Second Derivative Test - AP Calculus BC

1. Suppose $b'(3)=0$ and $b''(3)=-\tfrac{1}{5}$; classify the critical point at $x=3$.
   Answer: Local maximum at $x=3$ 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.
2. If $s'(1)=0$ and $s''(1)=-4$, how is the critical point at $x=1$ classified?
   Answer: Local maximum at $x=1$ 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, s'(1)=0 marks a critical point, and s''(1)=-4, negative, shows concave down. This suggests a hilltop shape, indicating a local maximum at x=1. The negative value ensures the function decreases on both sides. A tempting distractor is choice D, which claims s''(1)<0 implies concave up, but it actually 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.
3. If $F'(1)=0$ and $F''(1)=-2$, what does the Second Derivative Test conclude at $x=1$?
   Answer: Local maximum at $x=1$ 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, F'(1)=0 is a critical point, and F''(1)=-2, negative, shows concave down. This concavity implies a local maximum, like a hilltop. The function decreases on either side due to this curvature. A tempting distractor is choice C, which assumes it must be an inflection point, but negative f'' does not require a sign 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.
4. At $x=\tfrac{9}{2}$, $q'(\tfrac{9}{2})=0$ and $q''(\tfrac{9}{2})=-2$; classify the critical point.
   Answer: Local maximum at $x=\tfrac{9}{2}$ The Second Derivative Test uses signs for classification. Negative q''(9/2) = -2 < 0 shows concave down, local maximum. Peak occurs here. Thus, x = 9/2 is a local maximum. Negative implying concave up is backward, failing as distractor. Verify f' = 0 for transferable use.
5. For $w$, $w'(\!-4)=0$ and $w''(\!-4)=11$; what is the classification at $x=-4$?
   Answer: Local minimum at $x=-4$ 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=-4, w'(-4)=0 marks a critical point, and w''(-4)=11, positive, indicates concave up. This suggests a valley, confirming a local minimum at x=-4. The positive value means the function increases away from the point. A tempting distractor is choice A, which claims w''(-4)>0 implies concave down, but it actually means concave up. 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.
6. A differentiable function satisfies $f'(\pi)=0$ and $f''(\pi)=-1$; classify the critical point at $x=\pi$.
   Answer: Local maximum at $x=\pi$ 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.
7. A function $s$ has a critical point at $x=-4$ and is concave down near $x=-4$. What is $x=-4$?
   Answer: Local maximum at $x=-4$ This problem applies the Second Derivative Test using concavity terminology. We know x = -4 is a critical point (so s'(-4) = 0) and s is concave down near x = -4. Concave down means the second derivative is negative, so s''(-4) < 0. The Second Derivative Test states that when f'(c) = 0 and f''(c) < 0, the function has a local maximum at x = c. Therefore, x = -4 is a local maximum. Some students might select option E (inflection point) by associating any concavity information with inflection points, but constant concavity at a critical point indicates an extremum. Remember: concave down at a critical point means local maximum—picture an upside-down parabola.
8. Let $h$ be twice differentiable with $h'(0)=0$ and $h''(0)=0$. What can be concluded about $x=0$?
   Answer: The Second Derivative Test is inconclusive at $x=0$. This problem illustrates when the Second Derivative Test fails to provide information. We have h'(0) = 0 (critical point) and h''(0) = 0. The Second Derivative Test specifically requires f''(c) ≠ 0 to draw conclusions; when f''(c) = 0, the test is inconclusive. This means we cannot determine whether x = 0 is a local maximum, local minimum, or neither using this test alone. Choice C incorrectly assumes that f'(0) = 0 automatically creates an inflection point, but inflection points require a change in concavity, not just a critical point. When f''(c) = 0 in the Second Derivative Test, you must use other methods like the First Derivative Test or analyzing higher derivatives.
9. At $x=5$, $p'(5)=0$ and $p''(5)=12$; classify the critical point at $x=5$.
   Answer: Local minimum at $x=5$ 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=5, p'(5)=0 denotes a critical point, and p''(5)=12, positive, indicates concave up. Concave up suggests a valley shape, so there is a local minimum at x=5. This is because the function increases as we depart from the point in both directions. A tempting distractor is choice D, which claims the test is inconclusive because p''(5)>0, but the test is conclusive for non-zero second derivatives. 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.
10. Given $t'(6)=0$ and $t''(6)=\tfrac{1}{3}$, what is the classification of the critical point at $x=6$?
    Answer: Local minimum at $x=6$ 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, t'(6)=0 is a critical point, and t''(6)=1/3, positive, indicates concave up. Concave up means the graph forms a minimum point. Thus, there is a local minimum at x=6, with values increasing away. A tempting distractor is choice A, which assumes it must be an inflection point, but positive f'' confirms a 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.
11. For $f(x)=x^3-3x$, $f'(1)=0$ and $f''(1)=6$. What is the classification of $x=1$?
    Answer: Local minimum at $x=1$ because $f''(1)>0$. This question applies the Second Derivative Test to a specific function. Given f'(1) = 0 and f''(1) = 6, we can classify the critical point at x = 1. Since f''(1) = 6 > 0, the Second Derivative Test tells us that f has a local minimum at x = 1. The positive second derivative indicates the function is concave up at this point, creating a valley shape. Choice B incorrectly suggests an inflection point, but inflection points occur where concavity changes (f'' changes sign), not where f' = 0. For quick test application: check f'(c) = 0 first, then if f''(c) > 0 → local min, if f''(c) < 0 → local max.
12. For $r$, $r'(5)=0$ and $r''(5)=-\tfrac{1}{2}$; which statement best classifies $x=5$?
    Answer: A relative maximum at $x=5$ This problem requires the Second Derivative Test to classify critical points of a function. The test concludes a relative maximum if r'(c) = 0 and r''(c) < 0, indicating concave down behavior. For r, r'(5) = 0 is the critical point, and r''(5) = -1/2 < 0 confirms downward concavity. This negative value means the function peaks at x = 5, curving downward around it. A tempting distractor is choice C, a relative minimum, but this requires a positive second derivative, not negative. For effective use, verify the second derivative's sign after confirming the first derivative is zero at the critical point.
13. A function $g$ has $g'(-1)=0$ and $g''(-1)=-4$. How should $x=-1$ be classified?
    Answer: Local maximum at $x=-1$ because $g''(-1)<0$. This question tests the Second Derivative Test for classifying critical points. We have g'(-1) = 0, confirming 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 g''(-1) = -4 < 0, the function is concave down at x = -1, forming a hill shape and thus a local maximum. Choice B incorrectly associates negative second derivative with a minimum, reversing the relationship between concavity and extrema. For the AP exam, memorize: f''(c) < 0 means concave down → local max; f''(c) > 0 means concave up → local min.
14. If $m'(4)=0$ and $m''(4)=-9$, what does the Second Derivative Test conclude at $x=4$?
    Answer: Local maximum at $x=4$ 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=4, m'(4)=0 is a critical point, and m''(4)=-9, negative, shows concave down. This concavity suggests a peak, confirming a local maximum at x=4. The function decreases on both sides due to this downward curvature. A tempting distractor is choice D, which claims it must be an inflection point, but negative f'' indicates a maximum without requiring a sign change in concavity. 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.
15. A differentiable function has $S'(\!-\tfrac{1}{2})=0$ and $S''(\!-\tfrac{1}{2})=\tfrac{1}{9}$; classify the critical point.
    Answer: Local minimum at $x=-\tfrac{1}{2}$ 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.
16. A differentiable function satisfies $v'(9)=0$ and $v''(9)=-\tfrac{3}{4}$; classify the critical point at $x=9$.
    Answer: Local maximum at $x=9$ 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=9, v'(9)=0 is a critical point, and v''(9)=-3/4, negative, shows concave down. This downward curve indicates a local maximum. The function peaks there, decreasing on both sides. A tempting distractor is choice D, which says it must be an inflection point, but negative f'' does not imply that. 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.
17. For $t$, $t'(0)=0$ and $t''(0)=0$, but $t''(x)>0$ for $x\ne 0$ near $0$. Classify $x=0$.
    Answer: Neither; the Second Derivative Test is inconclusive This question presents a subtle case where the Second Derivative Test is inconclusive. We have t'(0) = 0 (critical point) and t''(0) = 0, which makes the Second Derivative Test inconclusive. Even though t''(x) > 0 for x ≠ 0 near 0, the test specifically requires examining t''(0), not nearby values. The fact that t is concave up everywhere except at x = 0 suggests this might be a local minimum (like f(x) = x⁴), but the Second Derivative Test cannot confirm this. Students often incorrectly choose option E by using information about concavity near the point rather than at the point. The Second Derivative Test requires f''(c) ≠ 0; when f''(c) = 0, always report "inconclusive" regardless of nearby behavior.
18. A function $p$ has a critical point at $x=0$ with $p''(0)=12$. What is the classification of $x=0$?
    Answer: Local minimum at $x=0$ This question applies the Second Derivative Test when given that x = 0 is a critical point. The problem states p has a critical point at x = 0, which means p'(0) = 0. With p''(0) = 12 > 0, the Second Derivative Test tells us that p has a local minimum at x = 0 because the function is concave up there. The positive second derivative creates a U-shaped curve around the critical point. Students might be tempted by option A (inflection point) if they misremember that inflection points require f''(x) = 0, not f''(x) > 0. For quick recall: at a critical point, positive second derivative means local minimum—think "positive = U-shape = minimum."
19. For differentiable $f$, $f'(2)=0$ and $f''(2)=-5$; how is the critical point at $x=2$ classified?
    Answer: Local maximum at $x=2$ 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.
20. Let $f$ be twice differentiable with $f'(0)=0$ and $f''(0)=0$. What can be concluded at $x=0$?​​
    Answer: The Second Derivative Test is inconclusive at $x=0$ This question examines when the Second Derivative Test fails to provide a conclusion. We have f'(0) = 0 (critical point) and f''(0) = 0. The Second Derivative Test only works when f''(c) ≠ 0; specifically, it classifies critical points as relative maxima when f''(c) < 0 or relative minima when f''(c) > 0. When f''(c) = 0, the test gives no information about whether the critical point is a maximum, minimum, or neither. Choice D (inflection point) is incorrect because while f''(0) = 0 is necessary for an inflection point, we'd need to verify that f'' changes sign. Remember: f''(c) = 0 makes the Second Derivative Test inconclusive.

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