Question

Let S be the set of points where the function, ƒ(x) = |2-|x-3||, x \( \in \) R is not differentiable. Then \( \sum\limits_{x \in S} {f(f(x))} \) is equal to_____.

Hint:

f is non-differentiable where an inner or outer modulus argument becomes zero (corners): find these points to form S, then evaluate f(f(x)) at each.

Answer 1

Correct Answer: 3

\(f(x)=|2-|x-3||\). Corners occur where \(x-3=0\) (x = 3) and where \(2-|x-3|=0\), i.e. \(|x-3|=2\), x = 1 or 5. So \(S=\{1,3,5\}\).
f(1) = |2-2| = 0, f(3) = |2-0| = 2, f(5) = |2-2| = 0.
f(f(1)) = f(0) = |2-3| = 1; f(f(3)) = f(2) = |2-1| = 1; f(f(5)) = f(0) = 1.
Sum = 1 + 1 + 1 = 3.
Answer: 3