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AP Calculus AB Flashcards: Determining Intervals On Increasing Decreasing Functions

Study Determining Intervals On Increasing Decreasing Functions 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 Determining Intervals On Increasing Decreasing Functions, 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: Determining Intervals On Increasing Decreasing Functions

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QUESTION

What does it mean if f′(x)=0f'(x) = 0f′(x)=0 at some point?

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ANSWER

The function may have a local maximum, minimum, or saddle point. Critical points are candidates for local extrema or points of inflection.

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Flashcard 1: What does it mean if f′(x)=0f'(x) = 0f′(x)=0 at some point?

Answer: The function may have a local maximum, minimum, or saddle point. Critical points are candidates for local extrema or points of inflection.

Flashcard 2: State the derivative of f(x)=2x3−3x2+5x−4f(x) = 2x^3 - 3x^2 + 5x - 4f(x)=2x3−3x2+5x−4.

Answer: f′(x)=6x2−6x+5f'(x) = 6x^2 - 6x + 5f′(x)=6x2−6x+5. Apply power rule: derivative of axnax^naxn is naxn−1nax^{n-1}naxn−1.

Flashcard 3: What is the derivative of f(x)=x4f(x) = x^4f(x)=x4?

Answer: f′(x)=4x3f'(x) = 4x^3f′(x)=4x3. Apply power rule: (x4)′=4x3(x^4)' = 4x^3(x4)′=4x3.

Flashcard 4: What does the sign of f′(x)f'(x)f′(x) indicate?

Answer: The sign indicates whether the function is increasing or decreasing. Positive derivative means increasing, negative means decreasing.

Flashcard 5: Find f′(x)f'(x)f′(x) for f(x)=4x3−9x2+6x−1f(x) = 4x^3 - 9x^2 + 6x - 1f(x)=4x3−9x2+6x−1.

Answer: f′(x)=12x2−18x+6f'(x) = 12x^2 - 18x + 6f′(x)=12x2−18x+6. Apply power rule to each term of the polynomial.

Flashcard 6: Evaluate f′(x)f'(x)f′(x) for f(x)=13x3−2x2+xf(x) = \frac{1}{3}x^3 - 2x^2 + xf(x)=31​x3−2x2+x.

Answer: f′(x)=x2−4x+1f'(x) = x^2 - 4x + 1f′(x)=x2−4x+1. Apply power rule to each term: ddx[13x3]=x2\frac{d}{dx}[\frac{1}{3}x^3] = x^2dxd​[31​x3]=x2, etc.

Flashcard 7: What is the derivative of a constant function?

Answer: The derivative is 0. Constant functions have zero slope everywhere.

Flashcard 8: Compute the derivative of f(x)=x2+4x−5f(x) = x^2 + 4x - 5f(x)=x2+4x−5.

Answer: f′(x)=2x+4f'(x) = 2x + 4f′(x)=2x+4. Use power rule: (x2)′=2x(x^2)' = 2x(x2)′=2x and (4x)′=4(4x)' = 4(4x)′=4.

Flashcard 9: Explain the significance of the second derivative's sign.

Answer: Indicates concavity and potential inflection points. Second derivative determines concavity: positive means concave up, negative means concave down.

Flashcard 10: What is the first derivative test for determining intervals of decrease?

Answer: If f′(x)<0f'(x) < 0f′(x)<0, the function is decreasing on that interval. When the slope is negative, the function is falling.

Flashcard 11: What is the first derivative test for determining intervals of increase?

Answer: If f′(x)>0f'(x) > 0f′(x)>0, the function is increasing on that interval. This is the fundamental relationship between derivative sign and function behavior.

Flashcard 12: Find f′(x)f'(x)f′(x) given f(x)=5x3−2x+7f(x) = 5x^3 - 2x + 7f(x)=5x3−2x+7.

Answer: f′(x)=15x2−2f'(x) = 15x^2 - 2f′(x)=15x2−2. Apply power rule: (5x3)′=15x2(5x^3)' = 15x^2(5x3)′=15x2 and (−2x)′=−2(-2x)' = -2(−2x)′=−2.

Flashcard 13: What is the derivative of f(x)=12x2−3xf(x) = \frac{1}{2}x^2 - 3xf(x)=21​x2−3x?

Answer: f′(x)=x−3f'(x) = x - 3f′(x)=x−3. Apply power rule: (12x2)′=x(\frac{1}{2}x^2)' = x(21​x2)′=x and (−3x)′=−3(-3x)' = -3(−3x)′=−3.

Flashcard 14: What is the significance of f′(x)f'(x)f′(x) being zero at a point?

Answer: Possible local maximum, minimum, or inflection point. Zero derivative indicates horizontal tangent line and potential extremum.

Flashcard 15: How is the first derivative used to find intervals of monotonicity?

Answer: By evaluating f′(x)f'(x)f′(x) to determine where it is positive or negative. Sign of f′(x)f'(x)f′(x) determines whether function increases or decreases.

Flashcard 16: What is the purpose of finding the critical points?

Answer: To determine potential intervals of increase or decrease. Critical points divide the domain into intervals for monotonicity testing.

Flashcard 17: Find the critical points of f(x)=−x2+6x−9f(x) = -x^2 + 6x - 9f(x)=−x2+6x−9.

Answer: x=3x = 3x=3. Set f′(x)=−2x+6=0f'(x) = -2x + 6 = 0f′(x)=−2x+6=0 to find where slope equals zero.

Flashcard 18: What is the significance of f′(x)f'(x)f′(x) being zero at a point?

Answer: Possible local maximum, minimum, or inflection point. Zero derivative indicates horizontal tangent line and potential extremum.

Flashcard 19: Find f′(x)f'(x)f′(x) given f(x)=5x3−2x+7f(x) = 5x^3 - 2x + 7f(x)=5x3−2x+7.

Answer: f′(x)=15x2−2f'(x) = 15x^2 - 2f′(x)=15x2−2. Apply power rule: (5x3)′=15x2(5x^3)' = 15x^2(5x3)′=15x2 and (−2x)′=−2(-2x)' = -2(−2x)′=−2.

Flashcard 20: State the derivative of f(x)=2x3−3x2+5x−4f(x) = 2x^3 - 3x^2 + 5x - 4f(x)=2x3−3x2+5x−4.

Answer: f′(x)=6x2−6x+5f'(x) = 6x^2 - 6x + 5f′(x)=6x2−6x+5. Apply power rule: derivative of axnax^naxn is naxn−1nax^{n-1}naxn−1.

Flashcard 21: What does it mean if f′(x)=0f'(x) = 0f′(x)=0 at some point?

Answer: The function may have a local maximum, minimum, or saddle point. Critical points are candidates for local extrema or points of inflection.

Flashcard 22: Find f′(x)f'(x)f′(x) for f(x)=4x3−9x2+6x−1f(x) = 4x^3 - 9x^2 + 6x - 1f(x)=4x3−9x2+6x−1.

Answer: f′(x)=12x2−18x+6f'(x) = 12x^2 - 18x + 6f′(x)=12x2−18x+6. Apply power rule to each term of the polynomial.

Flashcard 23: Explain the significance of the second derivative's sign.

Answer: Indicates concavity and potential inflection points. Second derivative determines concavity: positive means concave up, negative means concave down.

Flashcard 24: What is the first derivative test for determining intervals of decrease?

Answer: If f′(x)<0f'(x) < 0f′(x)<0, the function is decreasing on that interval. When the slope is negative, the function is falling.

Flashcard 25: What is the first derivative test for determining intervals of increase?

Answer: If f′(x)>0f'(x) > 0f′(x)>0, the function is increasing on that interval. This is the fundamental relationship between derivative sign and function behavior.

Flashcard 26: What does the sign of f′(x)f'(x)f′(x) indicate?

Answer: The sign indicates whether the function is increasing or decreasing. Positive derivative means increasing, negative means decreasing.

Flashcard 27: Find the critical points of f(x)=−x2+6x−9f(x) = -x^2 + 6x - 9f(x)=−x2+6x−9.

Answer: x=3x = 3x=3. Set f′(x)=−2x+6=0f'(x) = -2x + 6 = 0f′(x)=−2x+6=0 to find where slope equals zero.

Flashcard 28: Evaluate f′(x)f'(x)f′(x) for f(x)=13x3−2x2+xf(x) = \frac{1}{3}x^3 - 2x^2 + xf(x)=31​x3−2x2+x.

Answer: f′(x)=x2−4x+1f'(x) = x^2 - 4x + 1f′(x)=x2−4x+1. Apply power rule to each term: ddx[13x3]=x2\frac{d}{dx}[\frac{1}{3}x^3] = x^2dxd​[31​x3]=x2, etc.

Flashcard 29: What is the derivative of a constant function?

Answer: The derivative is 0. Constant functions have zero slope everywhere.

Flashcard 30: Compute the derivative of f(x)=x2+4x−5f(x) = x^2 + 4x - 5f(x)=x2+4x−5.

Answer: f′(x)=2x+4f'(x) = 2x + 4f′(x)=2x+4. Use power rule: (x2)′=2x(x^2)' = 2x(x2)′=2x and (4x)′=4(4x)' = 4(4x)′=4.