Difference between revisions of "Mean Value Theorem"
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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> | 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 ![$f:[a,b]\rightarrow\mathbb R$](http://latex.artofproblemsolving.com/2/1/2/212b2745cf32bdb4ace7d1a8510395ecc6aee0aa.png)
be differentiable function on the interval 
. The Mean Value Theorem states there exists a point 
such that ![\[f'(c)=\frac{f(b)-f(a)}{b-a}.\]](http://latex.artofproblemsolving.com/b/e/3/be34c960742f6b639203c1c86c698c7dd2e81617.png)
Proof
Note that 
is the slope of the secant line which passes through the graph of 
at points 
and 
. Let 
. As 
is continuous on ![$[a,b]$](http://latex.artofproblemsolving.com/8/e/c/8ecbd1ba3da8f2adef66a63f2ab32c47e63fa734.png)
since it is the linear combinatoin of the continuous functions 
and 
, and that ![\[g'(x)=f'(x)-\frac{f(b)-f(a)}{b-a}.\]](http://latex.artofproblemsolving.com/a/0/c/a0c7ebe207e9b80b0ba88d5fac047820492481d5.png)
Since is differentiable on and 
, by Rolle's Theorem, there exists some such that 
. Thus ![\[f'(c)=g'(c)+\frac{f(b)-f(a)}{b-a}=\frac{f(b)-f(a)}{b-a}\]](http://latex.artofproblemsolving.com/f/a/2/fa2335dc82cd2b1c1a1cf4185c2c5fcc630030ef.png)
as desired. 
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