Let f (x) = 2x / (5x + 1) (x ∈ R, and X is not equal to - 1 / 5) Find: 1 inverse function F-1 (x); The range of 2F-1 (1 / 5) and F-1 (x)

Let f (x) = 2x / (5x + 1) (x ∈ R, and X is not equal to - 1 / 5) Find: 1 inverse function F-1 (x); The range of 2F-1 (1 / 5) and F-1 (x)

Let 2x / (5x + 1) = y, then y denotes X
2X = (5x + 1) y, so x = Y / (2-5y)
So the inverse function is to exchange X and y, that is, F-1 (x) = x / (2-5x)
So 2F-1 (1 / 5) = 2 * [(1 / 5) / (2-5 * 1 / 5)] = 2 / 5
The range of the inverse function is the domain of the original function, that is: (x ∈ R, and X is not equal to - 1 / 5)
(ha ha!