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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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    [Here's a thread](http://math.stackexchange.com/q/108388) on continuous functions differentiable only at one point.2012-09-11
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    Note that André's example below is a function that is only continuous at one point and is only differentiable at one point.2012-09-11
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    If you like such questions, you might want to check out the books by [Bruckner, Bruckner and Thompson](http://classicalrealanalysis.info/index.php) and pick the one that suits your level (the texts are available electronically for free).2012-09-11
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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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    Thank you! this gives me a perfect answer that I can be convinced!2012-09-11
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    @Emily: Regarding to $f(x)$ above, if you want a function differentiable at 2 points you can take $(x-1)^2f(x)$. It is differentiable at x=0 and x=1 only.2012-09-11
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    @Emily Another possible generalization: if you take $f(x) = x^np(x)$, with $p(x)$ as above, then $f$ is $(n-1)$ times differentiable at $0$, and not differentiable anywhere else. To make sense of that, you have to write a single limit corresponding to the $n$th derivative at $0$ (since you can't compute limits using $f'$, because $f'$ only exists at $0$).2012-09-11
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    Ah, but can you write a continuous function which is only differentiable in one point?2012-09-11
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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