Question

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

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 $(-2,2)$,since at all the points where the defination changes, the left hand limit is same as the right hand limit which is equal to the value of function at that point.

Now,

$g^{\prime }\left(x\right)=\{\begin{array}{cc}2x,& -2\le x<0\\ & \\ 0,& 0\le x<1\\ & \\ 4x,& 1\le x\le 2\end{array}$
At $x=0,$
$g^{\prime }\left(0^{-}\right)=0$ and $g^{\prime }\left(0^{+}\right)=0$
$\Rightarrow g$ is differentiable at $x=0$
At $x=1,$
$g^{\prime }\left(1^{-}\right)=0$ and $g^{\prime }\left(1^{+}\right)=4$
$\Rightarrow g$ is not differentiable at $x=1$
$\therefore g\left(x\right)$ is continuous but not differnetiable at one point.