Q. The function given by y = ||x| - 1| is differentiable for all real numbers except the poi
IIT JEEIIT JEE 2005Continuity and Differentiability
Solution:
Given function is y = | | x | - 1|
or $ y = \begin{cases}
- | x | + 1 & \quad \text{if } | x | < 1 \\
| x | - 1 & \quad \text{if } | x | \ge 1
\end{cases} $
$ = \begin{cases}
- | x | + 1 & \quad \text{if } - 1 < x < 1 \\
| x | - 1 & \quad \text{if } x \le - 1 \ or \ x \ge 1
\end{cases} $
$ = \begin{cases}
- | x | - 1 & \quad \text{if } x \le - 1 \\
| x | + 1 & \quad \text{if } - 1 < x < 0 \\
- x + 1 \quad \text{if } 0 \le x < 1 \\
x -1 \quad \text{if } x \ge 1
\end{cases} $
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.
