First Derivative Test - AP Calculus AB
Card 1 of 30
Find $f'(x)$ for $f(x) = 2x^3 - 3x^2 - 12x + 1$.
Find $f'(x)$ for $f(x) = 2x^3 - 3x^2 - 12x + 1$.
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$f'(x) = 6x^2 - 6x - 12$. Use power rule: $6x^2$ from $2x^3$, $-6x$ from $-3x^2$.
$f'(x) = 6x^2 - 6x - 12$. Use power rule: $6x^2$ from $2x^3$, $-6x$ from $-3x^2$.
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Identify the critical points of $f(x) = 2x^3 - 3x^2 - 12x + 1$.
Identify the critical points of $f(x) = 2x^3 - 3x^2 - 12x + 1$.
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Critical points: $x = -1, 2$. Solve $6x^2 - 6x - 12 = 6(x^2 - x - 2) = 0$.
Critical points: $x = -1, 2$. Solve $6x^2 - 6x - 12 = 6(x^2 - x - 2) = 0$.
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Determine the intervals where $f(x) = x^3 - 3x^2 + 2$ is increasing.
Determine the intervals where $f(x) = x^3 - 3x^2 + 2$ is increasing.
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Increasing on $(2, , \text{infinity})$. $f'(x) > 0$ when $x > 2$ (test a point like $x = 3$).
Increasing on $(2, , \text{infinity})$. $f'(x) > 0$ when $x > 2$ (test a point like $x = 3$).
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Determine the intervals where $f(x) = x^3 - 3x^2 + 2$ is decreasing.
Determine the intervals where $f(x) = x^3 - 3x^2 + 2$ is decreasing.
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Decreasing on $(-\text{infinity}, 0)$ and $(0, 2)$.. $f'(x) < 0$ when $x < 0$ or $0 < x < 2$.
Decreasing on $(-\text{infinity}, 0)$ and $(0, 2)$.. $f'(x) < 0$ when $x < 0$ or $0 < x < 2$.
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What does $f'(x) < 0$ indicate about a function on an interval?
What does $f'(x) < 0$ indicate about a function on an interval?
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The function is decreasing on that interval. Negative derivative means the function has negative slope.
The function is decreasing on that interval. Negative derivative means the function has negative slope.
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What is the relative extrema at $x = \pi$ for $f(x) = \cos x$?
What is the relative extrema at $x = \pi$ for $f(x) = \cos x$?
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Relative minimum. $f'$ changes from negative to positive at $x = \pi$.
Relative minimum. $f'$ changes from negative to positive at $x = \pi$.
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Identify the critical points of $f(x) = \cos x$ on $[0, 2\backslashpi]$.
Identify the critical points of $f(x) = \cos x$ on $[0, 2\backslashpi]$.
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Critical points: $x = \pi$. $-\sin x = 0$ when $x = \pi$ in the given interval.
Critical points: $x = \pi$. $-\sin x = 0$ when $x = \pi$ in the given interval.
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Identify the critical points of $f(x) = \frac{1}{4}x^4 - x^2 + 2$.
Identify the critical points of $f(x) = \frac{1}{4}x^4 - x^2 + 2$.
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Critical points: $x = 0, \pm 1$. Solve $x^3 - 2x = x(x^2 - 2) = 0$.
Critical points: $x = 0, \pm 1$. Solve $x^3 - 2x = x(x^2 - 2) = 0$.
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What is the relative extrema at $x = -\sqrt{2}$ for $f(x) = \frac{1}{3}x^3 - 2x + 1$?
What is the relative extrema at $x = -\sqrt{2}$ for $f(x) = \frac{1}{3}x^3 - 2x + 1$?
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Relative maximum. $f'$ changes from positive to negative at $x = -\sqrt{2}$.
Relative maximum. $f'$ changes from positive to negative at $x = -\sqrt{2}$.
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What is the relative extrema at $x = \frac{\pi}{2}$ for $f(x) = \sin x$?
What is the relative extrema at $x = \frac{\pi}{2}$ for $f(x) = \sin x$?
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Relative maximum. $f'$ changes from positive to negative at $x = \frac{\pi}{2}$.
Relative maximum. $f'$ changes from positive to negative at $x = \frac{\pi}{2}$.
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What is the relative extrema at $x = \frac{3\pi}{2}$ for $f(x) = \sin x$?
What is the relative extrema at $x = \frac{3\pi}{2}$ for $f(x) = \sin x$?
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Relative minimum. $f'$ changes from negative to positive at $x = \frac{3\pi}{2}$.
Relative minimum. $f'$ changes from negative to positive at $x = \frac{3\pi}{2}$.
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Find $f'(x)$ for $f(x) = \frac{1}{4}x^4 - x^2 + 2$.
Find $f'(x)$ for $f(x) = \frac{1}{4}x^4 - x^2 + 2$.
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$f'(x) = x^3 - 2x$. Derivative of $\frac{1}{4}x^4$ is $x^3$, of $x^2$ is $2x$.
$f'(x) = x^3 - 2x$. Derivative of $\frac{1}{4}x^4$ is $x^3$, of $x^2$ is $2x$.
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Determine the critical points of $f(x) = \frac{1}{3}x^3 - 2x + 1$.
Determine the critical points of $f(x) = \frac{1}{3}x^3 - 2x + 1$.
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Critical points: $x = \pm \sqrt{2}$. Solve $x^2 - 2 = 0$ to get $x^2 = 2$.
Critical points: $x = \pm \sqrt{2}$. Solve $x^2 - 2 = 0$ to get $x^2 = 2$.
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Find the critical points of $f(x) = x^2 + 4x + 4$.
Find the critical points of $f(x) = x^2 + 4x + 4$.
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Critical point: $x = -2$. $f'(x) = 2x + 4 = 0$ gives $x = -2$.
Critical point: $x = -2$. $f'(x) = 2x + 4 = 0$ gives $x = -2$.
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What is the relative extrema at $x = 2$ for $f(x) = x^3 - 3x^2 + 2$?
What is the relative extrema at $x = 2$ for $f(x) = x^3 - 3x^2 + 2$?
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Relative minimum. $f'$ changes from negative to positive at $x = 2$.
Relative minimum. $f'$ changes from negative to positive at $x = 2$.
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Find the derivative of $f(x) = x^3 - 3x^2 + 2$.
Find the derivative of $f(x) = x^3 - 3x^2 + 2$.
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$f'(x) = 3x^2 - 6x$. Use power rule: derivative of $x^3$ is $3x^2$, $x^2$ is $2x$.
$f'(x) = 3x^2 - 6x$. Use power rule: derivative of $x^3$ is $3x^2$, $x^2$ is $2x$.
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Find the critical points of $f(x) = x^3 - 3x^2 + 2$.
Find the critical points of $f(x) = x^3 - 3x^2 + 2$.
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Critical points: $x = 0, x = 2$. Set $f'(x) = 3x^2 - 6x = 3x(x-2) = 0$.
Critical points: $x = 0, x = 2$. Set $f'(x) = 3x^2 - 6x = 3x(x-2) = 0$.
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Identify the type of extrema at a critical point if $f'$ changes from positive to negative.
Identify the type of extrema at a critical point if $f'$ changes from positive to negative.
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Relative maximum. Function rises then falls, creating a local peak.
Relative maximum. Function rises then falls, creating a local peak.
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State the condition for a critical point of a function.
State the condition for a critical point of a function.
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The derivative $f'(x) = 0$ or $f'(x)$ is undefined. Critical points occur where the slope is zero or undefined.
The derivative $f'(x) = 0$ or $f'(x)$ is undefined. Critical points occur where the slope is zero or undefined.
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What is the First Derivative Test used for in calculus?
What is the First Derivative Test used for in calculus?
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To determine relative (local) extrema of a function. Identifies where functions reach local peaks and valleys.
To determine relative (local) extrema of a function. Identifies where functions reach local peaks and valleys.
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What is the relative extrema at $x = 2$ for $f(x) = 2x^3 - 3x^2 - 12x + 1$?
What is the relative extrema at $x = 2$ for $f(x) = 2x^3 - 3x^2 - 12x + 1$?
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Relative minimum. $f'$ changes from negative to positive at $x = 2$.
Relative minimum. $f'$ changes from negative to positive at $x = 2$.
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Identify the type of extrema at a critical point if $f'$ changes from negative to positive.
Identify the type of extrema at a critical point if $f'$ changes from negative to positive.
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Relative minimum. Function falls then rises, creating a local valley.
Relative minimum. Function falls then rises, creating a local valley.
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What is the relative extrema at $x = -1$ for $f(x) = 2x^3 - 3x^2 - 12x + 1$?
What is the relative extrema at $x = -1$ for $f(x) = 2x^3 - 3x^2 - 12x + 1$?
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Relative maximum. $f'$ changes from positive to negative at $x = -1$.
Relative maximum. $f'$ changes from positive to negative at $x = -1$.
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Find the derivative of $f(x) = \frac{1}{3}x^3 - 2x + 1$.
Find the derivative of $f(x) = \frac{1}{3}x^3 - 2x + 1$.
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$f'(x) = x^2 - 2$. Use power rule on each term of the polynomial.
$f'(x) = x^2 - 2$. Use power rule on each term of the polynomial.
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What conclusion is drawn if $f'(x)$ does not change sign at a critical point?
What conclusion is drawn if $f'(x)$ does not change sign at a critical point?
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No relative extrema at that point. No sign change means no local maximum or minimum.
No relative extrema at that point. No sign change means no local maximum or minimum.
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Find the derivative of $f(x) = \sin x$.
Find the derivative of $f(x) = \sin x$.
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$f'(x) = \cos x$. Derivative of sine function is cosine.
$f'(x) = \cos x$. Derivative of sine function is cosine.
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What does $f'(x) > 0$ indicate about a function on an interval?
What does $f'(x) > 0$ indicate about a function on an interval?
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The function is increasing on that interval. Positive derivative means the function has positive slope.
The function is increasing on that interval. Positive derivative means the function has positive slope.
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State the derivative of $f(x) = \cos x$.
State the derivative of $f(x) = \cos x$.
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$f'(x) = -\sin x$. Derivative of cosine function is negative sine.
$f'(x) = -\sin x$. Derivative of cosine function is negative sine.
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Determine the intervals where $f(x) = \cos x$ is decreasing on $[0, 2\pi]$.
Determine the intervals where $f(x) = \cos x$ is decreasing on $[0, 2\pi]$.
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Decreasing on $(0, \pi)$. $-\sin x < 0$ when $0 < x < \pi$.
Decreasing on $(0, \pi)$. $-\sin x < 0$ when $0 < x < \pi$.
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What is the relative extrema at $x = -2$ for $f(x) = x^2 + 4x + 4$?
What is the relative extrema at $x = -2$ for $f(x) = x^2 + 4x + 4$?
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Relative minimum. $f'$ changes from negative to positive at $x = -2$.
Relative minimum. $f'$ changes from negative to positive at $x = -2$.
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