Candidates Test - AP Calculus BC
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What condition must a function meet to apply the Candidates Test?
What condition must a function meet to apply the Candidates Test?
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Function must be continuous on a closed interval. Ensures the function has guaranteed absolute maximum and minimum values.
Function must be continuous on a closed interval. Ensures the function has guaranteed absolute maximum and minimum values.
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What function values are compared in the Candidates Test?
What function values are compared in the Candidates Test?
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Values at critical points and endpoints. These are all possible locations where absolute extrema can occur.
Values at critical points and endpoints. These are all possible locations where absolute extrema can occur.
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Evaluate $f(x) = x^2$ at $x = -1, 0, 1$ to find extrema.
Evaluate $f(x) = x^2$ at $x = -1, 0, 1$ to find extrema.
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Minimum at $x = 0$, Maximum at $x = -1$ or $x = 1$. $f(-1) = 1$, $f(0) = 0$, $f(1) = 1$, so min at $x = 0$, max at endpoints.
Minimum at $x = 0$, Maximum at $x = -1$ or $x = 1$. $f(-1) = 1$, $f(0) = 0$, $f(1) = 1$, so min at $x = 0$, max at endpoints.
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Evaluate $f(x) = 3x^2 - 12x$ at $x = 0, 2, 4$.
Evaluate $f(x) = 3x^2 - 12x$ at $x = 0, 2, 4$.
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$f(0) = 0$, $f(2) = -12$, $f(4) = 0$. Direct substitution shows function values at these candidate points.
$f(0) = 0$, $f(2) = -12$, $f(4) = 0$. Direct substitution shows function values at these candidate points.
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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 and solve for critical points. Critical points occur where the derivative equals zero or is undefined.
Find the derivative and solve for critical points. Critical points occur where the derivative equals zero or is undefined.
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How do you find critical points?
How do you find critical points?
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Solve $f'(x) = 0$ and check where $f'(x)$ is undefined. Both conditions identify all points where extrema can occur.
Solve $f'(x) = 0$ and check where $f'(x)$ is undefined. Both conditions identify all points where extrema can occur.
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Determine the absolute minimum of $f(x) = x^3 - 3x^2 + 2$ on $[0, 3]$.
Determine the absolute minimum of $f(x) = x^3 - 3x^2 + 2$ on $[0, 3]$.
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Minimum at $x = 2$. Critical point $x = 2$ gives $f(2) = -2$, the lowest value on the interval.
Minimum at $x = 2$. Critical point $x = 2$ gives $f(2) = -2$, the lowest value on the interval.
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Identify the absolute extrema for $f(x) = x^2 - 4$ on $[-3, 3]$.
Identify the absolute extrema for $f(x) = x^2 - 4$ on $[-3, 3]$.
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Maximum at $x = 3$, Minimum at $x = 0$. Compare $f(-3) = 5$, $f(0) = -4$, $f(3) = 5$ for extrema locations.
Maximum at $x = 3$, Minimum at $x = 0$. Compare $f(-3) = 5$, $f(0) = -4$, $f(3) = 5$ for extrema locations.
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Evaluate $f(x) = x^2 - 4x + 3$ at endpoints $x = 0$ and $x = 3$.
Evaluate $f(x) = x^2 - 4x + 3$ at endpoints $x = 0$ and $x = 3$.
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$f(0) = 3$, $f(3) = -3$. Direct substitution into the function at the boundary points.
$f(0) = 3$, $f(3) = -3$. Direct substitution into the function at the boundary points.
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How is the derivative used in the Candidates Test?
How is the derivative used in the Candidates Test?
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To find critical points. Sets $f'(x) = 0$ to locate potential extrema within the interval.
To find critical points. Sets $f'(x) = 0$ to locate potential extrema within the interval.
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Identify the second step in the Candidates Test.
Identify the second step in the Candidates Test.
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Evaluate the function at critical points and endpoints. All candidate points must be tested to find absolute extrema.
Evaluate the function at critical points and endpoints. All candidate points must be tested to find absolute extrema.
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Find the absolute extrema of $f(x) = |x|$ on $[-2, 2]$.
Find the absolute extrema of $f(x) = |x|$ on $[-2, 2]$.
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Minimum at $x = 0$, Maximum at $x = -2$ or $x = 2$. Critical point at $x = 0$ gives minimum, endpoints give maximum values.
Minimum at $x = 0$, Maximum at $x = -2$ or $x = 2$. Critical point at $x = 0$ gives minimum, endpoints give maximum values.
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State the condition for using limits in the Candidates Test.
State the condition for using limits in the Candidates Test.
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Limits are used if endpoints are undefined. When function is undefined at endpoints, use limit values instead.
Limits are used if endpoints are undefined. When function is undefined at endpoints, use limit values instead.
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Identify the absolute extrema for $f(x) = \frac{1}{x^2}$ on $[1, 3]$.
Identify the absolute extrema for $f(x) = \frac{1}{x^2}$ on $[1, 3]$.
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Minimum at $x = 3$, Maximum at $x = 1$. $f(x) = \frac{1}{x^2}$ decreases on $[1,3]$, so max at $x=1$, min at $x=3$.
Minimum at $x = 3$, Maximum at $x = 1$. $f(x) = \frac{1}{x^2}$ decreases on $[1,3]$, so max at $x=1$, min at $x=3$.
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Why is continuity important for the Candidates Test?
Why is continuity important for the Candidates Test?
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Ensures extrema exist on closed intervals. Without continuity, absolute extrema might not exist.
Ensures extrema exist on closed intervals. Without continuity, absolute extrema might not exist.
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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 to identify absolute extrema. The highest and lowest function values give the absolute max and min.
Compare function values to identify absolute extrema. The highest and lowest function values give the absolute max and min.
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Determine the absolute minimum of $f(x) = x^2$ on $[-1, 1]$.
Determine the absolute minimum of $f(x) = x^2$ on $[-1, 1]$.
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Minimum at $x = 0$. Critical point $x = 0$ gives $f(0) = 0$, the minimum value.
Minimum at $x = 0$. Critical point $x = 0$ gives $f(0) = 0$, the minimum value.
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What is the purpose of evaluating endpoints in the Candidates Test?
What is the purpose of evaluating endpoints in the Candidates Test?
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To ensure all potential extrema are considered. Extrema can occur at boundary points of the closed interval.
To ensure all potential extrema are considered. Extrema can occur at boundary points of the closed interval.
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Evaluate $f(x) = x^2 + 1$ at $x = -2, 0, 2$.
Evaluate $f(x) = x^2 + 1$ at $x = -2, 0, 2$.
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$f(-2) = 5$, $f(0) = 1$, $f(2) = 5$. Shows function values at critical point and endpoints for comparison.
$f(-2) = 5$, $f(0) = 1$, $f(2) = 5$. Shows function values at critical point and endpoints for comparison.
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Find the absolute minimum of $f(x) = x^3 - 3x$ on $[-2, 2]$.
Find the absolute minimum of $f(x) = x^3 - 3x$ on $[-2, 2]$.
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Minimum at $x = -2$. Critical point $x = 1$ gives local max, compare with endpoint values.
Minimum at $x = -2$. Critical point $x = 1$ gives local max, compare with endpoint values.
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Evaluate $f(x) = -x^3 + 3x^2$ at endpoints $x = 0$ and $x = 2$.
Evaluate $f(x) = -x^3 + 3x^2$ at endpoints $x = 0$ and $x = 2$.
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$f(0) = 0$, $f(2) = 0$. Shows function values at the boundary points of the interval.
$f(0) = 0$, $f(2) = 0$. Shows function values at the boundary points of the interval.
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What is the absolute minimum of $f(x) = 1 - x^2$ on $[-1, 1]$?
What is the absolute minimum of $f(x) = 1 - x^2$ on $[-1, 1]$?
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-1 at $x = -1$ or $x = 1$. Parabola opens downward with maximum at $x = 0$, minimum at endpoints.
-1 at $x = -1$ or $x = 1$. Parabola opens downward with maximum at $x = 0$, minimum at endpoints.
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Find the absolute extrema for $f(x) = \frac{1}{x}$ on $[1, 4]$.
Find the absolute extrema for $f(x) = \frac{1}{x}$ on $[1, 4]$.
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Minimum at $x = 4$, Maximum at $x = 1$. $f(x) = \frac{1}{x}$ decreases on $[1,4]$, so max at $x=1$, min at $x=4$.
Minimum at $x = 4$, Maximum at $x = 1$. $f(x) = \frac{1}{x}$ decreases on $[1,4]$, so max at $x=1$, min at $x=4$.
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What role do endpoints play in the Candidates Test?
What role do endpoints play in the Candidates Test?
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They are evaluated for potential extrema. Include boundary points as candidates since extrema can occur there.
They are evaluated for potential extrema. Include boundary points as candidates since extrema can occur there.
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What is the absolute maximum of $f(x) = x^4 - 4x^2 + 4$ on $[-2, 2]$?
What is the absolute maximum of $f(x) = x^4 - 4x^2 + 4$ on $[-2, 2]$?
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Maximum at $x = 0$. Critical points at $x = \pm\sqrt{2}$ give local minima, max at $x = 0$.
Maximum at $x = 0$. Critical points at $x = \pm\sqrt{2}$ give local minima, max at $x = 0$.
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What is the absolute maximum of $f(x) = x^3$ on $[-1, 2]$?
What is the absolute maximum of $f(x) = x^3$ on $[-1, 2]$?
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$f(2) = 8$. Evaluating at endpoints and critical point shows maximum at $x = 2$.
$f(2) = 8$. Evaluating at endpoints and critical point shows maximum at $x = 2$.
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What is the Candidates Test used for?
What is the Candidates Test used for?
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Determining absolute extrema on a closed interval. Finds global max/min on closed intervals using critical points and endpoints.
Determining absolute extrema on a closed interval. Finds global max/min on closed intervals using critical points and endpoints.
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Find the critical points for $f(x) = x^3 - 3x^2 + 4$.
Find the critical points for $f(x) = x^3 - 3x^2 + 4$.
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$x = 0, x = 2$. $f'(x) = 3x^2 - 6x = 3x(x-2)$, so $x = 0$ and $x = 2$.
$x = 0, x = 2$. $f'(x) = 3x^2 - 6x = 3x(x-2)$, so $x = 0$ and $x = 2$.
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Which theorem supports the use of the Candidates Test?
Which theorem supports the use of the Candidates Test?
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Extreme Value Theorem. Guarantees continuous functions have absolute extrema on closed intervals.
Extreme Value Theorem. Guarantees continuous functions have absolute extrema on closed intervals.
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Evaluate $f(x) = x^2 + x - 6$ at $x = -3, 0, 2$.
Evaluate $f(x) = x^2 + x - 6$ at $x = -3, 0, 2$.
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$f(-3) = 0$, $f(0) = -6$, $f(2) = 0$. Direct evaluation by substituting each x-value into the function.
$f(-3) = 0$, $f(0) = -6$, $f(2) = 0$. Direct evaluation by substituting each x-value into the function.
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