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This deck focuses on Second Derivative Test, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus BC.
AP Calculus BC Flashcards: Second Derivative Test
Study Second Derivative Test in AP Calculus BC with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
AP Calculus BC
Second Derivative Test
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QUESTION
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Evaluate f′′(2) for f(x)=x4−4x3+6x2.
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ANSWER
f′′(2)=12. f′(x)=4x3−12x2+12x, so f′′(x)=12x2−24x+12.
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Flashcard 1: Evaluate f′′(2) for f(x)=x4−4x3+6x2.
Answer: f′′(2)=12. f′(x)=4x3−12x2+12x, so f′′(x)=12x2−24x+12.
Flashcard 2: Evaluate f′′(0) for f(x)=x4−4x2+1.
Answer: f′′(0)=−8. f′(x)=4x3−8x, so f′′(x)=12x2−8 and f′′(0)=−8.
Flashcard 3: What is the second derivative of f(x)=−x4+2x2−x?
Answer: f′′(x)=−12x2+4. Differentiate f′(x)=−4x3+4x−1 to get the second derivative.
Flashcard 4: What is the sign of f′′(x) if f(x) is concave down?
Answer: Negative. Concave down functions have negative curvature everywhere.
Flashcard 5: What does f′′(c)<0 imply about f(x) at x=c?
Answer: Local maximum. Negative second derivative confirms the critical point is a maximum.
Flashcard 6: What is the sign of f′′(x) if f(x) is concave up?
Answer: Positive. Concave up functions have positive curvature everywhere.
Flashcard 7: What is the second derivative of f(x)=3x4−8x2+5?
Answer: f′′(x)=36x2−16. Differentiate f′(x)=12x3−16x to get the second derivative.
Flashcard 8: What does f′′(c)=0 indicate about the critical point c?
Answer: Test is inconclusive at c. Zero second derivative provides no information about the extremum type.
Flashcard 9: Find the second derivative of f(x)=x3+3x2+x+1.
Answer: f′′(x)=6x+6. Differentiate f′(x)=3x2+6x+1 to get the second derivative.
Flashcard 10: Determine the nature of the extremum at x=21 for f(x)=−x2+x.
Answer: Local maximum. f′′(1/2)=−2<0 at the critical point, indicating a maximum.
Flashcard 11: Determine if f(x)=x3−3x2+4 has a local extremum at x=2.
Answer: Local minimum at x=2. f′(2)=0 and f′′(2)=6>0, confirming a local minimum.
Flashcard 12: What does f′′(c)>0 imply about f(x) at x=c?
Answer: Local minimum. Positive second derivative confirms the critical point is a minimum.
Flashcard 13: What does f′′(c)>0 imply about f(x) at x=c?
Answer: Local minimum. Positive second derivative confirms the critical point is a minimum.
Flashcard 14: Evaluate f′′(1) for f(x)=x4−x2+x.
Answer: f′′(1)=10. f′(x)=4x3−2x+1 and f′′(x)=12x2−2, so f′′(1)=10.
Flashcard 15: Find the second derivative of f(x)=x5−5x3+7.
Answer: f′′(x)=20x3−30x. Differentiate f′(x)=5x4−15x2 to get the second derivative.
Flashcard 16: State the second derivative test condition for a local maximum.
Answer: f′′(c)<0 at critical point c. Negative second derivative indicates concave down, confirming a maximum.
Flashcard 17: Determine the nature of the extremum for f(x)=x2−4x+4 at x=2.
Answer: Local minimum. f′(2)=0 and f′′(2)=2>0, confirming a local minimum.
Flashcard 18: What does f′′(c)<0 imply about f(x) at x=c?
Answer: Local maximum. Negative second derivative confirms the critical point is a maximum.
Flashcard 19: Identify the nature of the critical point for f(x)=x3−3x at x=0.
Answer: Inconclusive. f′′(0)=0 at this critical point, so the test fails.
Flashcard 20: At which type of point is the second derivative test applied?
Answer: Critical points. Only points where f′(x)=0 can be tested for extrema.
Flashcard 21: What is the sign of f′′(x) if f(x) is concave down?
Answer: Negative. Concave down functions have negative curvature everywhere.
Flashcard 22: Find f′′(x) for f(x)=x4−4x2+2.
Answer: f′′(x)=12x2−8. Take the derivative of f′(x)=4x3−8x to get the second derivative.
Flashcard 23: What is the second derivative of f(x)=−x3+3x−5?
Answer: f′′(x)=−6x. Differentiate f′(x)=−3x2+3 to get the second derivative.
Flashcard 24: Calculate f′′(3) for f(x)=x3−9x2+27x−3.
Answer: f′′(3)=0. f′(x)=3x2−18x+27 and f′′(x)=6x−18, so f′′(3)=0.
Flashcard 25: What is the second derivative test used for?
Answer: Determining local extrema of a function. Uses concavity at critical points to classify local maxima and minima.
Flashcard 26: Determine the nature of the extremum for f(x)=x2−4x+4 at x=2.
Answer: Local minimum. f′(2)=0 and f′′(2)=2>0, confirming a local minimum.
Flashcard 27: What conclusion is drawn if f′′(c)=0 at a critical point c?
Answer: Test inconclusive. Zero second derivative means the test cannot determine extremum type.
Flashcard 28: Identify the sign of f′′(x) for a concave up interval.
Answer: f′′(x)>0. Positive second derivative always indicates upward concavity.
Flashcard 29: Find the second derivative of f(x)=31x3−2x2+3x.
Answer: f′′(x)=2x−4. f′(x)=x2−4x+3, so f′′(x)=2x−4.
Flashcard 30: What is the second derivative of f(x)=2x3−3x2+x?
Answer: f′′(x)=12x−6. Differentiate f′(x)=6x2−6x+1 to get the second derivative.
Flashcard 31: Calculate f′′(3) for f(x)=x3−9x2+27x−3.
Answer: f′′(3)=0. f′(x)=3x2−18x+27 and f′′(x)=6x−18, so f′′(3)=0.
Flashcard 32: What conclusion is drawn if f′′(c)=0 at a critical point c?
Answer: Test inconclusive. Zero second derivative means the test cannot determine extremum type.
Flashcard 33: Identify the sign of f′′(x) for a concave down interval.
Answer: f′′(x)<0. Negative second derivative always indicates downward concavity.
Flashcard 34: Identify the sign of f′′(x) for a concave up interval.
Answer: f′′(x)>0. Positive second derivative always indicates upward concavity.
Flashcard 35: Find the second derivative of f(x)=31x3−2x2+3x.
Answer: f′′(x)=2x−4. f′(x)=x2−4x+3, so f′′(x)=2x−4.
Flashcard 36: Find f′′(x) for f(x)=x4−4x2+2.
Answer: f′′(x)=12x2−8. Take the derivative of f′(x)=4x3−8x to get the second derivative.
Flashcard 37: What is the second derivative of f(x)=3x4−8x2+5?
Answer: f′′(x)=36x2−16. Differentiate f′(x)=12x3−16x to get the second derivative.
Flashcard 38: Determine the nature of the extremum for f(x)=x2−4 at x=0.
Answer: Local minimum. f′(0)=0 and f′′(0)=2>0, confirming a local minimum.
Flashcard 39: What is the condition for test inconclusiveness in the second derivative test?
Answer: f′′(c)=0 at c. This condition makes the second derivative test unable to determine extremum type.
Flashcard 40: What is the second derivative of f(x)=2x3−3x2+x?
Answer: f′′(x)=12x−6. Differentiate f′(x)=6x2−6x+1 to get the second derivative.
Flashcard 41: What is the second derivative of f(x)=−x3+3x−5?
Answer: f′′(x)=−6x. Differentiate f′(x)=−3x2+3 to get the second derivative.
Flashcard 42: What is the sign of f′′(x) if f(x) is concave up?
Answer: Positive. Concave up functions have positive curvature everywhere.
Flashcard 43: Find the second derivative of f(x)=x5−5x3+7.
Answer: f′′(x)=20x3−30x. Differentiate f′(x)=5x4−15x2 to get the second derivative.
Flashcard 44: State the second derivative test condition for a local minimum.
Answer: f′′(c)>0 at critical point c. Positive second derivative indicates concave up, confirming a minimum.
Flashcard 45: What is the second derivative of f(x)=−x4+2x2−x?
Answer: f′′(x)=−12x2+4. Differentiate f′(x)=−4x3+4x−1 to get the second derivative.
Flashcard 46: Determine concavity for f(x)=2x2−3x+1 at x=0.
Answer: Concave up. f′′(x)=4>0 everywhere, so the function is concave up.
Flashcard 47: Determine concavity for f(x)=2x2−3x+1 at x=0.
Answer: Concave up. f′′(x)=4>0 everywhere, so the function is concave up.
Flashcard 48: Evaluate f′′(0) for f(x)=x4−4x2+1.
Answer: f′′(0)=−8. f′(x)=4x3−8x, so f′′(x)=12x2−8 and f′′(0)=−8.
Flashcard 49: Determine the nature of the extremum for f(x)=x2−4 at x=0.
Answer: Local minimum. f′(0)=0 and f′′(0)=2>0, confirming a local minimum.
Flashcard 50: At which type of point is the second derivative test applied?
Answer: Critical points. Only points where f′(x)=0 can be tested for extrema.
Flashcard 51: What is the condition for test inconclusiveness in the second derivative test?
Answer: f′′(c)=0 at c. This condition makes the second derivative test unable to determine extremum type.
Flashcard 52: What does f′′(c)=0 indicate about the critical point c?
Answer: Test is inconclusive at c. Zero second derivative provides no information about the extremum type.
Flashcard 53: Evaluate f′′(2) for f(x)=x4−4x3+6x2.
Answer: f′′(2)=12. f′(x)=4x3−12x2+12x, so f′′(x)=12x2−24x+12.
Flashcard 54: State the second derivative test condition for a local maximum.
Answer: f′′(c)<0 at critical point c. Negative second derivative indicates concave down, confirming a maximum.
Flashcard 55: Determine the nature of the extremum at x=21 for f(x)=−x2+x.
Answer: Local maximum. f′′(1/2)=−2<0 at the critical point, indicating a maximum.
Flashcard 56: Find the second derivative of f(x)=x3+3x2+x+1.
Answer: f′′(x)=6x+6. Differentiate f′(x)=3x2+6x+1 to get the second derivative.
Flashcard 57: Evaluate f′′(1) for f(x)=x4−x2+x.
Answer: f′′(1)=10. f′(x)=4x3−2x+1 and f′′(x)=12x2−2, so f′′(1)=10.
Flashcard 58: Identify the nature of the critical point for f(x)=x3−3x at x=0.
Answer: Inconclusive. f′′(0)=0 at this critical point, so the test fails.
Flashcard 59: State the second derivative test condition for a local minimum.
Answer: f′′(c)>0 at critical point c. Positive second derivative indicates concave up, confirming a minimum.
Flashcard 60: Determine if f(x)=x3−3x2+4 has a local extremum at x=2.
Answer: Local minimum at x=2. f′(2)=0 and f′′(2)=6>0, confirming a local minimum.