# no of points of non diffrentibility for f(x) = max{ ||x|-1| , ½ } is253

Jitender Singh IIT Delhi
8 years ago
Ans: 5
$f(x) = max(||x|-1|, \frac{1}{2}})$
$f(x) = -x-1, x< -1$
$= \frac{1}{2}, -1\leq x< \frac{-1}{2}$
$= x+1, \frac{-1}{2}\leq x< 0$
$= 1-x, 0\leq x< \frac{1}{2}$
$= \frac{1}{2}, \frac{1}{2}\leq x< 1$
$= x-1, x\geq 1$
So, there are five points where slope will change. You can easily plot the graph of f(x).
Thanks & Regards
Jitender Singh
IIT Delhi