Let f : R – → R be a function defined as. Show that f : R – → range(f) is one-one and onto. Hence, find f^-1.

We have f : R – → R and

We need to prove f : R – → range(f) is invertible.

First, we will prove that f is one-one.

Let x₁, x₂ϵ A (domain) such that f(x₁) = f(x₂)

⇒ (4x₁)(3x₂ + 4) = (3x₁ + 4)(4x₂)

⇒ 12x₁x₂ + 16x₁ = 12x₁x₂ + 16x₂

⇒ 16x₁ = 16x₂

∴ x₁ = x₂

So, we have f(x₁) = f(x₂) ⇒ x₁ = x₂.

Thus, function f is one-one.

Now, we will prove that f is onto.

Let y ϵ range(f) (co-domain) such that f(x) = y

⇒ 4x = 3xy + 4y

⇒ 4x – 3xy = 4y

⇒ x(4 – 3y) = 4y

Clearly, for every y ϵ range(f), there exists x ϵ A (domain) such that f(x) = y and hence, function f is onto.

Thus, the function f has an inverse.

We have f(x) = y ⇒ x = f^-1(y)

But, we found f(x) = y ⇒

Hence,

Thus, f(x) is invertible and