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AP Calculus BC Flashcards: First Derivative Test

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 this deck covers

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.

How to use these flashcards

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: First Derivative Test

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QUESTION

What test helps confirm if a critical point is an extremum?

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ANSWER

The First Derivative Test. Distinguishes between maxima, minima, and inflection points.

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Flashcard 1: What test helps confirm if a critical point is an extremum?

Answer: The First Derivative Test. Distinguishes between maxima, minima, and inflection points.

Flashcard 2: What test helps confirm if a critical point is an extremum?

Answer: The First Derivative Test. Distinguishes between maxima, minima, and inflection points.

Flashcard 3: State the condition for a critical point.

Answer: f′(x)=0f'(x) = 0f′(x)=0 or f′(x)f'(x)f′(x) is undefined. Necessary condition for potential extrema to exist.

Flashcard 4: Which test determines where a function is increasing or decreasing?

Answer: The First Derivative Test. Sign of f′(x)f'(x)f′(x) determines increasing/decreasing intervals.

Flashcard 5: What indicates a relative minimum using the First Derivative Test?

Answer: Derivative changes from negative to positive. Sign change from −-− to +++ creates a valley in the graph.

Flashcard 6: What indicates a relative maximum using the First Derivative Test?

Answer: Derivative changes from positive to negative. Sign change from +++ to −-− creates a peak in the graph.

Flashcard 7: What is a relative extremum?

Answer: A point where f(x)f(x)f(x) changes from increasing to decreasing or vice versa. Local maximum or minimum where function direction reverses.

Flashcard 8: What is the derivative of f(x)=13x3−xf(x) = \frac{1}{3}x^3 - xf(x)=31​x3−x?

Answer: f′(x)=x2−1f'(x) = x^2 - 1f′(x)=x2−1. Apply power rule to each term separately.

Flashcard 9: Identify the relative extrema: f(x)=x3−3x2f(x) = x^3 - 3x^2f(x)=x3−3x2.

Answer: Relative maximum at x=0x = 0x=0; relative minimum at x=2x = 2x=2. f′(x)=3x2−6xf'(x) = 3x^2 - 6xf′(x)=3x2−6x; sign changes at x=0x = 0x=0 and x=2x = 2x=2.

Flashcard 10: What does f′(x)>0f'(x) > 0f′(x)>0 indicate about f(x)f(x)f(x)?

Answer: f(x)f(x)f(x) is increasing. Positive derivative means function values are rising.

Flashcard 11: Describe the behavior if f′(x)f'(x)f′(x) changes from negative to positive.

Answer: There's a relative minimum. Function reaches a valley; slope changes from downward to upward.

Flashcard 12: What is the derivative of f(x)=x5−5x3f(x) = x^5 - 5x^3f(x)=x5−5x3?

Answer: f′(x)=5x4−15x2f'(x) = 5x^4 - 15x^2f′(x)=5x4−15x2. Power rule applied to polynomial with multiple terms.

Flashcard 13: Find the derivative of f(x)=x3−3x2+2xf(x) = x^3 - 3x^2 + 2xf(x)=x3−3x2+2x.

Answer: f′(x)=3x2−6x+2f'(x) = 3x^2 - 6x + 2f′(x)=3x2−6x+2. Apply power rule term by term to the polynomial.

Flashcard 14: Find the derivative: f(x)=x4−4x3f(x) = x^4 - 4x^3f(x)=x4−4x3.

Answer: f′(x)=4x3−12x2f'(x) = 4x^3 - 12x^2f′(x)=4x3−12x2. Apply power rule: multiply by exponent, reduce exponent by 1.

Flashcard 15: Determine if there is a minimum at x=0x = 0x=0 for f(x)=x2f(x) = x^2f(x)=x2.

Answer: Yes, minimum at x=0x = 0x=0. f′(x)=2xf'(x) = 2xf′(x)=2x changes from negative to positive at origin.

Flashcard 16: Find f′(x)f'(x)f′(x) for f(x)=x3−9x2+27xf(x) = x^3 - 9x^2 + 27xf(x)=x3−9x2+27x.

Answer: f′(x)=3x2−18x+27f'(x) = 3x^2 - 18x + 27f′(x)=3x2−18x+27. Apply power rule to each term in the polynomial.

Flashcard 17: Evaluate f′(x)f'(x)f′(x) at x=0x = 0x=0 for f(x)=13x3−xf(x) = \frac{1}{3}x^3 - xf(x)=31​x3−x.

Answer: f′(0)=−1f'(0) = -1f′(0)=−1. Substitute x=0x = 0x=0 into f′(x)=x2−1f'(x) = x^2 - 1f′(x)=x2−1.

Flashcard 18: Determine relative extrema for f(x)=x3−3x+1f(x) = x^3 - 3x + 1f(x)=x3−3x+1 using the test.

Answer: No relative extrema. f′(x)=3x2−3=0f'(x) = 3x^2 - 3 = 0f′(x)=3x2−3=0 at x=±1x = \pm 1x=±1; no sign changes.

Flashcard 19: Describe the behavior if f′(x)f'(x)f′(x) changes from positive to negative.

Answer: There's a relative maximum. Function reaches a peak; slope changes from upward to downward.

Flashcard 20: Determine if there is an extremum at x=1x = 1x=1 for f(x)=2x3−3x2f(x) = 2x^3 - 3x^2f(x)=2x3−3x2.

Answer: No extremum at x=1x = 1x=1. f′(x)=6x2−6xf'(x) = 6x^2 - 6xf′(x)=6x2−6x; no sign change at x=1x = 1x=1.

Flashcard 21: Determine extrema for f(x)=x2−4x+3f(x) = x^2 - 4x + 3f(x)=x2−4x+3 using the test.

Answer: Minimum at x=2x = 2x=2. f′(x)=2x−4=0f'(x) = 2x - 4 = 0f′(x)=2x−4=0 at x=2x = 2x=2; sign changes from −-− to +++.

Flashcard 22: What is the next step after finding critical points in the test?

Answer: Evaluate f′(x)f'(x)f′(x) around critical points. Check sign changes of f′(x)f'(x)f′(x) on intervals around each critical point.

Flashcard 23: What must be true for a critical point to be a local extremum?

Answer: The derivative must change signs. Sign change of f′(x)f'(x)f′(x) distinguishes extrema from inflection points.

Flashcard 24: Identify extrema for f(x)=x2−2x+1f(x) = x^2 - 2x + 1f(x)=x2−2x+1 using the test.

Answer: Minimum at x=1x = 1x=1. f′(x)=2x−2=0f'(x) = 2x - 2 = 0f′(x)=2x−2=0 at x=1x = 1x=1; sign changes from −-− to +++.

Flashcard 25: Determine if there is a maximum at x=0x = 0x=0 for f(x)=−x2f(x) = -x^2f(x)=−x2.

Answer: Yes, maximum at x=0x = 0x=0. f′(x)=−2xf'(x) = -2xf′(x)=−2x changes from positive to negative at origin.

Flashcard 26: What is the derivative of f(x)=x2−4xf(x) = x^2 - 4xf(x)=x2−4x?

Answer: f′(x)=2x−4f'(x) = 2x - 4f′(x)=2x−4. Basic power rule: derivative of x2x^2x2 is 2x2x2x.

Flashcard 27: Find f′(x)f'(x)f′(x): f(x)=2x5−5x4f(x) = 2x^5 - 5x^4f(x)=2x5−5x4.

Answer: f′(x)=10x4−20x3f'(x) = 10x^4 - 20x^3f′(x)=10x4−20x3. Apply power rule: bring down exponent, reduce power by 1.

Flashcard 28: Evaluate f′(x)f'(x)f′(x) at x=2x = 2x=2 for f(x)=x3−3x2f(x) = x^3 - 3x^2f(x)=x3−3x2.

Answer: f′(2)=0f'(2) = 0f′(2)=0. Substitute x=2x = 2x=2 into f′(x)=3x2−6xf'(x) = 3x^2 - 6xf′(x)=3x2−6x.

Flashcard 29: Find f′(x)f'(x)f′(x): f(x)=x33−xf(x) = \frac{x^3}{3} - xf(x)=3x3​−x.

Answer: f′(x)=x2−1f'(x) = x^2 - 1f′(x)=x2−1. Standard power rule application with fractional coefficient.

Flashcard 30: Determine relative extrema for f(x)=x2f(x) = x^2f(x)=x2 using the test.

Answer: Minimum at x=0x = 0x=0. f′(x)=2xf'(x) = 2xf′(x)=2x; changes from −-− to +++ at x=0x = 0x=0.