# Checking discontinuity

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?

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