Extreme Value Theorem, Extrema, Critical Points - AP Calculus BC
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What is the First Derivative Test?
What is the First Derivative Test?
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Determines local extrema using $f'(x)$ sign change. Analyzes where $f'$ changes from positive to negative or vice versa.
Determines local extrema using $f'(x)$ sign change. Analyzes where $f'$ changes from positive to negative or vice versa.
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Determine if $x=1$ is a critical point for $f(x) = x^3 - 3x$.
Determine if $x=1$ is a critical point for $f(x) = x^3 - 3x$.
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Yes, $x=1$ is a critical point. $f'(1) = 3(1)^2 - 3 = 0$, so derivative equals zero.
Yes, $x=1$ is a critical point. $f'(1) = 3(1)^2 - 3 = 0$, so derivative equals zero.
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Find the derivative of $f(x) = 3x^2 - 6x + 1$.
Find the derivative of $f(x) = 3x^2 - 6x + 1$.
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$f'(x) = 6x - 6$. Power rule applied to each term of the polynomial.
$f'(x) = 6x - 6$. Power rule applied to each term of the polynomial.
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Find critical points of $f(x) = \frac{1}{3}x^3 - x^2$.
Find critical points of $f(x) = \frac{1}{3}x^3 - x^2$.
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Critical points: $x = 0, x = 2$. Set $f'(x) = x^2 - 2x = x(x-2) = 0$ to find solutions.
Critical points: $x = 0, x = 2$. Set $f'(x) = x^2 - 2x = x(x-2) = 0$ to find solutions.
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Identify the global extrema of $f(x) = x^2 - 4x$ on $[0, 3]$.
Identify the global extrema of $f(x) = x^2 - 4x$ on $[0, 3]$.
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Global min at $x=2$, Global max at $x=3$. Vertex at $x=2$ gives min; endpoint at $x=3$ gives max.
Global min at $x=2$, Global max at $x=3$. Vertex at $x=2$ gives min; endpoint at $x=3$ gives max.
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Find critical points of $f(x) = x^2 - 4x$.
Find critical points of $f(x) = x^2 - 4x$.
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Critical points: $x = 2$. Set $f'(x) = 2x - 4 = 0$ to solve for critical points.
Critical points: $x = 2$. Set $f'(x) = 2x - 4 = 0$ to solve for critical points.
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Identify extrema of $f(x) = x^2 - 2x + 1$ on $[0, 2]$.
Identify extrema of $f(x) = x^2 - 2x + 1$ on $[0, 2]$.
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Global min at $x=1$, Global max at $x=2$. Vertex of parabola at $x=1$ gives min; endpoint gives max.
Global min at $x=1$, Global max at $x=2$. Vertex of parabola at $x=1$ gives min; endpoint gives max.
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Identify the extrema of $f(x) = -x^2 + 4x - 3$ on $[0, 3]$.
Identify the extrema of $f(x) = -x^2 + 4x - 3$ on $[0, 3]$.
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Global max at $x=2$, Global min at $x=3$. Vertex at $x=2$ gives max; endpoint at $x=3$ gives min.
Global max at $x=2$, Global min at $x=3$. Vertex at $x=2$ gives max; endpoint at $x=3$ gives min.
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Find global extrema of $f(x) = 2x^2$ on $[-1, 2]$.
Find global extrema of $f(x) = 2x^2$ on $[-1, 2]$.
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Global min at $x=0$, Global max at $x=2$. Evaluate $f$ at critical point $x=0$ and endpoints $x=-1,2$.
Global min at $x=0$, Global max at $x=2$. Evaluate $f$ at critical point $x=0$ and endpoints $x=-1,2$.
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What is the derivative test for local extrema?
What is the derivative test for local extrema?
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Use $f'(x)$ sign changes to identify local extrema. Sign changes in $f'(x)$ indicate transitions between increasing/decreasing.
Use $f'(x)$ sign changes to identify local extrema. Sign changes in $f'(x)$ indicate transitions between increasing/decreasing.
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Find global extrema for $f(x) = -x^2$ on $[-1, 1]$.
Find global extrema for $f(x) = -x^2$ on $[-1, 1]$.
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Global max at $x=0$, Global min at $x=-1$ and $x=1$. Parabola opens downward with vertex at origin and symmetric endpoints.
Global max at $x=0$, Global min at $x=-1$ and $x=1$. Parabola opens downward with vertex at origin and symmetric endpoints.
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Find critical points of $f(x)=x^4-4x^2$.
Find critical points of $f(x)=x^4-4x^2$.
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Critical points: $x = 0, x = 2, x = -2$. Set $f'(x) = 4x^3 - 8x = 4x(x^2-2) = 0$ to solve.
Critical points: $x = 0, x = 2, x = -2$. Set $f'(x) = 4x^3 - 8x = 4x(x^2-2) = 0$ to solve.
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Identify global extrema of $f(x) = x^3 - 6x^2 + 9x + 1$ on $[0, 3]$.
Identify global extrema of $f(x) = x^3 - 6x^2 + 9x + 1$ on $[0, 3]$.
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Global min at $x=0$, Global max at $x=3$. Evaluate at critical points and endpoints to compare values.
Global min at $x=0$, Global max at $x=3$. Evaluate at critical points and endpoints to compare values.
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What conditions are necessary for EVT to apply?
What conditions are necessary for EVT to apply?
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Function must be continuous on closed interval. Closed interval ensures compactness for guaranteed extrema existence.
Function must be continuous on closed interval. Closed interval ensures compactness for guaranteed extrema existence.
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What is a local maximum?
What is a local maximum?
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$f(c) \geq f(x)$ for all $x$ near $c$. Function value at $c$ exceeds nearby values within some interval.
$f(c) \geq f(x)$ for all $x$ near $c$. Function value at $c$ exceeds nearby values within some interval.
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Find critical points of $f(x) = x^2 - 1$.
Find critical points of $f(x) = x^2 - 1$.
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Critical point: $x = 0$. $f'(x) = 2x = 0$ only at $x = 0$.
Critical point: $x = 0$. $f'(x) = 2x = 0$ only at $x = 0$.
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What is the definition of a continuous function?
What is the definition of a continuous function?
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No gaps, jumps, or holes in domain. Function exists at every point with no discontinuities or breaks.
No gaps, jumps, or holes in domain. Function exists at every point with no discontinuities or breaks.
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What is the Second Derivative Test?
What is the Second Derivative Test?
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Uses $f''(x)$ to determine concavity and local extrema. If $f''(c) > 0$, local min; if $f''(c) < 0$, local max.
Uses $f''(x)$ to determine concavity and local extrema. If $f''(c) > 0$, local min; if $f''(c) < 0$, local max.
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Identify the critical point for $f(x) = x^2$.
Identify the critical point for $f(x) = x^2$.
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Critical point: $x = 0$. $f'(x) = 2x = 0$ only when $x = 0$.
Critical point: $x = 0$. $f'(x) = 2x = 0$ only when $x = 0$.
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Is $f(x) = \frac{1}{x}$ on $[1, 3]$ continuous?
Is $f(x) = \frac{1}{x}$ on $[1, 3]$ continuous?
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Yes, $f(x)$ is continuous on $[1, 3]$. Rational functions are continuous where denominator is nonzero.
Yes, $f(x)$ is continuous on $[1, 3]$. Rational functions are continuous where denominator is nonzero.
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Determine if $x=0$ is a critical point for $f(x)=x^4$.
Determine if $x=0$ is a critical point for $f(x)=x^4$.
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Yes, $x=0$ is a critical point. $f'(0) = 4(0)^3 = 0$, so derivative equals zero at origin.
Yes, $x=0$ is a critical point. $f'(0) = 4(0)^3 = 0$, so derivative equals zero at origin.
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Is $f(x) = x^3$ on $[-1, 1]$ continuous?
Is $f(x) = x^3$ on $[-1, 1]$ continuous?
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Yes, $f(x)$ is continuous on $[-1, 1]$. Polynomial functions are continuous everywhere on their domain.
Yes, $f(x)$ is continuous on $[-1, 1]$. Polynomial functions are continuous everywhere on their domain.
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Describe how to find global extrema on $[a, b]$.
Describe how to find global extrema on $[a, b]$.
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Evaluate $f$ at critical points and endpoints. Compare function values at all candidate points to find extrema.
Evaluate $f$ at critical points and endpoints. Compare function values at all candidate points to find extrema.
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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. Potential locations for local extrema based on derivative behavior.
Where $f'(x) = 0$ or $f'(x)$ is undefined. Potential locations for local extrema based on derivative behavior.
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Identify the critical points of $f(x) = x^3 - 3x^2 + 2$.
Identify the critical points of $f(x) = x^3 - 3x^2 + 2$.
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Critical points: $x = 0, x = 2$. Set $f'(x) = 3x^2 - 6x = 3x(x-2) = 0$ to find critical points.
Critical points: $x = 0, x = 2$. Set $f'(x) = 3x^2 - 6x = 3x(x-2) = 0$ to find critical points.
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What is a local minimum?
What is a local minimum?
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$f(c) \leq f(x)$ for all $x$ near $c$. Function value at $c$ is below nearby values within some interval.
$f(c) \leq f(x)$ for all $x$ near $c$. Function value at $c$ is below nearby values within some interval.
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State Fermat's Theorem.
State Fermat's Theorem.
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If $f$ has local extremum at $c$, $f'(c)=0$ or undefined. Local extrema only occur where the derivative is zero or undefined.
If $f$ has local extremum at $c$, $f'(c)=0$ or undefined. Local extrema only occur where the derivative is zero or undefined.
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Calculate $f'(x)$ for $f(x) = x^3 - 6x^2 + 9x$.
Calculate $f'(x)$ for $f(x) = x^3 - 6x^2 + 9x$.
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$f'(x) = 3x^2 - 12x + 9$. Power rule applied: $\frac{d}{dx}[x^3 - 6x^2 + 9x] = 3x^2 - 12x + 9$.
$f'(x) = 3x^2 - 12x + 9$. Power rule applied: $\frac{d}{dx}[x^3 - 6x^2 + 9x] = 3x^2 - 12x + 9$.
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Define a global maximum.
Define a global maximum.
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$f(c) \geq f(x)$ for all $x$ in domain of $f$. Function value at $c$ exceeds all others in the entire domain.
$f(c) \geq f(x)$ for all $x$ in domain of $f$. Function value at $c$ exceeds all others in the entire domain.
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What is the Extreme Value Theorem?
What is the Extreme Value Theorem?
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If $f$ is continuous on $[a, b]$, it has a max and min. Guarantees absolute extrema exist on closed, bounded intervals.
If $f$ is continuous on $[a, b]$, it has a max and min. Guarantees absolute extrema exist on closed, bounded intervals.
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