Second Derivative Test - AP Calculus BC
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State the second derivative test condition for a local maximum.
State the second derivative test condition for a local maximum.
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$f''(c) < 0$ at critical point $c$. Negative second derivative indicates concave down, confirming a maximum.
$f''(c) < 0$ at critical point $c$. Negative second derivative indicates concave down, confirming a maximum.
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Evaluate $f''(1)$ for $f(x) = x^4 - x^2 + x$.
Evaluate $f''(1)$ for $f(x) = x^4 - x^2 + x$.
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$f''(1) = 10$. $f'(x) = 4x^3 - 2x + 1$ and $f''(x) = 12x^2 - 2$, so $f''(1) = 10$.
$f''(1) = 10$. $f'(x) = 4x^3 - 2x + 1$ and $f''(x) = 12x^2 - 2$, so $f''(1) = 10$.
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What does $f''(c) < 0$ imply about $f(x)$ at $x=c$?
What does $f''(c) < 0$ imply about $f(x)$ at $x=c$?
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Local maximum. Negative second derivative confirms the critical point is a maximum.
Local maximum. Negative second derivative confirms the critical point is a maximum.
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Determine the nature of the extremum at $x = \frac{1}{2}$ for $f(x) = -x^2 + x$.
Determine the nature of the extremum at $x = \frac{1}{2}$ for $f(x) = -x^2 + x$.
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Local maximum. $f''(1/2) = -2 < 0$ at the critical point, indicating a maximum.
Local maximum. $f''(1/2) = -2 < 0$ at the critical point, indicating a maximum.
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Determine if $f(x) = x^3 - 3x^2 + 4$ has a local extremum at $x = 2$.
Determine if $f(x) = x^3 - 3x^2 + 4$ has a local extremum at $x = 2$.
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Local minimum at $x = 2$. $f'(2) = 0$ and $f''(2) = 6 > 0$, confirming a local minimum.
Local minimum at $x = 2$. $f'(2) = 0$ and $f''(2) = 6 > 0$, confirming a local minimum.
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What conclusion is drawn if $f''(c) = 0$ at a critical point $c$?
What conclusion is drawn if $f''(c) = 0$ at a critical point $c$?
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Test inconclusive. Zero second derivative means the test cannot determine extremum type.
Test inconclusive. Zero second derivative means the test cannot determine extremum type.
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What is the second derivative of $f(x) = 3x^4 - 8x^2 + 5$?
What is the second derivative of $f(x) = 3x^4 - 8x^2 + 5$?
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$f''(x) = 36x^2 - 16$. Differentiate $f'(x) = 12x^3 - 16x$ to get the second derivative.
$f''(x) = 36x^2 - 16$. Differentiate $f'(x) = 12x^3 - 16x$ to get the second derivative.
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At which type of point is the second derivative test applied?
At which type of point is the second derivative test applied?
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Critical points. Only points where $f'(x) = 0$ can be tested for extrema.
Critical points. Only points where $f'(x) = 0$ can be tested for extrema.
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Determine the nature of the extremum for $f(x) = x^2 - 4x + 4$ at $x = 2$.
Determine the nature of the extremum for $f(x) = x^2 - 4x + 4$ at $x = 2$.
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Local minimum. $f'(2) = 0$ and $f''(2) = 2 > 0$, confirming a local minimum.
Local minimum. $f'(2) = 0$ and $f''(2) = 2 > 0$, confirming a local minimum.
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Identify the nature of the critical point for $f(x) = x^3 - 3x$ at $x = 0$.
Identify the nature of the critical point for $f(x) = x^3 - 3x$ at $x = 0$.
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Inconclusive. $f''(0) = 0$ at this critical point, so the test fails.
Inconclusive. $f''(0) = 0$ at this critical point, so the test fails.
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What is the sign of $f''(x)$ if $f(x)$ is concave down?
What is the sign of $f''(x)$ if $f(x)$ is concave down?
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Negative. Concave down functions have negative curvature everywhere.
Negative. Concave down functions have negative curvature everywhere.
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What is the second derivative of $f(x) = -x^4 + 2x^2 - x$?
What is the second derivative of $f(x) = -x^4 + 2x^2 - x$?
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$f''(x) = -12x^2 + 4$. Differentiate $f'(x) = -4x^3 + 4x - 1$ to get the second derivative.
$f''(x) = -12x^2 + 4$. Differentiate $f'(x) = -4x^3 + 4x - 1$ to get the second derivative.
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Evaluate $f''(2)$ for $f(x) = x^4 - 4x^3 + 6x^2$.
Evaluate $f''(2)$ for $f(x) = x^4 - 4x^3 + 6x^2$.
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$f''(2) = 12$. $f'(x) = 4x^3 - 12x^2 + 12x$, so $f''(x) = 12x^2 - 24x + 12$.
$f''(2) = 12$. $f'(x) = 4x^3 - 12x^2 + 12x$, so $f''(x) = 12x^2 - 24x + 12$.
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Find the second derivative of $f(x) = \frac{1}{3}x^3 - 2x^2 + 3x$.
Find the second derivative of $f(x) = \frac{1}{3}x^3 - 2x^2 + 3x$.
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$f''(x) = 2x - 4$. $f'(x) = x^2 - 4x + 3$, so $f''(x) = 2x - 4$.
$f''(x) = 2x - 4$. $f'(x) = x^2 - 4x + 3$, so $f''(x) = 2x - 4$.
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What does $f''(c) = 0$ indicate about the critical point $c$?
What does $f''(c) = 0$ indicate about the critical point $c$?
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Test is inconclusive at $c$. Zero second derivative provides no information about the extremum type.
Test is inconclusive at $c$. Zero second derivative provides no information about the extremum type.
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State the second derivative test condition for a local minimum.
State the second derivative test condition for a local minimum.
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$f''(c) > 0$ at critical point $c$. Positive second derivative indicates concave up, confirming a minimum.
$f''(c) > 0$ at critical point $c$. Positive second derivative indicates concave up, confirming a minimum.
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What is the second derivative of $f(x) = 2x^3 - 3x^2 + x$?
What is the second derivative of $f(x) = 2x^3 - 3x^2 + x$?
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$f''(x) = 12x - 6$. Differentiate $f'(x) = 6x^2 - 6x + 1$ to get the second derivative.
$f''(x) = 12x - 6$. Differentiate $f'(x) = 6x^2 - 6x + 1$ to get the second derivative.
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Determine concavity for $f(x) = 2x^2 - 3x + 1$ at $x = 0$.
Determine concavity for $f(x) = 2x^2 - 3x + 1$ at $x = 0$.
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Concave up. $f''(x) = 4 > 0$ everywhere, so the function is concave up.
Concave up. $f''(x) = 4 > 0$ everywhere, so the function is concave up.
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Identify the sign of $f''(x)$ for a concave down interval.
Identify the sign of $f''(x)$ for a concave down interval.
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$f''(x) < 0$. Negative second derivative always indicates downward concavity.
$f''(x) < 0$. Negative second derivative always indicates downward concavity.
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What is the condition for test inconclusiveness in the second derivative test?
What is the condition for test inconclusiveness in the second derivative test?
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$f''(c) = 0$ at $c$. This condition makes the second derivative test unable to determine extremum type.
$f''(c) = 0$ at $c$. This condition makes the second derivative test unable to determine extremum type.
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What is the second derivative of $f(x) = -x^3 + 3x - 5$?
What is the second derivative of $f(x) = -x^3 + 3x - 5$?
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$f''(x) = -6x$. Differentiate $f'(x) = -3x^2 + 3$ to get the second derivative.
$f''(x) = -6x$. Differentiate $f'(x) = -3x^2 + 3$ to get the second derivative.
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Determine the nature of the extremum for $f(x) = x^2 - 4$ at $x = 0$.
Determine the nature of the extremum for $f(x) = x^2 - 4$ at $x = 0$.
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Local minimum. $f'(0) = 0$ and $f''(0) = 2 > 0$, confirming a local minimum.
Local minimum. $f'(0) = 0$ and $f''(0) = 2 > 0$, confirming a local minimum.
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What does $f''(c) > 0$ imply about $f(x)$ at $x=c$?
What does $f''(c) > 0$ imply about $f(x)$ at $x=c$?
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Local minimum. Positive second derivative confirms the critical point is a minimum.
Local minimum. Positive second derivative confirms the critical point is a minimum.
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Calculate $f''(3)$ for $f(x) = x^3 - 9x^2 + 27x - 3$.
Calculate $f''(3)$ for $f(x) = x^3 - 9x^2 + 27x - 3$.
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$f''(3) = 0$. $f'(x) = 3x^2 - 18x + 27$ and $f''(x) = 6x - 18$, so $f''(3) = 0$.
$f''(3) = 0$. $f'(x) = 3x^2 - 18x + 27$ and $f''(x) = 6x - 18$, so $f''(3) = 0$.
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Evaluate $f''(0)$ for $f(x) = x^4 - 4x^2 + 1$.
Evaluate $f''(0)$ for $f(x) = x^4 - 4x^2 + 1$.
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$f''(0) = -8$. $f'(x) = 4x^3 - 8x$, so $f''(x) = 12x^2 - 8$ and $f''(0) = -8$.
$f''(0) = -8$. $f'(x) = 4x^3 - 8x$, so $f''(x) = 12x^2 - 8$ and $f''(0) = -8$.
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Find the second derivative of $f(x) = x^5 - 5x^3 + 7$.
Find the second derivative of $f(x) = x^5 - 5x^3 + 7$.
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$f''(x) = 20x^3 - 30x$. Differentiate $f'(x) = 5x^4 - 15x^2$ to get the second derivative.
$f''(x) = 20x^3 - 30x$. Differentiate $f'(x) = 5x^4 - 15x^2$ to get the second derivative.
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What is the sign of $f''(x)$ if $f(x)$ is concave up?
What is the sign of $f''(x)$ if $f(x)$ is concave up?
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Positive. Concave up functions have positive curvature everywhere.
Positive. Concave up functions have positive curvature everywhere.
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Find $f''(x)$ for $f(x) = x^4 - 4x^2 + 2$.
Find $f''(x)$ for $f(x) = x^4 - 4x^2 + 2$.
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$f''(x) = 12x^2 - 8$. Take the derivative of $f'(x) = 4x^3 - 8x$ to get the second derivative.
$f''(x) = 12x^2 - 8$. Take the derivative of $f'(x) = 4x^3 - 8x$ to get the second derivative.
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Identify the sign of $f''(x)$ for a concave up interval.
Identify the sign of $f''(x)$ for a concave up interval.
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$f''(x) > 0$. Positive second derivative always indicates upward concavity.
$f''(x) > 0$. Positive second derivative always indicates upward concavity.
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Determine the nature of the extremum for $f(x) = x^2 - 4x + 4$ at $x = 2$.
Determine the nature of the extremum for $f(x) = x^2 - 4x + 4$ at $x = 2$.
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Local minimum. $f'(2) = 0$ and $f''(2) = 2 > 0$, confirming a local minimum.
Local minimum. $f'(2) = 0$ and $f''(2) = 2 > 0$, confirming a local minimum.
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