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This deck focuses on First Derivative Test, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus BC.
Study First 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 does f′(x)=0 imply?
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Possible extremum; critical point. Potential location for maximum, minimum, or inflection point.
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This deck focuses on First Derivative Test, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus BC.
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: Possible extremum; critical point. Potential location for maximum, minimum, or inflection point.
Answer: To determine relative (local) extrema of a function. Identifies local maxima and minima by analyzing sign changes of f′(x).
Answer: f′(x)=5x4−15x2. Power rule applied to polynomial with multiple terms.
Answer: The First Derivative Test. Distinguishes between maxima, minima, and inflection points.
Answer: f′(2)=0. Substitute x=2 into f′(x)=3x2−6x.
Answer: No extremum at x=1. f′(x)=6x2−6x; no sign change at x=1.
Answer: There's a relative minimum. Function reaches a valley; slope changes from downward to upward.
Answer: Where f′(x)=0 or f′(x) is undefined. These are the only points where extrema can occur.
Answer: f′(2)=0. Substitute x=2 into f′(x)=3x2−6x.
Answer: There's a relative maximum. Function reaches a peak; slope changes from upward to downward.
Answer: f′(x)=0 or f′(x) is undefined. Necessary condition for potential extrema to exist.
Answer: f(x) is increasing. Positive derivative means function values are rising.
Answer: The derivative must change signs. Sign change of f′(x) distinguishes extrema from inflection points.
Answer: The derivative must change signs. Sign change of f′(x) distinguishes extrema from inflection points.
Answer: f′(x)=10x4−20x3. Apply power rule: bring down exponent, reduce power by 1.
Answer: Relative maximum at x=0; relative minimum at x=2. f′(x)=3x2−6x; sign changes at x=0 and x=2.
Answer: f′(0)=−1. Substitute x=0 into f′(x)=x2−1.
Answer: f′(x)=3x2−6x+2. Apply power rule term by term to the polynomial.
Answer: f′(x)=2x−4. Basic power rule: derivative of x2 is 2x.
Answer: There's a relative minimum. Function reaches a valley; slope changes from downward to upward.
Answer: Minimum at x=2. f′(x)=2x−4=0 at x=2; sign changes from − to +.
Answer: f′(x)=0 or f′(x) is undefined. Necessary condition for potential extrema to exist.
Answer: Yes, maximum at x=0. f′(x)=−2x changes from positive to negative at origin.
Answer: A point where f(x) changes from increasing to decreasing or vice versa. Local maximum or minimum where function direction reverses.
Answer: f′(0)=−1. Substitute x=0 into f′(x)=x2−1.
Answer: f′(x)=5x4−15x2. Power rule applied to polynomial with multiple terms.
Answer: f′(x)=3x2−18x+27. Apply power rule to each term in the polynomial.
Answer: Minimum at x=0. f′(x)=2x; changes from − to + at x=0.
Answer: f′(x)=10x4−20x3. Apply power rule: bring down exponent, reduce power by 1.
Answer: Evaluate f′(x) around critical points. Check sign changes of f′(x) on intervals around each critical point.
Answer: The First Derivative Test. Sign of f′(x) determines increasing/decreasing intervals.
Answer: Minimum at x=1. f′(x)=2x−2=0 at x=1; sign changes from − to +.
Answer: No relative extrema. f′(x)=3x2−3=0 at x=±1; no sign changes.
Answer: Derivative changes from negative to positive. Sign change from − to + creates a valley in the graph.
Answer: f′(x)=x2−1. Standard power rule application with fractional coefficient.
Answer: Yes, maximum at x=0. f′(x)=−2x changes from positive to negative at origin.
Answer: There's a relative maximum. Function reaches a peak; slope changes from upward to downward.
Answer: Relative maximum at x=0; relative minimum at x=2. f′(x)=3x2−6x; sign changes at x=0 and x=2.
Answer: Derivative changes from positive to negative. Sign change from + to − creates a peak in the graph.
Answer: Derivative changes from positive to negative. Sign change from + to − creates a peak in the graph.
Answer: The First Derivative Test. Sign of f′(x) determines increasing/decreasing intervals.
Answer: f(x) is increasing. Positive derivative means function values are rising.
Answer: A point where f(x) changes from increasing to decreasing or vice versa. Local maximum or minimum where function direction reverses.
Answer: f′(x)=4x3−12x2. Apply power rule: multiply by exponent, reduce exponent by 1.
Answer: f′(x)=4x3−12x2. Apply power rule: multiply by exponent, reduce exponent by 1.
Answer: f′(x)=x2−1. Standard power rule application with fractional coefficient.
Answer: Possible extremum; critical point. Potential location for maximum, minimum, or inflection point.
Answer: f′(x)=3x2−6x+2. Apply power rule term by term to the polynomial.
Answer: Evaluate f′(x) around critical points. Check sign changes of f′(x) on intervals around each critical point.
Answer: Minimum at x=1. f′(x)=2x−2=0 at x=1; sign changes from − to +.
Answer: Derivative changes from negative to positive. Sign change from − to + creates a valley in the graph.
Answer: Yes, minimum at x=0. f′(x)=2x changes from negative to positive at origin.
Answer: The First Derivative Test. Distinguishes between maxima, minima, and inflection points.
Answer: f(x) is decreasing. Negative derivative means function values are falling.
Answer: No relative extrema. f′(x)=3x2−3=0 at x=±1; no sign changes.
Answer: f′(x)=x2−1. Apply power rule to each term separately.
Answer: No extremum at x=1. f′(x)=6x2−6x; no sign change at x=1.
Answer: Minimum at x=0. f′(x)=2x; changes from − to + at x=0.
Answer: Minimum at x=2. f′(x)=2x−4=0 at x=2; sign changes from − to +.
Answer: f′(x)=3x2−18x+27. Apply power rule to each term in the polynomial.
Answer: Yes, minimum at x=0. f′(x)=2x changes from negative to positive at origin.
Answer: f′(x)=2x−4. Basic power rule: derivative of x2 is 2x.
Answer: f′(x)=x2−1. Apply power rule to each term separately.
Answer: f(x) is decreasing. Negative derivative means function values are falling.