Difference between revisions of "Lagrange's Mean Value Theorem"
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| − | '''Lagrange's mean value theorem''' or LMVT is considered one of the most important results in real analysis. An elegant proof of the [[Fundamental Theorem of Calculus]] can be given using LMVT. | + | '''Lagrange's mean value theorem''' (often called "the mean value theorem," and abbreviated MVT or LMVT) is considered one of the most important results in [[real analysis]]. An elegant proof of the [[Fundamental Theorem of Calculus]] can be given using LMVT. |
==Statement== | ==Statement== | ||
| − | Let <math>f:[a,b]\rightarrow\mathbb{R}</math> | + | Let <math>f:[a,b]\rightarrow\mathbb{R}</math> be a [[continuous function]], [[differentiable]] on the [[open interval]] <math>(a,b)</math>. Then there exists some <math>c\in (a,b)</math> such that <math>f'(c)=\frac{f(b)-f(a)}{b-a}</math>. |
| − | + | Informally, this says that a differentiable function must at some point grow with instantaneous velocity equal to its average velocity over an interval. | |
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==Proof== | ==Proof== | ||
| − | We reduce the problem to | + | We reduce the problem to [[Rolle's theorem]] by using an auxiliary function. |
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| − | or < | + | Consider <math>g(x)=f(x)-\frac{f(b)-f(a)}{b-a}(x-a).</math> Note that <math>g(a)=g(b)=f(a).</math> By [[Rolle's theorem]], there exists <math>c</math> in <math>(a,b)</math> such that <math>g'(c)=0,</math> or $<cmath>f'(c)-\frac{f(b)-f(a)}{b-a}=0,</cmath> which simplifies to <cmath>f'(c)=\frac{f(b)-f(a)}{b-a},</cmath> as desired. |
<p align=right>QED</p> | <p align=right>QED</p> | ||
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[[Category:Calculus]] | [[Category:Calculus]] | ||
| + | [[Category:Theorems]] | ||
Latest revision as of 11:53, 20 February 2024
Lagrange's mean value theorem (often called "the mean value theorem," and abbreviated MVT or LMVT) is considered one of the most important results in real analysis. An elegant proof of the Fundamental Theorem of Calculus can be given using LMVT.
Statement
Let ![$f:[a,b]\rightarrow\mathbb{R}$](http://latex.artofproblemsolving.com/6/7/c/67c320754a36aaec178e61f04bfcbac41c0bf675.png)
be a continuous function, differentiable on the open interval 
. Then there exists some 
such that 
.
Informally, this says that a differentiable function must at some point grow with instantaneous velocity equal to its average velocity over an interval.
Proof
We reduce the problem to Rolle's theorem by using an auxiliary function.
Consider 
Note that 
By Rolle's theorem, there exists 
in such that 
or $![\[f'(c)-\frac{f(b)-f(a)}{b-a}=0,\]](http://latex.artofproblemsolving.com/c/5/1/c51c72b8fb36f24e0341478c9be003a4c07ef2ae.png)
which simplifies to ![\[f'(c)=\frac{f(b)-f(a)}{b-a},\]](http://latex.artofproblemsolving.com/8/2/3/823dea528f1d909c3514e321b5338a928bf8aed7.png)
as desired.
QED
