We have f: R → (0, 2) and

Given that f^-1 exists.

Let y ϵ (0, 2) such that f(x) = y

⇒ 2e^2x = y (e^2x + 1)

⇒ 2e^2x = e^2xy + y

⇒ 2e^2x – e^2xy = y

⇒ e^2x (2 – y) = y

Taking ln on both sides, we get

We have f(x) = y ⇒ x = f^-1(y)

But, we found f(x) = y ⇒

Hence,

Thus,