) > 0, the function is concave up, classifying the critical point as a local minimum. This upward concavity means the graph bottoms out at that point. In this case, v'(8) = 0 and v''(8) = 4, positive, so there is a local minimum at x = 8. A distractor like local maximum fails because it requires a negative second derivative, not positive. To apply the test, ensure the point is critical and the second derivative is non-zero; if zero, consider using sign changes in the first derivative instead.

Question 16

1. Local minimum at $x=3$
2. Local maximum at $x=3$
3. Neither; it must be an inflection point
4. The test is inconclusive at $x=3$ (correct answer)
5. Cannot be determined because $g'(3)=0$

Explanation: The Second Derivative Test is a method to classify critical points of a function by examining the sign of the second derivative at those points. At x=3, g'(3)=0 indicates a critical point, but g''(3)=0 means the test cannot determine the nature based on concavity. When the second derivative is zero, the graph's curvature is flat at that point, so it could be a maximum, minimum, or neither. Therefore, the Second Derivative Test is inconclusive, and alternative methods are needed for classification. A tempting distractor is choice C, which assumes it must be an inflection point, but f''=0 does not guarantee a change in concavity; higher derivatives or sign changes around the point are required. To apply the Second Derivative Test effectively, always ensure the first derivative is zero and check if the second derivative is non-zero; if it is zero, switch to the First Derivative Test for classification.

Question 17

1. Local maximum at $x=2$ (correct answer)
2. Local minimum at $x=2$
3. Neither; $f''(2)<0$ implies an inflection point
4. Neither; the Second Derivative Test is inconclusive because $f''(2)\neq 0$
5. Cannot be determined without $f(2)$

Explanation: The Second Derivative Test is a method to classify critical points of a function by examining the sign of the second derivative at those points. At x=2, since f'(2)=0, it is a critical point, and f''(2)=-5, which is negative, indicates the function is concave down there. Concave down means the graph curves downward, resembling the peak of a hill, so there is a local maximum at x=2. This classification holds because the negative second derivative confirms the function values decrease as we move away from the point in either direction. A tempting distractor is choice C, which suggests it's an inflection point, but a negative second derivative does not imply a change in concavity required for an inflection point. To apply the Second Derivative Test effectively, always ensure the first derivative is zero and check if the second derivative is non-zero; if it is zero, switch to the First Derivative Test for classification.

Question 18

1. Neither; $f''(\pi)=-1$ makes the test inconclusive
2. Local minimum at $x=\pi$
3. Local maximum at $x=\pi$ (correct answer)
4. Neither; $f''(\pi)<0$ implies concave up
5. Cannot be determined without $f(\pi)$

Explanation: The Second Derivative Test is a method to classify critical points of a function by examining the sign of the second derivative at those points. At x=π, f'(π)=0 is a critical point, and f''(π)=-1, negative, shows concave down. This means the graph has a downward curve, indicating a local maximum at x=π. The classification is based on the function decreasing away from the point due to this concavity. A tempting distractor is choice D, which says f''(π)<0 implies concave up, but actually, negative second derivative means concave down. To apply the Second Derivative Test effectively, always ensure the first derivative is zero and check if the second derivative is non-zero; if it is zero, switch to the First Derivative Test for classification.

Question 19

1. A local maximum at $x=2$ (correct answer)
2. A local minimum at $x=2$
3. The Second Derivative Test is inconclusive at $x=2$
4. An inflection point at $x=2$
5. Neither a local maximum nor a local minimum at $x=2$

Explanation: This problem assesses your understanding of the Second Derivative Test for classifying critical points. The test specifies that a negative second derivative at a critical point implies concave-down shape, indicating a local maximum. This means the graph peaks at that point, with values decreasing on either side. For r'(2) = 0 and r''(2) = -1, negative, it confirms a local maximum at x = 2. A distractor like inconclusive might appeal if one confuses the sign, but the test is conclusive when the second derivative is non-zero. Always verify test eligibility by ensuring the function is twice differentiable and the second derivative is not zero at the critical point.

Question 20

1. A local minimum at $x=3$
2. A local maximum at $x=3$ (correct answer)
3. Neither a local maximum nor a local minimum at $x=3$
4. The Second Derivative Test is inconclusive at $x=3$
5. An inflection point at $x=3$

Explanation: This problem assesses your understanding of the Second Derivative Test for classifying critical points. The Second Derivative Test states that if f'(c) = 0 and f''(c) > 0, then there is a local minimum at x = c because the graph is concave up, resembling a valley. Conversely, if f''(c) < 0, the graph is concave down, resembling a hill, which indicates a local maximum. In this case, f'(3) = 0 and f''(3) = -5, which is negative, confirming a local maximum at x = 3. A tempting distractor might be choosing the inconclusive option, but that only applies when f''(c) = 0, not when it is definitively negative. To ensure the Second Derivative Test is eligible, verify that the first derivative is zero and the second derivative is defined and non-zero at the critical point.