Question:
If $f(x)=|x|^{3}$, show that $f^{\prime \prime}(x)$ exists for all real $x$, and find it.
Solution:
It is known that, $|x|= \begin{cases}x, & \text { if } x \geq 0 \\ -x, & \text { if } x<0\end{cases}$
Therefore, when $x \geq 0, f(x)=|x|^{3}=x^{3}$
In this case, $f^{\prime}(x)=3 x^{2}$ and hence, $f^{\prime \prime}(x)=6 x$
When $x<0, f(x)=|x|^{3}=(-x)^{3}=-x^{3}$
In this case, $f^{\prime}(x)=-3 x^{2}$ and hence, $f^{\prime \prime}(x)=-6 x$
Thus, for $f(x)=|x|^{3}, f^{\prime \prime}(x)$ exists for all real $x$ and is given by,
$f^{\prime \prime}(x)= \begin{cases}6 x, & \text { if } x \geq 0 \\ -6 x, & \text { if } x<0\end{cases}$