Second Derivative Test - AP Calculus AB

At $x=-17$ , $F'(-17)=0$ and $F''(-17)<0$ ; classify the critical point at $x=-17$ .
Answer: Local maximum at $x=-17$ . The Second Derivative Test determines critical point nature through concavity analysis at those points. Since $F'(-17)=0$ establishes a critical point and $F''(-17)<0$ indicates negative concavity, the function is concave down at $x=-17$ . 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 the second derivative.
If $s'(1)=0$ and $s''(1)=0.7$ , what does the Second Derivative Test imply at $x=1$ ?
Answer: A local minimum at $x=1$ . To apply the Second Derivative Test, we examine both derivative conditions at x = 1. Since s'(1) = 0, we have a critical point, and with s''(1) = 0.7 > 0, the function is concave up at this point. The positive second derivative indicates the graph forms a U-shape or valley at x = 1, creating a local minimum. The function decreases as it approaches x = 1 and increases as it moves away, with the lowest point occurring at x = 1. Choice B incorrectly suggests that we cannot draw conclusions from a positive second derivative, misunderstanding that the Second Derivative Test explicitly states f''(c) > 0 at a critical point indicates a local minimum. Remember: the magnitude of f''(c) doesn't matter for classification—only its sign matters (positive = minimum, negative = maximum).
At $x=-15$ , $p'(-15)=0$ and $p$ is concave down at $x=-15$ ; classify the point.
Answer: Local maximum at $x=-15$ . The Second Derivative Test classifies critical points by examining the concavity behavior at those locations. Given $p'(-15)=0$ (critical point) and concave down behavior at $x=-15$ (meaning $p''(-15)<0$ ), the function exhibits downward curvature. Downward concavity at a critical point produces a hill-shaped curve, confirming a local maximum. Choice C could attract students who associate downward direction with minimums, but concave down specifically creates the hill formation characteristic of maximums. Use this test when both critical point status and concavity direction are clearly established.
For differentiable $f$ , if $f'(3)=0$ and $f''(3)>0$ , what is the classification of $x=3$ ?
Answer: Local minimum at $x=3$ because $f''(3)>0$ . 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 if f'(c) = 0 and f''(c) > 0, then f has a local minimum at x = c because the function is concave up at that point. Here, f''(3) > 0 indicates the graph is shaped like a U near x = 3, creating a valley. Choice A incorrectly claims a maximum despite positive concavity, which would require f''(3) < 0. When using the Second Derivative Test, remember: positive second derivative means concave up (local min), negative second derivative means concave down (local max).
A differentiable function satisfies $p'(3)=0$ and is concave up at $x=3$ ; classify $x=3$ .
Answer: Local minimum at $x=3$ . 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.
For twice-differentiable $t$ , $t'(\tfrac12)=0$ and $t''(\tfrac12)>0$ . Classify the critical point at $x=\tfrac12$ .
Answer: Local minimum at $x=\tfrac12$ . 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.
For $h$ , $h'(8)=0$ and $h''(8)>0$ ; what is the classification of $x=8$ ?
Answer: Local minimum at $x=8$ . The Second Derivative Test determines the nature of critical points through concavity analysis. Since $h'(8)=0$ establishes a critical point and $h''(8)>0$ indicates positive concavity, the function is concave up at $x=8$ . 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.
For $p$ , suppose $p'(0)=0$ and $p''(0)=-0.4$ ; what is true about $x=0$ ?
Answer: A local maximum at $x=0$ . This problem applies the Second Derivative Test to classify a critical point with a negative second derivative. Since p'(0) = 0, we have a critical point at x = 0, and with p''(0) = -0.4 < 0, the function is concave down at this location. When a function has zero slope and negative concavity at a point, it forms a local maximum—picture an upside-down bowl or hilltop. The negative value -0.4 tells us the function curves downward, creating a peak at x = 0. Choice C incorrectly suggests that having p'(0) = 0 prevents classification, but this is precisely the condition needed to apply the Second Derivative Test, which then uses the sign of p''(0) to determine the nature of the critical point. Test strategy: negative second derivative at a critical point always means local maximum.
At $x=17$ , $f'(17)=0$ and $f''(17)>0$ ; what does the Second Derivative Test imply?
Answer: Local minimum at $x=17$ . The Second Derivative Test examines concavity at critical points to determine their classification. Given $f'(17)=0$ (critical point) and $f''(17)>0$ (positive second derivative), the function is concave up at $x=17$ . Positive concavity at a critical point produces upward curvature resembling a valley, confirming a local minimum. Choice B 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.
A function $q$ is twice differentiable and has $q'(1)=0$ with $q''(1)=-9$ . Classify $q$ at $x=1$ .
Answer: Local maximum at $x=1$ . 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.
A function has $q'(19)=0$ and $q''(19)>0$ ; how is the critical point at $x=19$ classified?
Answer: Local minimum at $x=19$ . The Second Derivative Test determines the nature of critical points through concavity analysis. Since $q'(19)=0$ establishes a critical point and $q''(19)>0$ indicates positive concavity, the function is concave up at $x=19$ . Upward concavity at a critical point forms a valley-shaped curve, indicating a local minimum. Choice B might confuse students who think positive values suggest maximums, but positive second derivative specifically describes the valley formation of minimums. Apply this test when you can verify both critical point existence and the sign of the second derivative.
For differentiable $f$ , $f'(2)=0$ and $f''(2)>0$ ; what is the classification of $x=2$ ?
Answer: Local minimum at $x=2$ . The Second Derivative Test classifies critical points by examining the concavity at those points. Given that $f'(2)=0$ (critical point) and $f''(2)>0$ (positive second derivative), the function is concave up at $x=2$ . When a function is concave up at a critical point, the point represents a local minimum since the curve forms a valley shape. Choice A is tempting because students might confuse positive concavity with maximum, but positive second derivative indicates upward concavity (minimum). Apply the test only when you have both a critical point ( $f'=0$ ) and known sign of the second derivative.
A function $m$ has $m'(-1)=0$ and is concave down on $(-2,0)$ . What can be concluded at $x=-1$ ?
Answer: Local maximum at $x=-1$ because concave down with a critical point. This question tests the Second Derivative Test using concavity information. We have m'(-1) = 0, establishing a critical point at x = -1. Since m is concave down on (-2,0), we know m''(x) < 0 throughout this interval, including m''(-1) < 0. By the Second Derivative Test, when f'(c) = 0 and f''(c) < 0, there is a local maximum at x = c. The concave down shape creates an inverted U with a peak at x = -1. Choice A incorrectly claims concave down implies m''(-1) > 0, which reverses the definition of concavity. Test tip: "concave down" always means negative second derivative and local maximum at critical points.
At $x=6$ , $s'(6)=0$ and $s''(6)<0$ ; what does the Second Derivative Test imply?
Answer: Local maximum at $x=6$ . The Second Derivative Test uses concavity information to classify the nature of critical points. Since $s'(6)=0$ establishes a critical point and $s''(6)<0$ 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 associate negative signs with minimums, but negative second derivative specifically means downward concavity (maximum). Apply this test when you have confirmed both the critical point condition and the sign of the second derivative.
For $q$ , $q'(10)=0$ and $q''(10)>0$ ; what does the Second Derivative Test conclude?
Answer: Local minimum at $x=10$ . The Second Derivative Test uses concavity information to classify the type of critical point present. With $q'(10)=0$ confirming a critical point and $q''(10)>0$ indicating positive concavity, the function exhibits upward curvature at $x=10$ . Upward concavity at a critical point creates a valley-like shape, establishing a local minimum. Choice A might mislead students who link positive direction with maximums, but positive second derivative specifically describes the upward curvature of minimums. Remember to confirm both critical point existence and definitive second derivative sign before classification.
Given $g'(5)=0$ and $g''(5)>0$ , what does the Second Derivative Test conclude at $x=5$ ?
Answer: Local minimum at $x=5$ . 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.
At $x=-19$ , $g'(-19)=0$ and $g$ is concave down at $x=-19$ ; classify the critical point.
Answer: Local maximum at $x=-19$ . 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.
At $x=-10$ , $q'(-10)=0$ and $q''(-10)>0$ ; how is $x=-10$ classified?
Answer: Local minimum at $x=-10$ . The Second Derivative Test examines concavity at critical points to determine their classification. Given $q'(-10)=0$ (critical point) and $q''(-10)>0$ (positive second derivative), the function is concave up at $x=-10$ . Positive concavity at a critical point produces upward curvature resembling a valley, confirming a local minimum. Choice A 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.
A twice-differentiable function $q$ satisfies $q'(5)=0$ and $q''(5)=2$ ; classify $x=5$ .
Answer: A local minimum at $x=5$ . To classify this critical point, we systematically apply the Second Derivative Test. With q'(5) = 0, we confirm x = 5 is a critical point, and since q''(5) = 2 > 0, the function exhibits upward concavity at this point. The combination of a horizontal tangent and positive concavity creates a local minimum—imagine a smooth valley or bowl shape centered at x = 5. The function decreases as it approaches x = 5 from either side, then increases as it moves away. Choice C incorrectly suggests that a positive second derivative prevents drawing conclusions, but the Second Derivative Test explicitly states that f''(c) > 0 at a critical point indicates a local minimum. Remember: at critical points, positive concavity (f''(c) > 0) means minimum, negative concavity (f''(c) < 0) means maximum.
At $x=15$ , $s'(15)=0$ and $s''(15)>0$ ; classify the critical point at $x=15$ .
Answer: Local minimum at $x=15$ . The Second Derivative Test uses concavity information to classify the type of critical point present. With $s'(15)=0$ confirming a critical point and $s''(15)>0$ indicating positive concavity, the function exhibits upward curvature at $x=15$ . Upward concavity at a critical point creates a valley-like shape, establishing a local minimum. Choice B might mislead students who link positive direction with maximums, but positive second derivative specifically describes the upward curvature of minimums. Remember to confirm both critical point existence and definitive second derivative sign before classification.

Start this quiz