Question

If $f\left(x\right)={\left|x\right|}^3$. show that $f^{\prime \prime }\left(x\right)$ exists for all real $x$ and find it.

Answer 1

Given, $f\left(x\right)={|x|}^3$
Removing mod, we get
$f\left(x\right)=\{\begin{array}{c}{x}^{3}ifx\ge 0\\ \\ {-x}^{3}ifx<0\end{array}$
For case: when $x>0$
$f^{{}^{\prime }}\left(x\right)=3x^2$
$\Rightarrow f^{{}^{\prime \prime }}\left(x\right)=6x$
Now lets see for $x<0$
$f^{{}^{\prime }}\left(x\right)=-3x^2$
$\Rightarrow f^{{}^{\prime \prime }}\left(x\right)=-6x$
This shows that $f^{{}^{\prime \prime }}\left(x\right)$ exist and is given by
$f^{\prime \prime }\left(x\right)=\{\begin{array}{c}6xifx\ge 0\\ \\ -6xifx<0\end{array}$