AP Calculus AB Flashcards: Determining Intervals On Increasing Decreasing Functions

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.

Flashcard 1: Evaluate $f'(x)$ for $f(x) = \frac{1}{3}x^3 - 2x^2 + x$.

Answer: $f'(x) = x^2 - 4x + 1$. Apply power rule to each term: $\frac{d}{dx}[\frac{1}{3}x^3] = x^2$, etc.

Flashcard 2: Find $f'(x)$ given $f(x) = 5x^3 - 2x + 7$.

Answer: $f'(x) = 15x^2 - 2$. Apply power rule: $(5x^3)' = 15x^2$ and $(-2x)' = -2$.

Flashcard 3: What does it mean if $f'(x) = 0$ 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 4: Find $f'(x)$ for $f(x) = 4x^3 - 9x^2 + 6x - 1$.

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

Flashcard 5: What does it mean if $f'(x)$ changes sign at a point?

Answer: There is a local extremum at that point. Sign changes in the derivative indicate local maxima or minima.

Flashcard 6: What is the derivative of $f(x) = x^4$?

Answer: $f'(x) = 4x^3$. Apply power rule: $(x^4)' = 4x^3$.

Flashcard 7: 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 8: What is the first derivative test for determining intervals of decrease?

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

Flashcard 9: What is the derivative of $f(x) = x^4$?

Answer: $f'(x) = 4x^3$. Apply power rule: $(x^4)' = 4x^3$.

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

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

Flashcard 11: Find $f'(x)$ given $f(x) = 5x^3 - 2x + 7$.

Answer: $f'(x) = 15x^2 - 2$. Apply power rule: $(5x^3)' = 15x^2$ and $(-2x)' = -2$.

Flashcard 12: What does the sign of $f'(x)$ indicate?

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

Flashcard 13: Compute the derivative of $f(x) = x^2 + 4x - 5$.

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

Flashcard 14: Find $f'(x)$ for $f(x) = 4x^3 - 9x^2 + 6x - 1$.

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

Flashcard 15: What does the sign of $f'(x)$ indicate?

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

Flashcard 16: What is the derivative of $f(x) = \frac{1}{2}x^2 - 3x$?

Answer: $f'(x) = x - 3$. Apply power rule: $(\frac{1}{2}x^2)' = x$ and $(-3x)' = -3$.

Flashcard 17: What is the significance of a critical point?

Answer: It is where $f'(x) = 0$ or $f'(x)$ is undefined. These points are where the derivative equals zero or doesn't exist.

Flashcard 18: 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 19: 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 20: Compute the derivative of $f(x) = x^2 + 4x - 5$.

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

Flashcard 21: 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 22: State the derivative of $f(x) = 2x^3 - 3x^2 + 5x - 4$.

Answer: $f'(x) = 6x^2 - 6x + 5$. Apply power rule: derivative of $ax^n$ is $nax^{n-1}$.

Flashcard 23: What is the significance of $f'(x)$ being zero at a point?

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

Flashcard 24: What is the derivative of a constant function?

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

Flashcard 25: Find the critical points of $f(x) = -x^2 + 6x - 9$.

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

Flashcard 26: What is the significance of $f'(x)$ being zero at a point?

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

Flashcard 27: What is the first derivative test for determining intervals of increase?

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

Flashcard 28: State the derivative of $f(x) = 2x^3 - 3x^2 + 5x - 4$.

Answer: $f'(x) = 6x^2 - 6x + 5$. Apply power rule: derivative of $ax^n$ is $nax^{n-1}$.

Flashcard 29: How is the first derivative used to find intervals of monotonicity?

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

Flashcard 30: How is the first derivative used to find intervals of monotonicity?

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

Flashcard 31: What is the significance of a critical point?

Answer: It is where $f'(x) = 0$ or $f'(x)$ is undefined. These points are where the derivative equals zero or doesn't exist.

Flashcard 32: What does it mean if $f'(x) = 0$ 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 33: What is the derivative of $f(x) = \frac{1}{2}x^2 - 3x$?

Answer: $f'(x) = x - 3$. Apply power rule: $(\frac{1}{2}x^2)' = x$ and $(-3x)' = -3$.

Flashcard 34: Evaluate $f'(x)$ for $f(x) = \frac{1}{3}x^3 - 2x^2 + x$.

Answer: $f'(x) = x^2 - 4x + 1$. Apply power rule to each term: $\frac{d}{dx}[\frac{1}{3}x^3] = x^2$, etc.

Flashcard 35: Find the critical points of $f(x) = -x^2 + 6x - 9$.

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

Flashcard 36: What is the derivative of a constant function?

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

Flashcard 37: What is the first derivative test for determining intervals of increase?

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

Flashcard 38: What does it mean if $f'(x)$ changes sign at a point?

Answer: There is a local extremum at that point. Sign changes in the derivative indicate local maxima or minima.