We have f : A → B where A = R – {3} and B = R – {1}

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

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

⇒ (x₁ – 2)(x₂ – 3) = (x₁ – 3)(x₂ – 2)

⇒ x₁x₂ – 3x₁ – 2x₂ + 6 = x₁x₂ – 2x₁ – 3x₂ + 6

⇒ –3x₁ – 2x₂ = –2x₁ – 3x₂

⇒ –3x₁ + 2x₁ = 2x₂ – 3x₂

⇒ –x₁ = –x₂

∴ 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 B (co-domain) such that f(x) = y

Clearly, for every y ϵ B, 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