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Math Help - Rolle's theorem and open interval

  1. #1
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    Rolle's theorem and open interval

    Prove: if f(x) is differentiable on the open interval (a,b) and \lim_{x\to a+}f(x) = \lim_{x\to b-}f(x) then there exists c in the interval such that 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 (a,b) is an open interval?
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    Re: Rolle's theorem and open interval

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

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

    Define a new function g(x) as follows:

    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 g is continuous on [a, b] and differentiable on (a, b), and since g(a) = g(b), the requirements for Rolle's theorem are satisfied. And since f'(x) = g'(x) for a<x<b, g'(c) = 0\Rightarrow f'(c) = 0.
    Thanks from loui1410
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