Question

Let $f\left(x\right)$ be defined in the interval $[-2,2]$ such that $f\left(x\right)=\{\begin{array}{c}-1,-2\le x\le 0\\ x-1,0<x\le 2\end{array}$ and $g\left(x\right)=f\left(\left|x\right|\right)+\left|f\right(x\left)\right|$ Test the differentiablity of $g\left(x\right)$ in $(-2,2)$.

• A. not derivable at $x=0$ and $x=1$
• B. derivable at all points
• C. not derivable at $x=0$
• D. not derivable at $x=1$

Answer 1

The correct option is A not derivable at $x=0$ and $x=1$

$f\left(x\right)=\{\begin{array}{c}-1,-2\le x\le 0\\ x-1,0<x\le 2\end{array}$
$f\left(|x|\right)=\{\begin{array}{c}-x-1,-2\le x\le 0\\ x-1,0<x\le 2\end{array}$
$|f\left(x\right)|=\{\begin{array}{c}1,-2\le x\le 0\\ 1-x0<x\le 1\\ x-1,1<x\le 2\end{array}$
$\Rightarrow g\left(x\right)=\{\begin{array}{c}-x,-2\le x\le 0\\ 00<x\le 1\\ 2(x-1),1<x\le 2\end{array}$
Now, $g^{\prime }\left(x\right)=\{\begin{array}{c}-1,-2\le x\le 0\\ 00<x\le 1\\ 2,1<x\le 2\end{array}$
Hence, $g\left(x\right)$ isn't differentiable at $x=0$ and $x=1.$