Answer
$$
f(x)=x|x|=\left\{\begin{array}{ll}{x^{2},} & {x \geq 0} \\ {-x^{2},} & {x<0}\end{array}\right.
$$
we can find
$$
\begin{aligned}
f^{\prime}(x)& =\left\{\begin{array}{ll}{2 x,} & {x \geq 0} \\ {-2 x,} & {x<0}\end{array}\right.\\
&=2 |x|.
\end{aligned}
$$
and
$$
f^{\prime \prime}(x)=\left\{\begin{array}{l}{2, \quad x \geq 0} \\ {-2, x<0}\end{array}\right.
$$
Thus we can say that $f^{\prime \prime}(x) $ does not exist because the left and right derivatives do not agree at 0.
Work Step by Step
$$
f(x)=x|x|=\left\{\begin{array}{ll}{x^{2},} & {x \geq 0} \\ {-x^{2},} & {x<0}\end{array}\right.
$$
we can find
$$
\begin{aligned}
f^{\prime}(x)& =\left\{\begin{array}{ll}{2 x,} & {x \geq 0} \\ {-2 x,} & {x<0}\end{array}\right.\\
&=2 |x|.
\end{aligned}
$$
and
$$
f^{\prime \prime}(x)=\left\{\begin{array}{l}{2, \quad x \geq 0} \\ {-2, x<0}\end{array}\right.
$$
Thus we can say that $f^{\prime \prime}(x) $ does not exist because the left and right derivatives do not agree at 0.
