Question

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

Answer 1

Given function : $f\left(x\right)=|x{|}^3$

$|x|=\{\begin{array}{cc}x& x\ge 0\\ -x& x<0\end{array}$

Therefore,

$f\left(x\right)=|x{|}^3=\{\begin{array}{cc}{x}^3& x\ge 0\\ -x^3& x<0\end{array}$

When $x>0$

$f\left(x\right)=x^3$
Differentiating w.r.t. $x$ , we get

$f^{\prime }\left(x\right)=3x^2$

Again , differentiating w.r.t. $x$ , we get
$f^{\prime \prime }\left(x\right)=(3x^2{)}^{\prime }$
$f^{\prime \prime }\left(x\right)=6x$

Hence, $f^{\prime \prime }\left(x\right)$ exists for $x>0$

When $x<0$

$f\left(x\right)=-x^3$
Differentiating w.r.t. $x$ , we get

$f^{\prime }\left(x\right)=-3x^2$

Again , differentiating w.r.t. $x$ , we get
$f^{\prime \prime }\left(x\right)=(-3x^2{)}^{\prime }$
$f^{\prime \prime }\left(x\right)=-6x$

Hence, $f^{\prime \prime }\left(x\right)$ exists for $x<0$
At $x=0,$

To check if $f^{\prime \prime }\left(x\right)$ exists for $x=0$

We need to check differentiability of $f^{\prime \prime }\left(x\right)$ at $x=0$

Here,
$f\left(x\right)=\{\begin{array}{cc}{x}^3& ,x\ge 0\\ -x^3& ,x<0\end{array}$

$f^{\prime }\left(x\right)=\{\begin{array}{cc}3x^2& ,x\ge 0\\ -3x^2& ,x<0\end{array}$

We know that $f^{\prime }\left(x\right)$ is differentiate at $x=0$ if its
$L.H.D=R.H.D$.

$\Rightarrow L.H.D=\underset{h\to 0}{lim}\frac{f^{\prime }(0-h)-f^{\prime }\left(0\right)}{-h}$

$\Rightarrow L.H.D=\underset{h\to 0}{lim}\frac{f^{\prime }(-h)-f^{\prime }\left(0\right)}{-h}$

$\Rightarrow L.H.D=\underset{h\to 0}{lim}\frac{-3(-h{)}^2-3(0{)}^2}{-h}$

​​​​​​​$\Rightarrow L.H.D=\underset{h\to 0}{lim}\frac{-3h^2}{-h}$

​​​​​​​$\Rightarrow L.H.D=\underset{h\to 0}{lim}3h$

Substituting $h=0$

​​​​​​​$\Rightarrow L.H.D=3\left(0\right)=0$

$\Rightarrow R.H.D=\underset{h\to 0}{lim}\frac{f^{\prime }(0+h)-f^{\prime }\left(0\right)}{h}$

$\Rightarrow R.H.D=\underset{h\to 0}{lim}\frac{f^{\prime }\left(h\right)-f^{\prime }\left(0\right)}{h}$

$\Rightarrow R.H.D=\underset{h\to 0}{lim}\frac{3(h{)}^2-3(0{)}^2}{h}$

$\Rightarrow R.H.D=\underset{h\to 0}{lim}\frac{3h^2}{h}$

$\Rightarrow R.H.D=\underset{h\to 0}{lim}3h$

Substituting $h=0$
$\Rightarrow R.H.D=3\left(0\right)=0$

Thus, $L.H.D=R.H.D$ of $f^{\prime }\left(x\right)$ at $x=0$

Therefore, $f^{\prime }\left(x\right)$ is differentiable at $x=0$

So, $f^{\prime \prime }\left(x\right)$ exists for $x=0$.

Therefore, $f^{\prime \prime }\left(x\right)$ exists for all real values of $x$.

Hence, proved.