solution:
Given capacity: (x) = x/(x+2) Let x, y ∈ [–1, 1]
Let f(x) = f(y)
x/(x+2) = y/(y+2) xy + 2x = xy + 2y x = y
f is one-one. Once more,
Since f : [–1, 1] → Range f is onto say, y = x/(x+2)
yx + 2y = x x(1 – y) = 2y or x = 2y/(1-y)
x = f – 1 (y) = 2y/(1-y); y not equivalent to 1
f is onto capacity, and f – 1(x) = 2x/(1-x).