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 AB.
Study Second Derivative Test in AP Calculus AB with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
0% Complete
For f(x)=x3−6x2+12x−5, determine f′′(x).
Tap card or press Space to flip
f′′(x)=6x−12.. Second derivative of x3−6x2+12x−5 using power rule.
How well did you know it?
Card 1 / 65
Space to flip · ← / → to move · once flipped, → Got it · ← Still learning
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 AB.
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.
Answer: f′′(x)=6x−12.. Second derivative of x3−6x2+12x−5 using power rule.
Answer: Find the critical points where f′(x)=0 or f′(x) is undefined. Critical points are necessary candidates for local extrema.
Answer: Concave up since f′′(1)=10>0. Substituting x=1 gives f′′(1)=15(1)−5=10>0.
Answer: Local minimum since f′′(2)>0. Since f′′(2)=2>0, the critical point is a minimum.
Answer: Critical points are x=0 and x=2. Found by solving f′(x)=3x2−6x=0.
Answer: The function is concave up on the interval. Positive second derivative means upward curvature.
Answer: A point where the curve changes from decreasing to increasing. The lowest point in a local neighborhood of the function.
Answer: The concavity cannot be determined. Zero second derivative gives no concavity information.
Answer: The test is inconclusive. Zero second derivative provides no information about extremum type.
Answer: If f′′(c)<0, f(c) is a local maximum. Negative second derivative indicates concave down, creating a maximum.
Answer: f′′(x)=20x3−60x2. Taking derivative twice of x5−5x4 using power rule.
Answer: Inconclusive. Zero second derivative provides no classification information.
Answer: The concavity cannot be determined. Zero second derivative gives no concavity information.
Answer: The test is inconclusive. Zero second derivative provides no information about extremum type.
Answer: When f′′(c)=0 at a critical point c. Zero second derivative at critical points gives no information.
Answer: The function is concave down on the interval. Negative second derivative means downward curvature.
Answer: A point where the curve changes from increasing to decreasing. The highest point in a local neighborhood of the function.
Answer: Local minimum. Positive second derivative at a critical point indicates a minimum.
Answer: f′′(x)=4. Second derivative of quadratic function is constant.
Answer: f′′(x)=20x3−60x2. Taking derivative twice of x5−5x4 using power rule.
Answer: f′′(x)=12x2−8. Second derivative of x4−4x2 using power rule.
Answer: f′′(2)=6. Substituting x=2 into f′′(x)=6x−12.
Answer: A point where the curve changes from increasing to decreasing. The highest point in a local neighborhood of the function.
Answer: If f′′(c)>0, f(c) is a local minimum. Positive second derivative indicates concave up, creating a minimum.
Answer: A point where the curve changes from decreasing to increasing. The lowest point in a local neighborhood of the function.
Answer: f(x) is concave up on that interval. Positive second derivative means the graph curves upward.
Answer: Inconclusive; f′′(0)=0. For f(x)=x3, f′′(0)=0 makes the test inconclusive.
Answer: A point where f′(x)=0 or f′(x) is undefined. Where the first derivative equals zero or doesn't exist.
Answer: f′′(3)=12. Substituting x=3 into f′′(x)=6x−12.
Answer: f′′(x)=18x−18. Second derivative of 3x3−9x2 using power rule.
Answer: f′′(x)=4. Second derivative of quadratic function is constant.
Answer: f′′(x)=6x. Second derivative of x3−3x+1 using power rule.
Answer: Concave up since f′′(1)=10>0. Substituting x=1 gives f′′(1)=15(1)−5=10>0.
Answer: Whether f(c) is a local max, min, or inconclusive. Classifies critical points as maxima, minima, or undetermined.
Answer: Local minimum since f′′(2)>0. Since f′′(2)=2>0, the critical point is a minimum.
Answer: f(x) is concave down on that interval. Negative second derivative means the graph curves downward.
Answer: f′′(x)=6x−12.. Second derivative of x3−6x2+12x−5 using power rule.
Answer: Concavity is determined by the sign of the second derivative. Second derivative sign determines upward or downward curvature.
Answer: The direction of the curve of a function. Describes whether a graph curves upward or downward.
Answer: Find the critical points where f′(x)=0 or f′(x) is undefined. Critical points are necessary candidates for local extrema.
Answer: If f′′(c)>0, f(c) is a local minimum. Positive second derivative indicates concave up, creating a minimum.
Answer: To determine whether a critical point is a local extremum. Uses second derivative sign at critical points to classify extrema.
Answer: Inconclusive. Zero second derivative provides no classification information.
Answer: f(x) is concave up on that interval. Positive second derivative means the graph curves upward.
Answer: When f′′(c)=0 at a critical point c. Zero second derivative at critical points gives no information.
Answer: The function is concave up on the interval. Positive second derivative means upward curvature.
Answer: Concavity is determined by the sign of the second derivative. Second derivative sign determines upward or downward curvature.
Answer: f′′(3)=12. Substituting x=3 into f′′(x)=6x−12.
Answer: The direction of the curve of a function. Describes whether a graph curves upward or downward.
Answer: f′′(x)=18x−18. Second derivative of 3x3−9x2 using power rule.
Answer: If f′′(c)<0, f(c) is a local maximum. Negative second derivative indicates concave down, creating a maximum.
Answer: The function is concave down on the interval. Negative second derivative means downward curvature.
Answer: A point where f′(x)=0 or f′(x) is undefined. Where the first derivative equals zero or doesn't exist.
Answer: Local minimum since f′′(1)=6>0. Since f′′(1)=6>0, the critical point is a minimum.
Answer: Local maximum. Negative second derivative at a critical point indicates a maximum.
Answer: f′′(x)=6x. Second derivative of x3−3x+1 using power rule.
Answer: Whether f(c) is a local max, min, or inconclusive. Classifies critical points as maxima, minima, or undetermined.
Answer: f′′(x)=36x−6. Second derivative of 6x3−3x2+2 using power rule.
Answer: f′′(x)=12x2−8. Second derivative of x4−4x2 using power rule.
Answer: Local minimum since f′′(1)=6>0. Since f′′(1)=6>0, the critical point is a minimum.
Answer: To determine whether a critical point is a local extremum. Uses second derivative sign at critical points to classify extrema.
Answer: Local maximum. Negative second derivative at a critical point indicates a maximum.
Answer: f(x) is concave down on that interval. Negative second derivative means the graph curves downward.
Answer: Inconclusive; f′′(0)=0. For f(x)=x3, f′′(0)=0 makes the test inconclusive.
Answer: Local minimum. Positive second derivative at a critical point indicates a minimum.