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Difference between revisions of "Rolle's Theorem"

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Without loss of generality, we can assume that <math>M\neq f(a)</math>
 
Without loss of generality, we can assume that <math>M\neq f(a)</math>
  
−
By the [[Extremum value theorem]], <math>\exists c\in (a,b)</math> such that <math>f(c)=M</math>
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By the [[Maximum-minimum theorem]], <math>\exists c\in (a,b)</math> such that <math>f(c)=M</math>
  
 
Assume if possible <math>f'(c)>0</math>
 
Assume if possible <math>f'(c)>0</math>

Revision as of 22:01, 14 February 2008

Rolle's theorem is an important theorem among the class of results regarding the value of the derivative on an interval.

Statement

Let $f:[a,b]\rightarrow\mathbb{R}$

Let $f$ be continous on $[a,b]$ and differentiable on $(a,b)$

Let $f(a)=f(b)$

Then $\exists$ $c\in (a,b)$ such that $f'(c)=0$

Proof

The result is trivial for the case $f([a,b])=\{f(a)\}$. Hence, let us assume that $f$ is a non-constant function.

Let $M=\sup\{f([a,b])\}$ and $m=\inf\{f([a,b])\}$ Without loss of generality, we can assume that $M\neq f(a)$

By the Maximum-minimum theorem, $\exists c\in (a,b)$ such that $f(c)=M$

Assume if possible $f'(c)>0$

Let $\epsilon=\frac{f'(c)}{2}$

Hence, $\exists \delta>0$ such that $x\in V_{\delta}(c)\implies |\frac{f(x)-f(c)}{x-c}-f'(c)|<\epsilon$

i.e. $\forall x\in V_{\delta}(c)$, $\frac{f(x)-f(c)}{x-c}>0$

Thus we have that $f(x)>f(c)$ if $x\in (c,c+\delta)$, contradicting the assumption that $f(c)$ is a maximum.

Similarly we can show that $f'(c)<0$ leads to contradiction.

Therefore, $f'(c)=0$

QED

See Also

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