Question

If $f\left(x\right)=||x{|}^2-2|x|-3|$, then $f\left(x\right)$ is non differentiable at $x$ equal to

• A. $-1,0,1$
• B. $1,2$
• C. $-3,-1,0,1,3$
• D. $-3,0,3$

Answer 1

The correct option is D $-3,0,3$

$f\left(x\right)=|{\left|x\right|}^2-2\left|x\right|-3|$

$f^{\prime }\left(x\right)=\frac{|{\left|x\right|}^2-2\left|x\right|-3|}{{\left|x\right|}^2-2\left|x\right|-3}\times (2\left|x\right|\frac{\left|x\right|}{x}-2\frac{\left|x\right|}{x})$

$=\frac{|{\left|x\right|}^2-2\left|x\right|-3|}{({\left|x\right|}^2-2\left|x\right|-3)}\times \frac{(2{\left|x\right|}^2-2\left|x\right|)}{x}$

For $f\left(x\right)$ to be non differentiable

$x({\left|x\right|}^2-2\left|x\right|-3)=0$

$\Rightarrow x=0,{\left|x\right|}^2-2\left|x\right|-3=0$

From graph

${\left|x\right|}^2-2\left|x\right|-3=0$at $x=3,-3$

$\therefore$ at $x=0,3,-3$ f is not differentiable