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AP Calculus AB Flashcards: First Derivative Test

Study First 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.

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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 AB.

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 AB Flashcards: First Derivative Test

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QUESTION

Find f′(x)f'(x)f′(x) for f(x)=2x3−3x2−12x+1f(x) = 2x^3 - 3x^2 - 12x + 1f(x)=2x3−3x2−12x+1.

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ANSWER

f′(x)=6x2−6x−12f'(x) = 6x^2 - 6x - 12f′(x)=6x2−6x−12. Use power rule: 6x26x^26x2 from 2x32x^32x3, −6x-6x−6x from −3x2-3x^2−3x2.

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Flashcard 1: Find f′(x)f'(x)f′(x) for f(x)=2x3−3x2−12x+1f(x) = 2x^3 - 3x^2 - 12x + 1f(x)=2x3−3x2−12x+1.

Answer: f′(x)=6x2−6x−12f'(x) = 6x^2 - 6x - 12f′(x)=6x2−6x−12. Use power rule: 6x26x^26x2 from 2x32x^32x3, −6x-6x−6x from −3x2-3x^2−3x2.

Flashcard 2: Identify the critical points of f(x)=2x3−3x2−12x+1f(x) = 2x^3 - 3x^2 - 12x + 1f(x)=2x3−3x2−12x+1.

Answer: Critical points: x=−1,2x = -1, 2x=−1,2. Solve 6x2−6x−12=6(x2−x−2)=06x^2 - 6x - 12 = 6(x^2 - x - 2) = 06x2−6x−12=6(x2−x−2)=0.

Flashcard 3: Determine the intervals where f(x)=x3−3x2+2f(x) = x^3 - 3x^2 + 2f(x)=x3−3x2+2 is increasing.

Answer: Increasing on (2, infinity)(2, \, \text{infinity})(2,infinity). f′(x)>0f'(x) > 0f′(x)>0 when x>2x > 2x>2 (test a point like x=3x = 3x=3).

Flashcard 4: Determine the intervals where f(x)=x3−3x2+2f(x) = x^3 - 3x^2 + 2f(x)=x3−3x2+2 is decreasing.

Answer: Decreasing on (−infinity,0)(-\text{infinity}, 0)(−infinity,0) and (0,2)(0, 2)(0,2).. f′(x)<0f'(x) < 0f′(x)<0 when x<0x < 0x<0 or 0<x<20 < x < 20<x<2.

Flashcard 5: What does f′(x)<0f'(x) < 0f′(x)<0 indicate about a function on an interval?

Answer: The function is decreasing on that interval. Negative derivative means the function has negative slope.

Flashcard 6: What is the relative extrema at x=πx = \pix=π for f(x)=cos⁡xf(x) = \cos xf(x)=cosx?

Answer: Relative minimum. f′f'f′ changes from negative to positive at x=πx = \pix=π.

Flashcard 7: Identify the critical points of f(x)=cos⁡xf(x) = \cos xf(x)=cosx on [0,2\backslashpi][0, 2\backslashpi][0,2\backslashpi].

Answer: Critical points: x=πx = \pix=π. −sin⁡x=0-\sin x = 0−sinx=0 when x=πx = \pix=π in the given interval.

Flashcard 8: Identify the critical points of f(x)=14x4−x2+2f(x) = \frac{1}{4}x^4 - x^2 + 2f(x)=41​x4−x2+2.

Answer: Critical points: x=0,±1x = 0, \pm 1x=0,±1. Solve x3−2x=x(x2−2)=0x^3 - 2x = x(x^2 - 2) = 0x3−2x=x(x2−2)=0.

Flashcard 9: What is the relative extrema at x=−2x = -\sqrt{2}x=−2​ for f(x)=13x3−2x+1f(x) = \frac{1}{3}x^3 - 2x + 1f(x)=31​x3−2x+1?

Answer: Relative maximum. f′f'f′ changes from positive to negative at x=−2x = -\sqrt{2}x=−2​.

Flashcard 10: What is the relative extrema at x=π2x = \frac{\pi}{2}x=2π​ for f(x)=sin⁡xf(x) = \sin xf(x)=sinx?

Answer: Relative maximum. f′f'f′ changes from positive to negative at x=π2x = \frac{\pi}{2}x=2π​.

Flashcard 11: What is the relative extrema at x=3π2x = \frac{3\pi}{2}x=23π​ for f(x)=sin⁡xf(x) = \sin xf(x)=sinx?

Answer: Relative minimum. f′f'f′ changes from negative to positive at x=3π2x = \frac{3\pi}{2}x=23π​.

Flashcard 12: Find f′(x)f'(x)f′(x) for f(x)=14x4−x2+2f(x) = \frac{1}{4}x^4 - x^2 + 2f(x)=41​x4−x2+2.

Answer: f′(x)=x3−2xf'(x) = x^3 - 2xf′(x)=x3−2x. Derivative of 14x4\frac{1}{4}x^441​x4 is x3x^3x3, of x2x^2x2 is 2x2x2x.

Flashcard 13: Determine the critical points of f(x)=13x3−2x+1f(x) = \frac{1}{3}x^3 - 2x + 1f(x)=31​x3−2x+1.

Answer: Critical points: x=±2x = \pm \sqrt{2}x=±2​. Solve x2−2=0x^2 - 2 = 0x2−2=0 to get x2=2x^2 = 2x2=2.

Flashcard 14: Find the critical points of f(x)=x2+4x+4f(x) = x^2 + 4x + 4f(x)=x2+4x+4.

Answer: Critical point: x=−2x = -2x=−2. f′(x)=2x+4=0f'(x) = 2x + 4 = 0f′(x)=2x+4=0 gives x=−2x = -2x=−2.

Flashcard 15: What is the relative extrema at x=2x = 2x=2 for f(x)=x3−3x2+2f(x) = x^3 - 3x^2 + 2f(x)=x3−3x2+2?

Answer: Relative minimum. f′f'f′ changes from negative to positive at x=2x = 2x=2.

Flashcard 16: Find the derivative of f(x)=x3−3x2+2f(x) = x^3 - 3x^2 + 2f(x)=x3−3x2+2.

Answer: f′(x)=3x2−6xf'(x) = 3x^2 - 6xf′(x)=3x2−6x. Use power rule: derivative of x3x^3x3 is 3x23x^23x2, x2x^2x2 is 2x2x2x.

Flashcard 17: Find the critical points of f(x)=x3−3x2+2f(x) = x^3 - 3x^2 + 2f(x)=x3−3x2+2.

Answer: Critical points: x=0,x=2x = 0, x = 2x=0,x=2. Set f′(x)=3x2−6x=3x(x−2)=0f'(x) = 3x^2 - 6x = 3x(x-2) = 0f′(x)=3x2−6x=3x(x−2)=0.

Flashcard 18: Identify the type of extrema at a critical point if f′f'f′ changes from positive to negative.

Answer: Relative maximum. Function rises then falls, creating a local peak.

Flashcard 19: State the condition for a critical point of a function.

Answer: The derivative f′(x)=0f'(x) = 0f′(x)=0 or f′(x)f'(x)f′(x) is undefined. Critical points occur where the slope is zero or undefined.

Flashcard 20: What is the First Derivative Test used for in calculus?

Answer: To determine relative (local) extrema of a function. Identifies where functions reach local peaks and valleys.

Flashcard 21: What is the relative extrema at x=2x = 2x=2 for f(x)=2x3−3x2−12x+1f(x) = 2x^3 - 3x^2 - 12x + 1f(x)=2x3−3x2−12x+1?

Answer: Relative minimum. f′f'f′ changes from negative to positive at x=2x = 2x=2.

Flashcard 22: Identify the type of extrema at a critical point if f′f'f′ changes from negative to positive.

Answer: Relative minimum. Function falls then rises, creating a local valley.

Flashcard 23: What is the relative extrema at x=−1x = -1x=−1 for f(x)=2x3−3x2−12x+1f(x) = 2x^3 - 3x^2 - 12x + 1f(x)=2x3−3x2−12x+1?

Answer: Relative maximum. f′f'f′ changes from positive to negative at x=−1x = -1x=−1.

Flashcard 24: Find the derivative of f(x)=13x3−2x+1f(x) = \frac{1}{3}x^3 - 2x + 1f(x)=31​x3−2x+1.

Answer: f′(x)=x2−2f'(x) = x^2 - 2f′(x)=x2−2. Use power rule on each term of the polynomial.

Flashcard 25: What conclusion is drawn if f′(x)f'(x)f′(x) does not change sign at a critical point?

Answer: No relative extrema at that point. No sign change means no local maximum or minimum.

Flashcard 26: Find the derivative of f(x)=sin⁡xf(x) = \sin xf(x)=sinx.

Answer: f′(x)=cos⁡xf'(x) = \cos xf′(x)=cosx. Derivative of sine function is cosine.

Flashcard 27: What does f′(x)>0f'(x) > 0f′(x)>0 indicate about a function on an interval?

Answer: The function is increasing on that interval. Positive derivative means the function has positive slope.

Flashcard 28: State the derivative of f(x)=cos⁡xf(x) = \cos xf(x)=cosx.

Answer: f′(x)=−sin⁡xf'(x) = -\sin xf′(x)=−sinx. Derivative of cosine function is negative sine.

Flashcard 29: Determine the intervals where f(x)=cos⁡xf(x) = \cos xf(x)=cosx is decreasing on [0,2π][0, 2\pi][0,2π].

Answer: Decreasing on (0,π)(0, \pi)(0,π). −sin⁡x<0-\sin x < 0−sinx<0 when 0<x<π0 < x < \pi0<x<π.

Flashcard 30: What is the relative extrema at x=−2x = -2x=−2 for f(x)=x2+4x+4f(x) = x^2 + 4x + 4f(x)=x2+4x+4?

Answer: Relative minimum. f′f'f′ changes from negative to positive at x=−2x = -2x=−2.