If f: R → (–1, 1) defined by is invertible, find f-1.
We have f: R → (–1, 1) and
Given that f-1 exists.
Let y ϵ (–1, 1) such that f(x) = y
⇒ 102x – 1 = y (102x + 1)
⇒ 102x – 1 = 102xy + y
⇒ 102x – 102xy = 1 + y
⇒ 102x (1 – y) = 1 + y
Taking log10 on both sides, we get
We have f(x) = y ⇒ x = f-1(y)
But, we found f(x) = y ⇒
Hence,
Thus,