AP Calculus AB Flashcards: First Derivative Test

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.

Flashcard 1: Find the derivative of $f(x) = \frac{1}{3}x^3 - 2x + 1$.

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

Flashcard 2: Find the derivative of $f(x) = \sin x$.

Answer: $f'(x) = \cos x$. Derivative of sine function is cosine.

Flashcard 3: Identify the type of extrema at a critical point if $f'$ changes from positive to negative.

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

Flashcard 4: Identify the type of extrema at a critical point if $f'$ changes from positive to negative.

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

Flashcard 5: What is the First Derivative Test conclusion for $f'(x) = 0$ and no sign change?

Answer: No relative extrema. When $f'$ doesn't change sign, there's no local extremum.

Flashcard 6: What is the relative extrema at $x = \pi$ for $f(x) = \cos x$?

Answer: Relative minimum. $f'$ changes from negative to positive at $x = \pi$.

Flashcard 7: What is the relative extrema at $x = 2$ for $f(x) = x^3 - 3x^2 + 2$?

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

Flashcard 8: Determine the intervals where $f(x) = x^3 - 3x^2 + 2$ is decreasing.

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

Flashcard 9: What conclusion is drawn if $f'(x)$ 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 10: Determine the critical points of $f(x) = \sin x$ on $[0, 2\pi]$.

Answer: Critical points: $x = \frac{\pi}{2}, \frac{3\pi}{2}$. $\cos x = 0$ when $x = \frac{\pi}{2}, \frac{3\pi}{2}$.

Flashcard 11: What is the relative extrema at $x = -2$ for $f(x) = x^2 + 4x + 4$?

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

Flashcard 12: Identify the critical points of $f(x) = \frac{1}{4}x^4 - x^2 + 2$.

Answer: Critical points: $x = 0, \pm 1$. Solve $x^3 - 2x = x(x^2 - 2) = 0$.

Flashcard 13: Find the critical points of $f(x) = x^2 + 4x + 4$.

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

Flashcard 14: Determine the critical points of $f(x) = \frac{1}{3}x^3 - 2x + 1$.

Answer: Critical points: $x = \pm \sqrt{2}$. Solve $x^2 - 2 = 0$ to get $x^2 = 2$.

Flashcard 15: Find $f'(x)$ for $f(x) = 2x^3 - 3x^2 - 12x + 1$.

Answer: $f'(x) = 6x^2 - 6x - 12$. Use power rule: $6x^2$ from $2x^3$, $-6x$ from $-3x^2$.

Flashcard 16: What is the relative extrema at $x = -\sqrt{2}$ for $f(x) = \frac{1}{3}x^3 - 2x + 1$?

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

Flashcard 17: Identify the type of extrema at a critical point if $f'$ changes from negative to positive.

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

Flashcard 18: State the condition for a critical point of a function.

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

Flashcard 19: Find the derivative of $f(x) = x^3 - 3x^2 + 2$.

Answer: $f'(x) = 3x^2 - 6x$. Use power rule: derivative of $x^3$ is $3x^2$, $x^2$ is $2x$.

Flashcard 20: Find $f'(x)$ for $f(x) = \frac{1}{4}x^4 - x^2 + 2$.

Answer: $f'(x) = x^3 - 2x$. Derivative of $\frac{1}{4}x^4$ is $x^3$, of $x^2$ is $2x$.

Flashcard 21: Determine the intervals where $f(x) = x^3 - 3x^2 + 2$ is decreasing.

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

Flashcard 22: What does $f'(x) < 0$ indicate about a function on an interval?

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

Flashcard 23: Determine the intervals where $f(x) = \cos x$ is decreasing on $[0, 2\pi]$.

Answer: Decreasing on $(0, \pi)$. $-\sin x < 0$ when $0 < x < \pi$.

Flashcard 24: Identify the critical points of $f(x) = \frac{1}{4}x^4 - x^2 + 2$.

Answer: Critical points: $x = 0, \pm 1$. Solve $x^3 - 2x = x(x^2 - 2) = 0$.

Flashcard 25: Find the critical points of $f(x) = x^3 - 3x^2 + 2$.

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

Flashcard 26: Identify the critical points of $f(x) = \cos x$ on $[0, 2\backslashpi]$.

Answer: Critical points: $x = \pi$. $-\sin x = 0$ when $x = \pi$ in the given interval.

Flashcard 27: What does $f'(x) > 0$ 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: Determine the intervals where $f(x) = x^3 - 3x^2 + 2$ is increasing.

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

Flashcard 29: What is the relative extrema at $x = \frac{\pi}{2}$ for $f(x) = \sin x$?

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

Flashcard 30: 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 31: Find the derivative of $f(x) = \sin x$.

Answer: $f'(x) = \cos x$. Derivative of sine function is cosine.

Flashcard 32: Identify the critical points of $f(x) = 2x^3 - 3x^2 - 12x + 1$.

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

Flashcard 33: What is the relative extrema at $x = -1$ for $f(x) = 2x^3 - 3x^2 - 12x + 1$?

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

Flashcard 34: What is the relative extrema at $x = \sqrt{2}$ for $f(x) = \frac{1}{3}x^3 - 2x + 1$?

Answer: Relative minimum. $f'$ changes from negative to positive at $x = \sqrt{2}$.

Flashcard 35: Identify the type of extrema at a critical point if $f'$ changes from negative to positive.

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

Flashcard 36: What is the relative extrema at $x = -\sqrt{2}$ for $f(x) = \frac{1}{3}x^3 - 2x + 1$?

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

Flashcard 37: Find the derivative of $f(x) = x^3 - 3x^2 + 2$.

Answer: $f'(x) = 3x^2 - 6x$. Use power rule: derivative of $x^3$ is $3x^2$, $x^2$ is $2x$.

Flashcard 38: What is the relative extrema at $x = 2$ for $f(x) = 2x^3 - 3x^2 - 12x + 1$?

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

Flashcard 39: What is the relative extrema at $x = 2$ for $f(x) = x^3 - 3x^2 + 2$?

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

Flashcard 40: Determine the critical points of $f(x) = \sin x$ on $[0, 2\pi]$.

Answer: Critical points: $x = \frac{\pi}{2}, \frac{3\pi}{2}$. $\cos x = 0$ when $x = \frac{\pi}{2}, \frac{3\pi}{2}$.

Flashcard 41: What is the relative extrema at $x = \frac{3\pi}{2}$ for $f(x) = \sin x$?

Answer: Relative minimum. $f'$ changes from negative to positive at $x = \frac{3\pi}{2}$.

Flashcard 42: 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 43: What does $f'(x) > 0$ indicate about a function on an interval?

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

Flashcard 44: What does $f'(x) < 0$ indicate about a function on an interval?

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

Flashcard 45: Determine the critical points of $f(x) = \frac{1}{3}x^3 - 2x + 1$.

Answer: Critical points: $x = \pm \sqrt{2}$. Solve $x^2 - 2 = 0$ to get $x^2 = 2$.

Flashcard 46: What is the relative extrema at $x = -2$ for $f(x) = x^2 + 4x + 4$?

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

Flashcard 47: What is the relative extrema at $x = 2$ for $f(x) = 2x^3 - 3x^2 - 12x + 1$?

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

Flashcard 48: Find the critical points of $f(x) = x^2 + 4x + 4$.

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

Flashcard 49: What is the relative extrema at $x = \frac{3\pi}{2}$ for $f(x) = \sin x$?

Answer: Relative minimum. $f'$ changes from negative to positive at $x = \frac{3\pi}{2}$.

Flashcard 50: Determine the intervals where $f(x) = \cos x$ is increasing on $[0, 2\pi]$.

Answer: Increasing on $(\pi, 2\pi)$. $-\sin x > 0$ when $\pi < x < 2\pi$.

Flashcard 51: What is the relative extrema at $x = \pi$ for $f(x) = \cos x$?

Answer: Relative minimum. $f'$ changes from negative to positive at $x = \pi$.

Flashcard 52: Find the derivative of $f(x) = \frac{1}{3}x^3 - 2x + 1$.

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

Flashcard 53: Determine the intervals where $f(x) = \cos x$ is increasing on $[0, 2\pi]$.

Answer: Increasing on $(\pi, 2\pi).$. $-\sin x > 0$ when $\pi < x < 2\pi$.

Flashcard 54: Determine the intervals where $f(x) = x^3 - 3x^2 + 2$ is increasing.

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

Flashcard 55: Find $f'(x)$ for $f(x) = 2x^3 - 3x^2 - 12x + 1$.

Answer: $f'(x) = 6x^2 - 6x - 12$. Use power rule: $6x^2$ from $2x^3$, $-6x$ from $-3x^2$.

Flashcard 56: State the condition for a critical point of a function.

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

Flashcard 57: Find the critical points of $f(x) = x^3 - 3x^2 + 2$.

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

Flashcard 58: Identify the critical points of $f(x) = \cos x$ on $[0, 2\pi]$.

Answer: Critical points: $x = \pi$. $-\sin x = 0$ when $x = \pi$ in the given interval.

Flashcard 59: Determine the intervals where $f(x) = \cos x$ is decreasing on $[0, 2\pi]$.

Answer: Decreasing on $(0, \pi)$. $-\sin x < 0$ when $0 < x < \pi$.

Flashcard 60: Find $f'(x)$ for $f(x) = \frac{1}{4}x^4 - x^2 + 2$.

Answer: $f'(x) = x^3 - 2x$. Derivative of $\frac{1}{4}x^4$ is $x^3$, of $x^2$ is $2x$.

Flashcard 61: What is the First Derivative Test conclusion for $f'(x) = 0$ and no sign change?

Answer: No relative extrema. When $f'$ doesn't change sign, there's no local extremum.

Flashcard 62: What is the relative extrema at $x = -1$ for $f(x) = 2x^3 - 3x^2 - 12x + 1$?

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

Flashcard 63: Identify the critical points of $f(x) = 2x^3 - 3x^2 - 12x + 1$.

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

Flashcard 64: What conclusion is drawn if $f'(x)$ 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 65: State the derivative of $f(x) = \cos x$.

Answer: $f'(x) = -\sin x$. Derivative of cosine function is negative sine.

Flashcard 66: State the derivative of $f(x) = \cos x$.

Answer: $f'(x) = -\sin x$. Derivative of cosine function is negative sine.

Flashcard 67: What is the relative extrema at $x = \frac{\pi}{2}$ for $f(x) = \sin x$?

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

Flashcard 68: What is the relative extrema at $x = \sqrt{2}$ for $f(x) = \frac{1}{3}x^3 - 2x + 1$?

Answer: Relative minimum. $f'$ changes from negative to positive at $x = \sqrt{2}$.