Question

Find the set of points where $f\left(x\right)=x^2\left|x\right|$ is thrice differentiable.

• 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, $f\left(x\right)=\{\begin{array}{cc}-x^3& x<0\\ 0& x=0\\ x^3& x>0\end{array}{f}^{\prime }\left(x\right)=\{\begin{array}{cc}-{3x}^2& x<0\\ 0& x=0\\ 3x^2& x>0\end{array}{f}^{\prime \prime }\left(x\right)=\{\begin{array}{cc}-6x& x<0\\ 0& x=0\\ 6x& x>0\end{array}{f}^{\prime \prime \prime }\left(x\right)=\{\begin{array}{cc}-6& x<0\\ 0& x=0\\ 6& x>0\end{array}$
Since, $f^{\prime \prime \prime }(0-)\ne f^{\prime \prime \prime }\left(0\right)\ne f^{\prime \prime \prime }(0+)$
Therefore, $f^{\prime \prime \prime }\left(x\right)$ is not differential at $x=0$

Hence the required set of points is given by $R-\left\{0\right\}.$
Ans: A