Question

The set of points where $x^2|x|$ is differentiable thrice is

• A. $R$
• B. $R-0,\pm 1$
• C. $R-0$
• D. None of the above

Answer 1

Answer: (C)

$R-0$

Let $f\left(x\right)=x^2|x|$ which can be expressed as $f\left(x\right)=\{\begin{array}{cc}-x^3,& x<0\\ 0,& x=0\\ x^3,& x\ge 0\end{array}\therefore f^{\prime }\left(x\right)=\{\begin{array}{cc}-3x^2,& x<0\\ 0,& x=0\\ 3x^2,& x\ge 0\end{array}$
So, $f^{\prime }\left(x\right)$ exists for all real $x$.
$f"\left(x\right)=\{\begin{array}{cc}-6x,& x<0\\ 0,& x=0\\ 6x,& x\ge 0\end{array}$
So, f"(x) exists for all real $x$. $f{"}^{\prime }\left(x\right)=\{\begin{array}{cc}-6,& x\le 0\\ 0,& x=0\\ 6,& x\ge 0\end{array}$However, f'''(0) does not exists since $f^{\prime \prime \prime }\left(0^{-}\right)=-6$ and $f^{\prime \prime \prime }\left(0^{+}\right)=6$ which are not equal. Thus, the set of points where $f\left(x\right)$ is thrice differentiable is $R-0$
Hence, option 'C' is correct.