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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.

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What this deck covers

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.

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Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

AP Calculus BC Flashcards: Second Derivative Test

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QUESTION

State the second derivative test condition for a local maximum.

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ANSWER

f′′(c)<0f''(c) < 0f′′(c)<0 at critical point ccc. Negative second derivative indicates concave down, confirming a maximum.

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Flashcard 1: State the second derivative test condition for a local maximum.

Answer: f′′(c)<0f''(c) < 0f′′(c)<0 at critical point ccc. Negative second derivative indicates concave down, confirming a maximum.

Flashcard 2: Evaluate f′′(1)f''(1)f′′(1) for f(x)=x4−x2+xf(x) = x^4 - x^2 + xf(x)=x4−x2+x.

Answer: f′′(1)=10f''(1) = 10f′′(1)=10. f′(x)=4x3−2x+1f'(x) = 4x^3 - 2x + 1f′(x)=4x3−2x+1 and f′′(x)=12x2−2f''(x) = 12x^2 - 2f′′(x)=12x2−2, so f′′(1)=10f''(1) = 10f′′(1)=10.

Flashcard 3: What does f′′(c)<0f''(c) < 0f′′(c)<0 imply about f(x)f(x)f(x) at x=cx=cx=c?

Answer: Local maximum. Negative second derivative confirms the critical point is a maximum.

Flashcard 4: Determine the nature of the extremum at x=12x = \frac{1}{2}x=21​ for f(x)=−x2+xf(x) = -x^2 + xf(x)=−x2+x.

Answer: Local maximum. f′′(1/2)=−2<0f''(1/2) = -2 < 0f′′(1/2)=−2<0 at the critical point, indicating a maximum.

Flashcard 5: Determine if f(x)=x3−3x2+4f(x) = x^3 - 3x^2 + 4f(x)=x3−3x2+4 has a local extremum at x=2x = 2x=2.

Answer: Local minimum at x=2x = 2x=2. f′(2)=0f'(2) = 0f′(2)=0 and f′′(2)=6>0f''(2) = 6 > 0f′′(2)=6>0, confirming a local minimum.

Flashcard 6: What conclusion is drawn if f′′(c)=0f''(c) = 0f′′(c)=0 at a critical point ccc?

Answer: Test inconclusive. Zero second derivative means the test cannot determine extremum type.

Flashcard 7: What is the second derivative of f(x)=3x4−8x2+5f(x) = 3x^4 - 8x^2 + 5f(x)=3x4−8x2+5?

Answer: f′′(x)=36x2−16f''(x) = 36x^2 - 16f′′(x)=36x2−16. Differentiate f′(x)=12x3−16xf'(x) = 12x^3 - 16xf′(x)=12x3−16x to get the second derivative.

Flashcard 8: At which type of point is the second derivative test applied?

Answer: Critical points. Only points where f′(x)=0f'(x) = 0f′(x)=0 can be tested for extrema.

Flashcard 9: Determine the nature of the extremum for f(x)=x2−4x+4f(x) = x^2 - 4x + 4f(x)=x2−4x+4 at x=2x = 2x=2.

Answer: Local minimum. f′(2)=0f'(2) = 0f′(2)=0 and f′′(2)=2>0f''(2) = 2 > 0f′′(2)=2>0, confirming a local minimum.

Flashcard 10: Identify the nature of the critical point for f(x)=x3−3xf(x) = x^3 - 3xf(x)=x3−3x at x=0x = 0x=0.

Answer: Inconclusive. f′′(0)=0f''(0) = 0f′′(0)=0 at this critical point, so the test fails.

Flashcard 11: What is the sign of f′′(x)f''(x)f′′(x) if f(x)f(x)f(x) is concave down?

Answer: Negative. Concave down functions have negative curvature everywhere.

Flashcard 12: What is the second derivative of f(x)=−x4+2x2−xf(x) = -x^4 + 2x^2 - xf(x)=−x4+2x2−x?

Answer: f′′(x)=−12x2+4f''(x) = -12x^2 + 4f′′(x)=−12x2+4. Differentiate f′(x)=−4x3+4x−1f'(x) = -4x^3 + 4x - 1f′(x)=−4x3+4x−1 to get the second derivative.

Flashcard 13: Evaluate f′′(2)f''(2)f′′(2) for f(x)=x4−4x3+6x2f(x) = x^4 - 4x^3 + 6x^2f(x)=x4−4x3+6x2.

Answer: f′′(2)=12f''(2) = 12f′′(2)=12. f′(x)=4x3−12x2+12xf'(x) = 4x^3 - 12x^2 + 12xf′(x)=4x3−12x2+12x, so f′′(x)=12x2−24x+12f''(x) = 12x^2 - 24x + 12f′′(x)=12x2−24x+12.

Flashcard 14: Find the second derivative of f(x)=13x3−2x2+3xf(x) = \frac{1}{3}x^3 - 2x^2 + 3xf(x)=31​x3−2x2+3x.

Answer: f′′(x)=2x−4f''(x) = 2x - 4f′′(x)=2x−4. f′(x)=x2−4x+3f'(x) = x^2 - 4x + 3f′(x)=x2−4x+3, so f′′(x)=2x−4f''(x) = 2x - 4f′′(x)=2x−4.

Flashcard 15: What does f′′(c)=0f''(c) = 0f′′(c)=0 indicate about the critical point ccc?

Answer: Test is inconclusive at ccc. Zero second derivative provides no information about the extremum type.

Flashcard 16: State the second derivative test condition for a local minimum.

Answer: f′′(c)>0f''(c) > 0f′′(c)>0 at critical point ccc. Positive second derivative indicates concave up, confirming a minimum.

Flashcard 17: What is the second derivative of f(x)=2x3−3x2+xf(x) = 2x^3 - 3x^2 + xf(x)=2x3−3x2+x?

Answer: f′′(x)=12x−6f''(x) = 12x - 6f′′(x)=12x−6. Differentiate f′(x)=6x2−6x+1f'(x) = 6x^2 - 6x + 1f′(x)=6x2−6x+1 to get the second derivative.

Flashcard 18: Determine concavity for f(x)=2x2−3x+1f(x) = 2x^2 - 3x + 1f(x)=2x2−3x+1 at x=0x = 0x=0.

Answer: Concave up. f′′(x)=4>0f''(x) = 4 > 0f′′(x)=4>0 everywhere, so the function is concave up.

Flashcard 19: Identify the sign of f′′(x)f''(x)f′′(x) for a concave down interval.

Answer: f′′(x)<0f''(x) < 0f′′(x)<0. Negative second derivative always indicates downward concavity.

Flashcard 20: What is the condition for test inconclusiveness in the second derivative test?

Answer: f′′(c)=0f''(c) = 0f′′(c)=0 at ccc. This condition makes the second derivative test unable to determine extremum type.

Flashcard 21: What is the second derivative of f(x)=−x3+3x−5f(x) = -x^3 + 3x - 5f(x)=−x3+3x−5?

Answer: f′′(x)=−6xf''(x) = -6xf′′(x)=−6x. Differentiate f′(x)=−3x2+3f'(x) = -3x^2 + 3f′(x)=−3x2+3 to get the second derivative.

Flashcard 22: Determine the nature of the extremum for f(x)=x2−4f(x) = x^2 - 4f(x)=x2−4 at x=0x = 0x=0.

Answer: Local minimum. f′(0)=0f'(0) = 0f′(0)=0 and f′′(0)=2>0f''(0) = 2 > 0f′′(0)=2>0, confirming a local minimum.

Flashcard 23: What does f′′(c)>0f''(c) > 0f′′(c)>0 imply about f(x)f(x)f(x) at x=cx=cx=c?

Answer: Local minimum. Positive second derivative confirms the critical point is a minimum.

Flashcard 24: Calculate f′′(3)f''(3)f′′(3) for f(x)=x3−9x2+27x−3f(x) = x^3 - 9x^2 + 27x - 3f(x)=x3−9x2+27x−3.

Answer: f′′(3)=0f''(3) = 0f′′(3)=0. f′(x)=3x2−18x+27f'(x) = 3x^2 - 18x + 27f′(x)=3x2−18x+27 and f′′(x)=6x−18f''(x) = 6x - 18f′′(x)=6x−18, so f′′(3)=0f''(3) = 0f′′(3)=0.

Flashcard 25: Evaluate f′′(0)f''(0)f′′(0) for f(x)=x4−4x2+1f(x) = x^4 - 4x^2 + 1f(x)=x4−4x2+1.

Answer: f′′(0)=−8f''(0) = -8f′′(0)=−8. f′(x)=4x3−8xf'(x) = 4x^3 - 8xf′(x)=4x3−8x, so f′′(x)=12x2−8f''(x) = 12x^2 - 8f′′(x)=12x2−8 and f′′(0)=−8f''(0) = -8f′′(0)=−8.

Flashcard 26: Find the second derivative of f(x)=x5−5x3+7f(x) = x^5 - 5x^3 + 7f(x)=x5−5x3+7.

Answer: f′′(x)=20x3−30xf''(x) = 20x^3 - 30xf′′(x)=20x3−30x. Differentiate f′(x)=5x4−15x2f'(x) = 5x^4 - 15x^2f′(x)=5x4−15x2 to get the second derivative.

Flashcard 27: What is the sign of f′′(x)f''(x)f′′(x) if f(x)f(x)f(x) is concave up?

Answer: Positive. Concave up functions have positive curvature everywhere.

Flashcard 28: Find f′′(x)f''(x)f′′(x) for f(x)=x4−4x2+2f(x) = x^4 - 4x^2 + 2f(x)=x4−4x2+2.

Answer: f′′(x)=12x2−8f''(x) = 12x^2 - 8f′′(x)=12x2−8. Take the derivative of f′(x)=4x3−8xf'(x) = 4x^3 - 8xf′(x)=4x3−8x to get the second derivative.

Flashcard 29: Identify the sign of f′′(x)f''(x)f′′(x) for a concave up interval.

Answer: f′′(x)>0f''(x) > 0f′′(x)>0. Positive second derivative always indicates upward concavity.

Flashcard 30: Determine the nature of the extremum for f(x)=x2−4x+4f(x) = x^2 - 4x + 4f(x)=x2−4x+4 at x=2x = 2x=2.

Answer: Local minimum. f′(2)=0f'(2) = 0f′(2)=0 and f′′(2)=2>0f''(2) = 2 > 0f′′(2)=2>0, confirming a local minimum.