# continuous function with precisely one point differentiability

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

• See also here. – Andrés E. Caicedo Dec 25 '13 at 18:13
• @Andres: the function you refer to has many discontinuity points. – Martin Argerami Dec 25 '13 at 18:32
• (@MartinArgerami Which is why this was a comment and not an answer.) – Andrés E. Caicedo Dec 25 '13 at 18:43

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.
• $(g(x) - g(0))/x = g(x)/x = f(x) \to f(0)$. I guess the first power works. The square makes the derivative zero. – ncmathsadist Dec 25 '13 at 20:24
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 $$t=0$$.