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Does there exist a continuous function $f:\mathbb{R}\to\mathbb{R}$ so that $f$ is differentiable exactly at one point?

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1 Answer 1

Yes. Idea: choose a function which is continuous but nowhere differentiable. Then multiply it with $x^2$, say.

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Following noone's idea, you may find a concrete example such as $$f(x) = \begin{cases} x^2 & x \in \Bbb{Q} \\ -x^2 & x \in \Bbb{R}\setminus\Bbb{Q} \end{cases}.$$ –  sos440 Aug 26 '12 at 8:58
@sos440 why would this be continuous? –  noone Aug 26 '12 at 8:58
I missed the condition that $f$ must be continuous. Here is a correct example: $f(x) = x^2 \mathrm{blanc}(x)$, where $\mathrm{blanc}(x)$ is the blancmange function. –  sos440 Aug 26 '12 at 9:14

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