Question

The set of points where $f\left(x\right)=x^2|x|$ is thrice differentiable, is:

• A. $R-\left\{0\right\}$
• B. $R-\left\{1\right\}$
• C. $R$
• D. $R-\{-1\}$

Answer 1

The correct option is A $R-\left\{0\right\}$

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

$\Rightarrow f^{\prime }\left(x\right)=\{\begin{array}{cc}-3x^2,& x<0\\ 3x^2,& x>0\end{array}$
$\because f^{\prime }\left(0^{-}\right)=f^{\prime }\left(0^{+}\right)$
So, $f\left(x\right)$ is differentiable function.

$f^{\prime \prime }\left(x\right)=\{\begin{array}{cc}-6x,& x<0\\ 6x,& x>0\end{array}$
$\because f^{\prime \prime }\left(0^{-}\right)=f^{\prime \prime }\left(0^{+}\right)$
So, $f\left(x\right)$ is twice differentiable function.

$f^{\prime \prime \prime }\left(x\right)=\{\begin{array}{cc}-6,& x<0\\ 6,& x>0\end{array}$
$\because f^{\prime \prime \prime }\left(0^{-}\right)\ne f^{\prime \prime \prime }\left(0^{+}\right)$
$\Rightarrow f\left(x\right)$ is not thrice differentiable at $x=0.$

Hence, the required set of points where $f$ is thrice differential is $R-\left\{0\right\}.$