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Let $f(x)=|x|x$ then $f''(0)$ does not exist.

Why?

If $x>0$, $f'(x)=2x$ and if $x<0$, $f'(x)=-2x$.

Then when $x=0$, does $f'(x)$ also not exist?

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    You are on the right track but you got confused at the end. f'(x) exists and is continuous at all points. It is: 2x for x>0 and -2x for x<0. Now think about the function f'(x) (this really is a function for all x and is different to f(x) and f''(x) ). As x approaches 0 from below, the gradient of f('x) is -1. As f'(x) approached 0 from above, the gradient of f'(x) is 1 which does not equal -1. Hence f'(x) is not differentiable at x=0.2012-11-27

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