Extreme Value Theorem, Extrema, Critical Points - AP Calculus AB
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What is a sufficient condition for a local maximum?
What is a sufficient condition for a local maximum?
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$f'(x)$ changes from positive to negative. Derivative changes from positive to negative.
$f'(x)$ changes from positive to negative. Derivative changes from positive to negative.
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State the First Derivative Test.
State the First Derivative Test.
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Use $f'(x)$ sign changes to determine local extrema. Check if derivative changes sign around critical points.
Use $f'(x)$ sign changes to determine local extrema. Check if derivative changes sign around critical points.
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Determine the global extrema of $f(x) = -x^2$ on $[-1, 1]$.
Determine the global extrema of $f(x) = -x^2$ on $[-1, 1]$.
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Global max at $x = 0$; global min at $x = -1, 1$. Parabola opens downward with vertex at origin.
Global max at $x = 0$; global min at $x = -1, 1$. Parabola opens downward with vertex at origin.
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Identify the local extrema of $f(x) = \frac{x^2}{2} - 2x$.
Identify the local extrema of $f(x) = \frac{x^2}{2} - 2x$.
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Local min at $x = 2$. $f'(x) = x - 2 = 0$ when $x = 2$.
Local min at $x = 2$. $f'(x) = x - 2 = 0$ when $x = 2$.
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How do you find critical points?
How do you find critical points?
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Solve $f'(x) = 0$ or find where $f'(x)$ is undefined. Find where the derivative equals zero or doesn't exist.
Solve $f'(x) = 0$ or find where $f'(x)$ is undefined. Find where the derivative equals zero or doesn't exist.
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What condition must be met for the EVT to apply?
What condition must be met for the EVT to apply?
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Function must be continuous on a closed interval. Requires both continuity and a closed, bounded interval.
Function must be continuous on a closed interval. Requires both continuity and a closed, bounded interval.
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Differentiate between global and local extrema.
Differentiate between global and local extrema.
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Global is on entire domain; local is on a neighborhood. Global considers entire domain; local considers neighborhoods.
Global is on entire domain; local is on a neighborhood. Global considers entire domain; local considers neighborhoods.
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What is a necessary condition for local extrema?
What is a necessary condition for local extrema?
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$f'(x) = 0$ or $f'(x)$ is undefined. Critical points are necessary but not sufficient conditions.
$f'(x) = 0$ or $f'(x)$ is undefined. Critical points are necessary but not sufficient conditions.
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What is the purpose of the Extreme Value Theorem?
What is the purpose of the Extreme Value Theorem?
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To ensure global extrema exist on a closed interval. Guarantees extrema exist for optimization problems.
To ensure global extrema exist on a closed interval. Guarantees extrema exist for optimization problems.
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Find critical points for $f(x) = \frac{1}{3}x^3 - x^2 + 2$.
Find critical points for $f(x) = \frac{1}{3}x^3 - x^2 + 2$.
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Critical points at $x = 0$ and $x = 2$. $f'(x) = x^2 - 2x = x(x-2) = 0$ when $x = 0, 2$.
Critical points at $x = 0$ and $x = 2$. $f'(x) = x^2 - 2x = x(x-2) = 0$ when $x = 0, 2$.
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What is a global maximum?
What is a global maximum?
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The highest value of a function on its entire domain. Also called the absolute maximum value.
The highest value of a function on its entire domain. Also called the absolute maximum value.
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What is the impact of a discontinuity on EVT?
What is the impact of a discontinuity on EVT?
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Discontinuity can prevent existence of global extrema. EVT doesn't apply if function isn't continuous.
Discontinuity can prevent existence of global extrema. EVT doesn't apply if function isn't continuous.
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What is the significance of critical points?
What is the significance of critical points?
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Potential locations of local extrema. Only at critical points can local extrema occur.
Potential locations of local extrema. Only at critical points can local extrema occur.
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How does $f'(x)$ determine extrema?
How does $f'(x)$ determine extrema?
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Sign changes in $f'(x)$ indicate local extrema. Sign changes in first derivative locate extrema.
Sign changes in $f'(x)$ indicate local extrema. Sign changes in first derivative locate extrema.
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How can you confirm a global extremum?
How can you confirm a global extremum?
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Compare function values at critical points and endpoints. Test all candidates to find absolute extrema.
Compare function values at critical points and endpoints. Test all candidates to find absolute extrema.
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Identify the global extrema of $f(x) = x^3 - 3x$ on $[-2, 2]$.
Identify the global extrema of $f(x) = x^3 - 3x$ on $[-2, 2]$.
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Global max at $x = 2$; global min at $x = -2$. $f(-2) = -2$, $f(1) = -2$, $f(2) = 2$.
Global max at $x = 2$; global min at $x = -2$. $f(-2) = -2$, $f(1) = -2$, $f(2) = 2$.
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State the Second Derivative Test.
State the Second Derivative Test.
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Use $f''(x)$ to determine concavity and local extrema. Evaluate second derivative at critical points.
Use $f''(x)$ to determine concavity and local extrema. Evaluate second derivative at critical points.
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Define a critical point of a function.
Define a critical point of a function.
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A point where $f'(x) = 0$ or $f'(x)$ is undefined. These are the only candidates for local extrema.
A point where $f'(x) = 0$ or $f'(x)$ is undefined. These are the only candidates for local extrema.
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What is a global minimum?
What is a global minimum?
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The lowest value of a function on its entire domain. Also called the absolute minimum value.
The lowest value of a function on its entire domain. Also called the absolute minimum value.
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Explain how to find global extrema on a closed interval.
Explain how to find global extrema on a closed interval.
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Evaluate $f(x)$ at critical points and endpoints. Check critical points and interval endpoints systematically.
Evaluate $f(x)$ at critical points and endpoints. Check critical points and interval endpoints systematically.
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What role do endpoints play in finding global extrema?
What role do endpoints play in finding global extrema?
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Endpoints are evaluated for possible global extrema. Endpoints must be checked for global extrema.
Endpoints are evaluated for possible global extrema. Endpoints must be checked for global extrema.
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Determine the global extrema of $f(x) = -x^2 + 4x$ on $[0, 4]$.
Determine the global extrema of $f(x) = -x^2 + 4x$ on $[0, 4]$.
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Global max at $x = 2$; global min at $x = 0, 4$. Vertex of parabola at $x = 2$ gives maximum.
Global max at $x = 2$; global min at $x = 0, 4$. Vertex of parabola at $x = 2$ gives maximum.
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Identify the critical points of $f(x) = \frac{x^3}{3} - x$.
Identify the critical points of $f(x) = \frac{x^3}{3} - x$.
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Critical points at $x = -1, 1$. $f'(x) = x^2 - 1 = 0$ when $x = \pm 1$.
Critical points at $x = -1, 1$. $f'(x) = x^2 - 1 = 0$ when $x = \pm 1$.
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Why are continuous functions important in EVT?
Why are continuous functions important in EVT?
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Continuity ensures the existence of extrema on $[a, b]$. Discontinuous functions may not have global extrema.
Continuity ensures the existence of extrema on $[a, b]$. Discontinuous functions may not have global extrema.
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Determine the global extrema of $f(x) = x^2 - 4$ on $[-2, 2]$.
Determine the global extrema of $f(x) = x^2 - 4$ on $[-2, 2]$.
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Global max at $x = -2, 2$; global min at $x = 0$. Evaluate at critical point $x = 0$ and endpoints.
Global max at $x = -2, 2$; global min at $x = 0$. Evaluate at critical point $x = 0$ and endpoints.
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Identify the critical points of $f(x) = x^2 + 3x + 2$.
Identify the critical points of $f(x) = x^2 + 3x + 2$.
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Critical point at $x = -\frac{3}{2}$. $f'(x) = 2x + 3 = 0$ when $x = -\frac{3}{2}$.
Critical point at $x = -\frac{3}{2}$. $f'(x) = 2x + 3 = 0$ when $x = -\frac{3}{2}$.
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What function property is tested with $f''(x) = 0$?
What function property is tested with $f''(x) = 0$?
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Possible inflection point; concavity change. Second derivative zero suggests possible inflection point.
Possible inflection point; concavity change. Second derivative zero suggests possible inflection point.
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What does $f''(x) > 0$ at a point indicate?
What does $f''(x) > 0$ at a point indicate?
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The function is concave up; possible local min. Positive second derivative indicates upward concavity.
The function is concave up; possible local min. Positive second derivative indicates upward concavity.
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Determine local extrema for $f(x) = x^3 - 3x^2 + 4$.
Determine local extrema for $f(x) = x^3 - 3x^2 + 4$.
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Local max at $x = 0$; local min at $x = 2$. $f'(x) = 3x^2 - 6x$ gives critical points at $x = 0, 2$.
Local max at $x = 0$; local min at $x = 2$. $f'(x) = 3x^2 - 6x$ gives critical points at $x = 0, 2$.
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What is a sufficient condition for a local minimum?
What is a sufficient condition for a local minimum?
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$f'(x)$ changes from negative to positive. Derivative changes from negative to positive.
$f'(x)$ changes from negative to positive. Derivative changes from negative to positive.
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