Difference between revisions of "Mean Value Theorem"

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In words, there is a number <math>c</math> in <math>(a,b)</math> such that <math>f(c)</math> equals the average value of the function in the interval <math>[a,b]</math>.
 
In words, there is a number <math>c</math> in <math>(a,b)</math> such that <math>f(c)</math> equals the average value of the function in the interval <math>[a,b]</math>.
  
[b]Proof:[/b]
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{{stub}}
  
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==Proof==
  
  
[b]Other:[/b]
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==Other==
  
 
Rolle's Theorem is a sub-case of this theorem. It states that if <math>f(a)=f(b)=0</math> for two real numbers a and b, then there is a real number c such that <math>a<c<b</math> and <math>f'(c)=0</math>.
 
Rolle's Theorem is a sub-case of this theorem. It states that if <math>f(a)=f(b)=0</math> for two real numbers a and b, then there is a real number c such that <math>a<c<b</math> and <math>f'(c)=0</math>.

Revision as of 11:27, 7 August 2018

The Mean Value Theorem states that if $a < b$ are real numbers and the function $f:[a,b] \to \mathbb{R}$ is differentiable on the interval $(a,b)$, then there exists a value $c$ in $(a,b)$ such that

\[f(c)=\dfrac{1}{b-a}\int_{a}^{b}f(x)dx.\]

In words, there is a number $c$ in $(a,b)$ such that $f(c)$ equals the average value of the function in the interval $[a,b]$.

This article is a stub. Help us out by expanding it.

Proof

Other

Rolle's Theorem is a sub-case of this theorem. It states that if $f(a)=f(b)=0$ for two real numbers a and b, then there is a real number c such that $a<c<b$ and $f'(c)=0$.

This article is a stub. Help us out by expanding it.

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