Candidates Test - AP Calculus AB
Card 1 of 30
What are the critical points of $f(x) = x^2 - 9x + 20$?
What are the critical points of $f(x) = x^2 - 9x + 20$?
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Critical points: $x = \frac{9}{2}$. Set $f'(x)=2x-9=0$ and solve for $x$.
Critical points: $x = \frac{9}{2}$. Set $f'(x)=2x-9=0$ and solve for $x$.
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Find $f(x)$ at $x = 3$ for $f(x) = x^3 - 6x^2 + 9x + 1$.
Find $f(x)$ at $x = 3$ for $f(x) = x^3 - 6x^2 + 9x + 1$.
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$f(3) = 1$. Substitute $x=3$ into the function.
$f(3) = 1$. Substitute $x=3$ into the function.
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If $f'(x)$ changes sign at $x = c$, what is $x = c$?
If $f'(x)$ changes sign at $x = c$, what is $x = c$?
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A critical point. Sign changes in derivative indicate critical points.
A critical point. Sign changes in derivative indicate critical points.
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What is a critical point?
What is a critical point?
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Where $f'(x) = 0$ or $f'(x)$ is undefined. Points where derivative equals zero or doesn't exist.
Where $f'(x) = 0$ or $f'(x)$ is undefined. Points where derivative equals zero or doesn't exist.
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Which function value determines the absolute maximum?
Which function value determines the absolute maximum?
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The largest value among evaluated points. Highest function output among all tested candidates.
The largest value among evaluated points. Highest function output among all tested candidates.
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What must be determined after finding the derivative in the Candidates Test?
What must be determined after finding the derivative in the Candidates Test?
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Critical points where the derivative is zero or undefined. These are locations where extrema can occur on the interval.
Critical points where the derivative is zero or undefined. These are locations where extrema can occur on the interval.
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Find the derivative of $f(x) = x^3 - 3x^2 + 4$.
Find the derivative of $f(x) = x^3 - 3x^2 + 4$.
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$f'(x) = 3x^2 - 6x$. Apply power rule: $\frac{d}{dx}[x^3]=3x^2$, etc.
$f'(x) = 3x^2 - 6x$. Apply power rule: $\frac{d}{dx}[x^3]=3x^2$, etc.
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Evaluate $f'(x) = 6x - 6$ for critical points.
Evaluate $f'(x) = 6x - 6$ for critical points.
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Critical point: $x = 1$. Set $6x-6=0$ and solve for $x$.
Critical point: $x = 1$. Set $6x-6=0$ and solve for $x$.
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Find $f(x)$ at $x = 5$ for $f(x) = x^2 - 5x$.
Find $f(x)$ at $x = 5$ for $f(x) = x^2 - 5x$.
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$f(5) = 0$. Substitution shows $x=5$ is a root of the function.
$f(5) = 0$. Substitution shows $x=5$ is a root of the function.
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Determine the endpoints for the interval $[a, b] = [-2, 5]$.
Determine the endpoints for the interval $[a, b] = [-2, 5]$.
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Endpoints are $x = -2$ and $x = 5$. These are the boundary values of the given interval.
Endpoints are $x = -2$ and $x = 5$. These are the boundary values of the given interval.
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Evaluate $f'(x) = 4x - 4$ for critical points.
Evaluate $f'(x) = 4x - 4$ for critical points.
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Critical point: $x = 1$. Set $4x-4=0$ and solve for $x$.
Critical point: $x = 1$. Set $4x-4=0$ and solve for $x$.
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Find $f(x)$ at $x = 0$ for $f(x) = x^4 - 2x^2$.
Find $f(x)$ at $x = 0$ for $f(x) = x^4 - 2x^2$.
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$f(0) = 0$. Direct substitution of $x=0$ into the function.
$f(0) = 0$. Direct substitution of $x=0$ into the function.
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Identify the type of interval required for the Candidates Test.
Identify the type of interval required for the Candidates Test.
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A closed interval $[a, b]$. Ensures function is continuous and extrema exist by EVT.
A closed interval $[a, b]$. Ensures function is continuous and extrema exist by EVT.
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State the condition for a point to be a candidate for extrema.
State the condition for a point to be a candidate for extrema.
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Point must be a critical point or endpoint. Only these points can produce absolute extrema on closed intervals.
Point must be a critical point or endpoint. Only these points can produce absolute extrema on closed intervals.
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State the first step in the Candidates Test.
State the first step in the Candidates Test.
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Find the derivative of the function. Need derivative to locate critical points where extrema may occur.
Find the derivative of the function. Need derivative to locate critical points where extrema may occur.
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What is the Candidates Test used for in calculus?
What is the Candidates Test used for in calculus?
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Determining the absolute extrema on a closed interval. Finds global max/min on closed intervals by testing all candidate points.
Determining the absolute extrema on a closed interval. Finds global max/min on closed intervals by testing all candidate points.
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Find $f(1)$ for $f(x) = x^2 - 2x + 1$.
Find $f(1)$ for $f(x) = x^2 - 2x + 1$.
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$f(1) = 0$. This is a perfect square: $(x-1)^2=0$ at $x=1$.
$f(1) = 0$. This is a perfect square: $(x-1)^2=0$ at $x=1$.
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Identify the formula for finding critical points.
Identify the formula for finding critical points.
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Set $f'(x) = 0$ and solve for $x$. Standard method to find where slope equals zero.
Set $f'(x) = 0$ and solve for $x$. Standard method to find where slope equals zero.
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Find $f(x)$ at $x = 5$ for $f(x) = x^2 - 5x$.
Find $f(x)$ at $x = 5$ for $f(x) = x^2 - 5x$.
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$f(5) = 0$. Substitution shows $x=5$ is a root of the function.
$f(5) = 0$. Substitution shows $x=5$ is a root of the function.
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What is $f(x)$ at $x = 0$ for $f(x) = \frac{1}{x} + x$ on $[1, 2]$?
What is $f(x)$ at $x = 0$ for $f(x) = \frac{1}{x} + x$ on $[1, 2]$?
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Not applicable; $x = 0$ is not in $[1, 2]$. Point outside domain cannot be evaluated.
Not applicable; $x = 0$ is not in $[1, 2]$. Point outside domain cannot be evaluated.
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Determine $f(x)$ at $x=2$ for $f(x) = x^3 - 3x^2 + 4x$.
Determine $f(x)$ at $x=2$ for $f(x) = x^3 - 3x^2 + 4x$.
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$f(2) = 4$. Direct substitution of $x=2$ into the cubic function.
$f(2) = 4$. Direct substitution of $x=2$ into the cubic function.
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Determine the endpoints for the interval $[a, b] = [-2, 5]$.
Determine the endpoints for the interval $[a, b] = [-2, 5]$.
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Endpoints are $x = -2$ and $x = 5$. These are the boundary values of the given interval.
Endpoints are $x = -2$ and $x = 5$. These are the boundary values of the given interval.
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Find $f(x)$ at $x = 0$ for $f(x) = x^4 - 2x^2$.
Find $f(x)$ at $x = 0$ for $f(x) = x^4 - 2x^2$.
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$f(0) = 0$. Direct substitution of $x=0$ into the function.
$f(0) = 0$. Direct substitution of $x=0$ into the function.
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Which function value determines the absolute minimum?
Which function value determines the absolute minimum?
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The smallest value among evaluated points. Lowest function output among all tested candidates.
The smallest value among evaluated points. Lowest function output among all tested candidates.
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Evaluate $f(x)$ at endpoints for $f(x) = \frac{1}{x}$ on $[1, 3]$.
Evaluate $f(x)$ at endpoints for $f(x) = \frac{1}{x}$ on $[1, 3]$.
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$f(1) = 1$, $f(3) = \frac{1}{3}$. Substitute endpoint values into $f(x)=\frac{1}{x}$.
$f(1) = 1$, $f(3) = \frac{1}{3}$. Substitute endpoint values into $f(x)=\frac{1}{x}$.
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Which values are evaluated in the Candidates Test?
Which values are evaluated in the Candidates Test?
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Endpoints and critical points of the interval. All possible locations where absolute extrema can occur.
Endpoints and critical points of the interval. All possible locations where absolute extrema can occur.
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What is the final step in the Candidates Test?
What is the final step in the Candidates Test?
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Compare function values at critical points and endpoints. Determines which candidate gives the absolute maximum/minimum.
Compare function values at critical points and endpoints. Determines which candidate gives the absolute maximum/minimum.
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Find $f(1)$ for $f(x) = x^2 - 2x + 1$.
Find $f(1)$ for $f(x) = x^2 - 2x + 1$.
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$f(1) = 0$. This is a perfect square: $(x-1)^2=0$ at $x=1$.
$f(1) = 0$. This is a perfect square: $(x-1)^2=0$ at $x=1$.
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Evaluate $f'(x) = 3x^2 - 12x + 9$ for critical points.
Evaluate $f'(x) = 3x^2 - 12x + 9$ for critical points.
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Critical points: $x = 1, x = 3$. Solve $3x^2-12x+9=0$ to get $x=1,3$.
Critical points: $x = 1, x = 3$. Solve $3x^2-12x+9=0$ to get $x=1,3$.
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Evaluate $f'(x) = 4x - 4$ for critical points.
Evaluate $f'(x) = 4x - 4$ for critical points.
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Critical point: $x = 1$. Set $4x-4=0$ and solve for $x$.
Critical point: $x = 1$. Set $4x-4=0$ and solve for $x$.
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