Does there exist a continuous function $f:\mathbb{R}\to\mathbb{R}$ so that $f$ is differentiable exactly at one point?
A question about differentiability
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calculus
1 Answers
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Yes. Idea: choose a function which is continuous but nowhere differentiable. Then multiply it with $x^2$, say.
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0Following noone's idea, you may find a concrete example such as $$f(x) = \begin{cases} x^2 & x \in \Bbb{Q} \\ -x^2 & x \in \Bbb{R}\setminus\Bbb{Q} \end{cases}.$$ – 2012-08-26
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0@sos440 why would this be continuous? – 2012-08-26
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0I 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](http://en.wikipedia.org/wiki/Blancmange_curve)*. – 2012-08-26