Question

Let $f\left(x\right)=\{\begin{array}{cc}-1;& -2\le x<0\\ & \\ x^2-1;& 1\le x\le 2\end{array}$

and $g\left(x\right)=|f\left(x\right)|+f\left(|x|\right)$. Then, in the interval $(-2,2)$, $g$ is :

• A. not continuous
• B. not differentiable at one point
• C. not differentiable at two points
• D. differentiable at all points

Answer 1

The correct option is B not differentiable at one point

$|f\left(x\right)|=\{\begin{array}{cc}1;& -2\le x<0\\ & \\ 1-x^2;& 0\le x<1\\ & \\ x^2-1;& 1\le x\le 2\end{array}$

$f\left(|x|\right)=x^2-1$ for $x\in [-2,2]$

$\therefore g\left(x\right)=\{\begin{array}{cc}1+x^2-1;& -2\le x<0\\ & \\ 1-x^2+x^2-1;& 0\le x<1\\ & \\ x^2-1+x^2-1;& 1\le x\le 2\end{array}$

$\Rightarrow g\left(x\right)=\{\begin{array}{cc}{x}^2;& -2\le x<0\\ & \\ 0;& 0\le x<1\\ & \\ 2(x^2-1);& 1\le x\le 2\end{array}$

Clearly, $g$ is continuous in $(-2,2)$

$g^{\prime }\left(x\right)=\{\begin{array}{cc}2x;& -2\le x<0\\ & \\ 0;& 0\le x<1\\ & \\ 4x;& 1\le x\le 2\end{array}$

$g^{\prime }\left(0^{-}\right)=0$ and $g^{\prime }\left(0^{+}\right)=0$
$\Rightarrow g$ is differentiable at $x=0$

$g^{\prime }\left(1^{-}\right)=0$ and $g^{\prime }\left(1^{+}\right)=4$
$\Rightarrow g$ is not differentiable at $x=1$