# Rolle's theorem and open interval

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• Jul 11th 2012, 01:55 PM
loui1410
Rolle's theorem and open interval
Prove: if $\displaystyle f(x)$ is differentiable on the open interval $\displaystyle (a,b)$ and $\displaystyle \lim_{x\to a+}f(x) = \lim_{x\to b-}f(x)$ then there exists $\displaystyle c$ in the interval such that $\displaystyle f'(c)=0$.

It's almost clear that I need to use Rolle's theorem, but how do I overcome the fact that the given $\displaystyle (a,b)$ is an open interval?
• Jul 11th 2012, 03:02 PM
Reckoner
Re: Rolle's theorem and open interval
Quote:

Originally Posted by loui1410
Prove: if $\displaystyle f(x)$ is differentiable on the open interval $\displaystyle (a,b)$ and $\displaystyle \lim_{x\to a+}f(x) = \lim_{x\to b-}f(x)$ then there exists $\displaystyle c$ in the interval such that $\displaystyle f'(c)=0$.

Let

$\displaystyle L = \lim_{x\to a^+}f(x) = \lim_{x\to b^-}f(x).$

Define a new function $\displaystyle g(x)$ as follows:

$\displaystyle g(x) = \left\{\begin{array}{ll}L,&x = a\\f(x),&a < x < b\\L,&x = b\end{array}\right.$

It should be easy to show that $\displaystyle g$ is continuous on $\displaystyle [a, b]$ and differentiable on $\displaystyle (a, b),$ and since $\displaystyle g(a) = g(b),$ the requirements for Rolle's theorem are satisfied. And since $\displaystyle f'(x) = g'(x)$ for $\displaystyle a<x<b,$ $\displaystyle g'(c) = 0\Rightarrow f'(c) = 0.$