Question

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

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

Answer 1

Answer: (A)

Given function is y = | | x | - 1|
or $y=\{\begin{array}{ll}-|x|+1& \text{if }|x|<1\\ |x|-1& \text{if }|x|\ge 1\end{array}$
$=\{\begin{array}{ll}-|x|+1& \text{if }-1<x<1\\ |x|-1& \text{if }x\le -1orx\ge 1\end{array}$
$=\{\begin{array}{ll}-|x|-1& \text{if }x\le -1\\ |x|+1& \text{if }-1<x<0\\ -x+1\text{if }0\le x<1& \\ x-1\text{if }x\ge 1& \end{array}$
Here Ly' (- 1) = - 1 and Ry' (- 1) = 1
Ly' (0) = 1 and Ry' (0) = - 1 and
Ly' (1) = - 1 and Ry' (1) = 1
$\Rightarrow$y is not differentiable at x = - 1, 0, 1.
ALTERNATE SOLUTION
Graph of y = | | x | - 1| is as follows :
The graph has sharp turnings at x = - 1, 0 and 1; and hence not differentiable at x = - 1, 0, 1.