Let $f(x)=x|x|$. The set of points where $f(x)$ is twice differentiable is

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$f(x)=x|x|=\left\{\begin{array}{l}x^{2}, x \geq 0 \ -x^{2}, x<0\end{array} ight.$\ $\Rightarrow f^{\prime}(x)=\left\lvert\, \begin{array}{cc}2 x, & x>0 \ 0, & x=0 \ -2 x, & x<0\end{array} ight.$\ Now $f^{\prime \prime}(x)=2$ when $x>0 \& f^{\prime \prime}(x)=-2$ when $x<0$\ $ herefore f(x)$ is not twice differentiable at $x=0$. At all other points $(x eq 0), f(x)$ is twice differentiable.