Mean Value Theorem - AP Calculus AB
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Identify $c$ for $f(x)=x^2$ on $[1, 4]$ using the Mean Value Theorem.
Identify $c$ for $f(x)=x^2$ on $[1, 4]$ using the Mean Value Theorem.
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$c = 2.5$. Set $f'(c) = 2c$ equal to $\frac{16-1}{4-1} = 5$, so $c = 2.5$.
$c = 2.5$. Set $f'(c) = 2c$ equal to $\frac{16-1}{4-1} = 5$, so $c = 2.5$.
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Find $c$ for $f(x)=x^4$ on $[0, 2]$ using Mean Value Theorem.
Find $c$ for $f(x)=x^4$ on $[0, 2]$ using Mean Value Theorem.
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$c = \frac{2}{\sqrt[3]{3}\frac{3}{4}}$. Set $4c^3$ equal to average rate $\frac{16-0}{2-0} = 8$.
$c = \frac{2}{\sqrt[3]{3}\frac{3}{4}}$. Set $4c^3$ equal to average rate $\frac{16-0}{2-0} = 8$.
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What is the relationship between Rolle's and the Mean Value Theorem?
What is the relationship between Rolle's and the Mean Value Theorem?
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Rolle's Theorem is a special case of the Mean Value Theorem. Rolle's applies when $f(a) = f(b)$ in the MVT.
Rolle's Theorem is a special case of the Mean Value Theorem. Rolle's applies when $f(a) = f(b)$ in the MVT.
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What theorem guarantees $f'(c) = 0$ if $f(a)=f(b)$ on $[a, b]$?
What theorem guarantees $f'(c) = 0$ if $f(a)=f(b)$ on $[a, b]$?
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Rolle's Theorem. MVT with $f(a) = f(b)$ gives horizontal tangent.
Rolle's Theorem. MVT with $f(a) = f(b)$ gives horizontal tangent.
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Explain why $f(x) = |x|$ does not satisfy Mean Value Theorem on $[-1, 1]$.
Explain why $f(x) = |x|$ does not satisfy Mean Value Theorem on $[-1, 1]$.
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Not differentiable at $x = 0$. The sharp corner prevents differentiability at the origin.
Not differentiable at $x = 0$. The sharp corner prevents differentiability at the origin.
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State the Mean Value Theorem for derivatives.
State the Mean Value Theorem for derivatives.
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If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then $\frac{f(b)-f(a)}{b-a} = f'(c)$ for some $c \in (a, b)$. The fundamental theorem connecting secant and tangent line slopes.
If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then $\frac{f(b)-f(a)}{b-a} = f'(c)$ for some $c \in (a, b)$. The fundamental theorem connecting secant and tangent line slopes.
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Identify a function that does not satisfy the Mean Value Theorem on $[0, 1]$.
Identify a function that does not satisfy the Mean Value Theorem on $[0, 1]$.
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$f(x) = |x|$ (not differentiable at $x = 0$). The absolute value function has a corner at $x = 0$.
$f(x) = |x|$ (not differentiable at $x = 0$). The absolute value function has a corner at $x = 0$.
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What does the Mean Value Theorem imply for linear functions?
What does the Mean Value Theorem imply for linear functions?
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$f'(c) = m$, the slope of the line is constant. For linear functions, the derivative is constant everywhere.
$f'(c) = m$, the slope of the line is constant. For linear functions, the derivative is constant everywhere.
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What does the Mean Value Theorem guarantee about the derivative?
What does the Mean Value Theorem guarantee about the derivative?
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There exists $c \in (a, b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$. The instantaneous rate equals the average rate at some point.
There exists $c \in (a, b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$. The instantaneous rate equals the average rate at some point.
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How does the Mean Value Theorem relate to average rate of change?
How does the Mean Value Theorem relate to average rate of change?
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It states $f'(c)$ equals the average rate of change over $[a, b]$. MVT guarantees instantaneous rate equals average rate somewhere.
It states $f'(c)$ equals the average rate of change over $[a, b]$. MVT guarantees instantaneous rate equals average rate somewhere.
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Determine $f'(c)$ for $f(x)=e^x$ on $[0, 1]$ using Mean Value Theorem.
Determine $f'(c)$ for $f(x)=e^x$ on $[0, 1]$ using Mean Value Theorem.
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$f'(c) = e - 1$. The average rate $\frac{e-1}{1-0} = e-1$ equals $f'(c)$.
$f'(c) = e - 1$. The average rate $\frac{e-1}{1-0} = e-1$ equals $f'(c)$.
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Find $c$ for $f(x)=x^2$ on $[2, 5]$ using Mean Value Theorem.
Find $c$ for $f(x)=x^2$ on $[2, 5]$ using Mean Value Theorem.
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$c = 3.5$. Set $f'(c) = 2c$ equal to $\frac{25-4}{5-2} = 7$, so $c = 3.5$.
$c = 3.5$. Set $f'(c) = 2c$ equal to $\frac{25-4}{5-2} = 7$, so $c = 3.5$.
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Determine $c$ for $f(x)=x^2$ on $[3, 6]$ using the Mean Value Theorem.
Determine $c$ for $f(x)=x^2$ on $[3, 6]$ using the Mean Value Theorem.
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$c = 4.5$. Set $f'(c) = 2c$ equal to $\frac{36-9}{6-3} = 9$, so $c = 4.5$.
$c = 4.5$. Set $f'(c) = 2c$ equal to $\frac{36-9}{6-3} = 9$, so $c = 4.5$.
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Find the $c$ for $f(x) = \frac{1}{x^2}$ on $[1, 3]$ using the Mean Value Theorem.
Find the $c$ for $f(x) = \frac{1}{x^2}$ on $[1, 3]$ using the Mean Value Theorem.
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$c = \frac{3}{2}$. Set $f'(c) = -\frac{2}{c^3}$ equal to $\frac{1/9-1}{3-1} = -\frac{4}{9}$
$c = \frac{3}{2}$. Set $f'(c) = -\frac{2}{c^3}$ equal to $\frac{1/9-1}{3-1} = -\frac{4}{9}$
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What is the geometric interpretation of the Mean Value Theorem?
What is the geometric interpretation of the Mean Value Theorem?
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A tangent line at $c$ is parallel to the secant line through $(a, f(a))$ and $(b, f(b))$. The tangent slope at $c$ equals the secant slope.
A tangent line at $c$ is parallel to the secant line through $(a, f(a))$ and $(b, f(b))$. The tangent slope at $c$ equals the secant slope.
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Find $c$ for $f(x) = \frac{1}{x}$ on $[2, 8]$ using Mean Value Theorem.
Find $c$ for $f(x) = \frac{1}{x}$ on $[2, 8]$ using Mean Value Theorem.
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$c = 4$. Set $f'(c) = -\frac{1}{c^2}$ equal to $\frac{1/8-1/2}{8-2} = -\frac{1}{16}$.
$c = 4$. Set $f'(c) = -\frac{1}{c^2}$ equal to $\frac{1/8-1/2}{8-2} = -\frac{1}{16}$.
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Does $f(x)=\frac{1}{x}$ satisfy the Mean Value Theorem on $[0, 1]$?
Does $f(x)=\frac{1}{x}$ satisfy the Mean Value Theorem on $[0, 1]$?
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No, $f(x)$ is not continuous on $[0, 1]$. The function is undefined at $x = 0$, breaking continuity.
No, $f(x)$ is not continuous on $[0, 1]$. The function is undefined at $x = 0$, breaking continuity.
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Find $c$ for $f(x)=\frac{1}{x}$ on $[1, e]$ using Mean Value Theorem.
Find $c$ for $f(x)=\frac{1}{x}$ on $[1, e]$ using Mean Value Theorem.
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$c = \frac{1}{e}$. Set $f'(c) = -\frac{1}{c^2}$ equal to $\frac{1/e-1}{e-1}$.
$c = \frac{1}{e}$. Set $f'(c) = -\frac{1}{c^2}$ equal to $\frac{1/e-1}{e-1}$.
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Calculate $f'(c)$ for $f(x) = \frac{x^3}{3}$ on $[0, 1]$ using the Mean Value Theorem.
Calculate $f'(c)$ for $f(x) = \frac{x^3}{3}$ on $[0, 1]$ using the Mean Value Theorem.
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$f'(c) = \frac{1}{3}$. The average rate $\frac{1/3-0}{1-0} = \frac{1}{3}$ equals $f'(c)$.
$f'(c) = \frac{1}{3}$. The average rate $\frac{1/3-0}{1-0} = \frac{1}{3}$ equals $f'(c)$.
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Does $f(x)=x^3$ satisfy the Mean Value Theorem on $[-1, 1]$?
Does $f(x)=x^3$ satisfy the Mean Value Theorem on $[-1, 1]$?
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Yes, $f(x)$ is continuous and differentiable. Polynomials are continuous and differentiable everywhere.
Yes, $f(x)$ is continuous and differentiable. Polynomials are continuous and differentiable everywhere.
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For $f(x)=\frac{1}{x}$, find the $c$ on $[1, 4]$ using Mean Value Theorem.
For $f(x)=\frac{1}{x}$, find the $c$ on $[1, 4]$ using Mean Value Theorem.
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$c = 2$. Set $f'(c) = -\frac{1}{c^2}$ equal to $\frac{1/4-1}{4-1} = -\frac{1}{4}$.
$c = 2$. Set $f'(c) = -\frac{1}{c^2}$ equal to $\frac{1/4-1}{4-1} = -\frac{1}{4}$.
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What role does differentiability play in the Mean Value Theorem?
What role does differentiability play in the Mean Value Theorem?
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Ensures $f'$ exists at every point in $(a, b)$.. Without derivatives, we can't find the required tangent slope.
Ensures $f'$ exists at every point in $(a, b)$.. Without derivatives, we can't find the required tangent slope.
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What is the importance of continuity in the Mean Value Theorem?
What is the importance of continuity in the Mean Value Theorem?
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Ensures no jumps or gaps on $[a, b]$. Prevents breaks that would invalidate the theorem.
Ensures no jumps or gaps on $[a, b]$. Prevents breaks that would invalidate the theorem.
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Determine $c$ for $f(x) = x^3 - x$ on $[-1, 1]$ using the Mean Value Theorem.
Determine $c$ for $f(x) = x^3 - x$ on $[-1, 1]$ using the Mean Value Theorem.
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$c = 0$. Set $f'(c) = 3c^2 - 1$ equal to $\frac{0-0}{1-(-1)} = 0$, so $c = 0$.
$c = 0$. Set $f'(c) = 3c^2 - 1$ equal to $\frac{0-0}{1-(-1)} = 0$, so $c = 0$.
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Find $f'(c)$ if $f(x)=x^3$ on $[1, 2]$ using the Mean Value Theorem.
Find $f'(c)$ if $f(x)=x^3$ on $[1, 2]$ using the Mean Value Theorem.
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$f'(c) = 7$. Average rate is $\frac{8-1}{2-1} = 7$, which equals $f'(c)$.
$f'(c) = 7$. Average rate is $\frac{8-1}{2-1} = 7$, which equals $f'(c)$.
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For $f(x)=x^2+3x+2$, find the $c$ in $[0, 3]$ using Mean Value Theorem.
For $f(x)=x^2+3x+2$, find the $c$ in $[0, 3]$ using Mean Value Theorem.
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$c = 1.5$. Set $f'(c) = 2c + 3$ equal to $\frac{20-2}{3-0} = 6$, so $c = 1.5$.
$c = 1.5$. Set $f'(c) = 2c + 3$ equal to $\frac{20-2}{3-0} = 6$, so $c = 1.5$.
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Identify the value $c$ guaranteed by the Mean Value Theorem for $f(x)=x^2$ on $[1, 3]$.
Identify the value $c$ guaranteed by the Mean Value Theorem for $f(x)=x^2$ on $[1, 3]$.
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$c = 2$. Set $f'(c) = 2c$ equal to $\frac{9-1}{3-1} = 4$, so $c = 2$.
$c = 2$. Set $f'(c) = 2c$ equal to $\frac{9-1}{3-1} = 4$, so $c = 2$.
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What conditions must be met to apply the Mean Value Theorem?
What conditions must be met to apply the Mean Value Theorem?
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$f$ must be continuous on $[a, b]$ and differentiable on $(a, b)$. These ensure the function is smooth enough for the theorem to apply.
$f$ must be continuous on $[a, b]$ and differentiable on $(a, b)$. These ensure the function is smooth enough for the theorem to apply.
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What is the derivative expression found using Mean Value Theorem?
What is the derivative expression found using Mean Value Theorem?
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$f'(c) = \frac{f(b)-f(a)}{b-a}$. This is the core equation of the Mean Value Theorem.
$f'(c) = \frac{f(b)-f(a)}{b-a}$. This is the core equation of the Mean Value Theorem.
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Which theorem is a special case of the Mean Value Theorem?
Which theorem is a special case of the Mean Value Theorem?
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Rolle's Theorem. When $f(a) = f(b)$, MVT gives $f'(c) = 0$ (Rolle's).
Rolle's Theorem. When $f(a) = f(b)$, MVT gives $f'(c) = 0$ (Rolle's).
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