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Difference between revisions of "Rolle's Theorem"

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'''Rolle's theorem'' is an important theorem among the class of results regarding the value of the derivative on an interval.
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'''Rolle's theorem''' is an important theorem among the class of results regarding the value of the derivative on an interval.
  
 
==Statement==
 
==Statement==
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Then <math>\exists</math> <math>c\in (a,b)</math> such that <math>f'(c)=0</math>
 
Then <math>\exists</math> <math>c\in (a,b)</math> such that <math>f'(c)=0</math>
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 +
==Proof==
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The result is trivial for the case <math>f([a,b])=\{f(a)\}</math>. Hence, let us assume that <math>f</math> is a non-constant function.
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Let <math>M=\sup\{f([a,b])\}</math> and <math>m=\inf\{f([a,b])\}</math>
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Without loss of generality, we can assume that <math>M\neq f(a)</math>
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By the [[Maximum-minimum theorem]], <math>\exists c\in (a,b)</math> such that <math>f(c)=M</math>
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Assume if possible <math>f'(c)>0</math>
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Let <math>\epsilon=\frac{f'(c)}{2}</math>
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Hence, <math>\exists \delta>0</math> such that <math>x\in V_{\delta}(c)\implies |\frac{f(x)-f(c)}{x-c}-f'(c)|<\epsilon</math>
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i.e. <math>\forall x\in V_{\delta}(c)</math>, <math>\frac{f(x)-f(c)}{x-c}>0</math>
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Thus we have that <math>f(x)>f(c)</math> if <math>x\in (c,c+\delta)</math>, contradicting the assumption that <math>f(c)</math> is a maximum.
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Similarly we can show that <math>f'(c)<0</math> leads to contradiction.
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Therefore, <math>f'(c)=0</math>
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<P ALIGN=RIGHT>QED</P>
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==See Also==
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<UL>
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<LI>[[Lagrange's mean value theorem]]</LI>
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<LI>[[Calculus]]</LI>
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</UL>
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[[Category:Calculus]]
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[[Category:Theorems]]

Latest revision as of 11:13, 30 May 2019

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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