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While studying complex variables, I could learn that $f(z)=|z|^{2}$ has only one point which is $z=0$ that $f$ being differentiable and $f$ being not differentiable at any other points.

Then, I was wondering if there is a function $f: \mathbb R \to \mathbb R$ that has only one point differentiable and not on any other points.

In intuition, it seems there are no such point! However, I have no idea how I can prove this...

Additional question is that would there be any function $f: \mathbb R \to \mathbb R$ that has only one point continuous and not on any other points.

I think this is pretty interesting things to think about! :-)

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    Thank you for all your comments! Now I know that the above property of complex functions is not a special thing that is different from real functions.2012-09-11

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Let $p(x)= \begin{cases} 0,& x\in\mathbb Q\\\\1,& x\in \mathbb R-\mathbb Q \end{cases}$ Now take $f(x)=x^2p(x)$.

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    @AsafKaragila: Yes, but need to quote a non-trivial result, that there is a bounded continuous nowhere differentiable function. Then the $x^2$ trick fixes things at $0$, and doesn't help at $x\ne 0$.2012-09-11