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Does there exists a real valued function which is everywhere continuous and differentiable at exactly one point on the real line?

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  • $\begingroup$ See also here. $\endgroup$ – Andrés E. Caicedo Dec 25 '13 at 18:13
  • $\begingroup$ @Andres: the function you refer to has many discontinuity points. $\endgroup$ – Martin Argerami Dec 25 '13 at 18:32
  • $\begingroup$ (@MartinArgerami Which is why this was a comment and not an answer.) $\endgroup$ – Andrés E. Caicedo Dec 25 '13 at 18:43
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You know that you can produce a function $f$ so that $f$ is continuous and differentiable nowhere. Pick $a\in\mathbb{R}$; put $g(x) = (x - a)^2 f(x).$ The function $g$ is differentiable exactly at $a$ and nowhere else.

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  • $\begingroup$ Why does it have to be squared? $\endgroup$ – Sebastian Garrido Dec 25 '13 at 19:46
  • $\begingroup$ Try using the definition of the derivative with and without the square. You will see. $\endgroup$ – ncmathsadist Dec 25 '13 at 19:57
  • $\begingroup$ I still cant see the difference :/ excuse my stupidity :( $\endgroup$ – Sebastian Garrido Dec 25 '13 at 20:10
  • $\begingroup$ $(g(x) - g(0))/x = g(x)/x = f(x) \to f(0) $. I guess the first power works. The square makes the derivative zero. $\endgroup$ – ncmathsadist Dec 25 '13 at 20:24
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Let $f(t) $ be the Weierstrass function, continuous everywhere and nowhere differentiable. Define $$ g(t)=t^2f(t). $$ Then $g$ is continuous everywhere and differentiable only at $t=0$.

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