Question

Let $f\left(x\right)=\{\begin{array}{cc}-1,& -2\le x<0\\ x^2-1,& 0<x\le 2\end{array}$ and $g\left(x\right)=|f\left(x\right)|+f|x|$ then the number of points which $g\left(x\right)$ is non differentiable, is

• A. at most one point
• B. $2$
• C. exactly one point
• D. infinite

Answer 1

The correct option is C exactly one point

Mistake : ans is C !

Given $f\left(x\right)$ and also $g\left(x\right)=\mid f\left(x\right)\mid +f\mid x\mid$ from this ;

$g\left(x\right)=x^2,-2\le x<0$

$=0,0<x\le 1$

$=2x^2-2,1\le x\le 2$

graphically differentiability means existence of non vertical tangent from the graph of $g\left(x\right)$ we can find the existence of the vertical tangent at only $x=0$. So we can say that only non differentiable at $x=0$.