It is given that f : [–1, 1] → R, given by

Now, Let f(x) = f(y)

⇒ xy + 2x = xy +2y

⇒ 2x = 2y

⇒ x = y

⇒ f is a one- one function.

Now, Let y = , xy = x + 2y so x =

So, for every y in the range there exists x in the domain such that f(x) = y

⇒ f is onto function.

⇒ f: [-1,1] → Range f is one-one and onto

⇒ the inverse of the function : f: [-1, 1] → Range f exists.

Let g: Range f → [-1, 1]be the inverse of range f.

Let y be an arbitrary element of range f.

Since, f :[-1, 1] → Range f is onto, we get:

y = f(x) for same x ϵ [-1, 1]

⇒ xy +2y = x

⇒ x(1 - y) = 2y

⇒ x = , y ≠ 1

Now, Let us define g: Range f → [-1, 1]

g(y) = , y ≠ 1

Now, (gof)(x) = g(f(x)) =

(fog)(y) = f(g(y)) =

Thus, gof = I_[-1,1] and fog = I_{Range f}

⇒ f^-1 = g

Therefore, f^-1(y) = , y ≠ 1