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no of points of non diffrentibility for f(x) = max{ ||x|-1| , ½ } is 2 5 3

no of points of non diffrentibility for f(x) = max{ ||x|-1| , ½ } is
2
5
3
 

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 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
askIITians Faculty

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