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If $f$ is a function defined on $[0,1]$ for which $ {f}'\left(\frac{1}{2}\right)$ exists but ${f}'\left(\frac{1}{2}\right) \notin [0,1]$, then $f$ is discontinuous at $x=\frac{1}{2} $.

Is this statement true and why?

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How can $f$ be discontinuous at a place where it's derivative exists? –  Hagen von Eitzen Jan 3 '13 at 15:13
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No. Take $f(x)=2x$. And actually, differentiable implies continuous. –  1015 Jan 3 '13 at 15:13
    
@julien: You could post that as an answer. –  Clive Newstead Jan 3 '13 at 15:20
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1 Answer

No, the function is not discontinuous at x= ${f}'\left(\frac{1}{2}\right)$

Since the derivative of the function exists at x= ${f}'\left(\frac{1}{2}\right)$, therefore, from the definition of continuity, the function is continuous at that point.

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I think it's good to mention that while differentiability implies continuity, continuity does not imply differentiability. –  Joe Jan 3 '13 at 17:10
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