Question

The function given by $y=||x|-1|$ is differentiable for all real numbers except the points.

• A. $\{0,1,-1\}$
• B. $\pm 1$
• C. 1
• D. $-1$

Answer 1

Answer: (A)

$\{0,1,-1\}$

$y=|\left|x\right|-1|$

$y^{\prime }=\frac{|\left|x\right|-1|}{\left|x\right|-1}(\frac{\left|x\right|}{x}-0)=\frac{\left|x\right||\left|x\right|-1|}{x\left[|\left|x\right|-1|\right]}$

$x\left[|\left|x\right|-1|\right]=0$

$x=0,\left|x\right|=1$

$x=0,x=\pm 1$

$\therefore$ At $x=0,1,-1$ $y^{\prime }$ is not defined

$\Rightarrow y^{\prime }$ is not differentiable at these points