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Difference between revisions of "Mean Value Theorem"
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The '''Mean Value Theorem''' states that if <math>a < b</math> are [[real number]]s and the [[function]] <math>f:[a,b] \to \mathbb{R}</math> is [[continuous]] on the [[interval]] <math>[a,b]</math>, then there exists a value <math>c</math> in <math>[a,b]</math> such that
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The Mean Value Theorem (often abbreviated MVT) is considered one of the most important results in [[real analysis]]. An elegant proof of the [[Fundamental Theorem of Calculus]] can be given using Mean Value Theorem.
  
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<cmath>f(c)=\dfrac{1}{b-a}\int_{a}^{b}f(x)dx.</cmath>
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== Statement ==
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Let <math>f:[a,b]\rightarrow\mathbb R</math> be differentiable function on the [[interval]] <math>(a,b)</math>. The '''Mean Value Theorem''' states there exists a point <math>c\in(a,b)</math> such that <cmath>f'(c)=\frac{f(b)-f(a)}{b-a}.</cmath>
  
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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>.
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== Proof ==
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Note that <math>\frac{f(b)-f(a)}{b-a}</math> is the slope of the [[secant]] [[line]] which passes through the graph of <math>f</math> at points <math>(a,f(a))</math> and <math>(b,f(b))</math>. Let <math>g(x)=f(x)-\frac{f(b)-f(a)}{b-a}(x-a)</math>. As <math>g</math> is continuous on <math>[a,b]</math> since it is the linear combinatoin of the continuous functions <math>f(x)</math> and <math>(x-a)</math>, and that <cmath>g'(x)=f'(x)-\frac{f(b)-f(a)}{b-a}.</cmath> Since <math>g</math> is differentiable on <math>(a,b)</math> and <math>g(a)=g(b)=f(a)</math>, by [[Rolle's Theorem]], there exists some <math>c\in(a,b)</math> such that <math>g'(c)=0</math>. Thus <cmath>f'(c)=g'(c)+\frac{f(b)-f(a)}{b-a}=\frac{f(b)-f(a)}{b-a}</cmath> as desired. <math>\fbox{}</math>
  
 
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Latest revision as of 17:04, 28 September 2024

The Mean Value Theorem (often abbreviated MVT) is considered one of the most important results in real analysis. An elegant proof of the Fundamental Theorem of Calculus can be given using Mean Value Theorem.

Statement

Let $f:[a,b]\rightarrow\mathbb R$ be differentiable function on the interval $(a,b)$. The Mean Value Theorem states there exists a point $c\in(a,b)$ such that \[f'(c)=\frac{f(b)-f(a)}{b-a}.\]

Proof

Note that $\frac{f(b)-f(a)}{b-a}$ is the slope of the secant line which passes through the graph of $f$ at points $(a,f(a))$ and $(b,f(b))$. Let $g(x)=f(x)-\frac{f(b)-f(a)}{b-a}(x-a)$. As $g$ is continuous on $[a,b]$ since it is the linear combinatoin of the continuous functions $f(x)$ and $(x-a)$, and that \[g'(x)=f'(x)-\frac{f(b)-f(a)}{b-a}.\] Since $g$ is differentiable on $(a,b)$ and $g(a)=g(b)=f(a)$, by Rolle's Theorem, there exists some $c\in(a,b)$ such that $g'(c)=0$. Thus \[f'(c)=g'(c)+\frac{f(b)-f(a)}{b-a}=\frac{f(b)-f(a)}{b-a}\] as desired. $\fbox{}$

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