# Is there a function $f: \mathbb R \to \mathbb R$ that has only one point differentiable?

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! :-)

• Here's a thread on continuous functions differentiable only at one point. – t.b. Sep 11 '12 at 15:36
• Note that André's example below is a function that is only continuous at one point and is only differentiable at one point. – copper.hat Sep 11 '12 at 15:41
• If you like such questions, you might want to check out the books by Bruckner, Bruckner and Thompson and pick the one that suits your level (the texts are available electronically for free). – t.b. Sep 11 '12 at 15:42
• 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. – Emily Sep 11 '12 at 16:06

Let $$p(x)= \begin{cases} 0,& x\in\mathbb Q\\\\1,& x\in \mathbb R-\mathbb Q \end{cases}$$ Now take $f(x)=x^2p(x)$.
• @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. – mrs Sep 11 '12 at 17:28
• @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 $0$). – student Sep 11 '12 at 17:31
• @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 $x\ne 0$. – André Nicolas Sep 11 '12 at 18:05