We have f : Q → Q and f(x) = 3x + 5.

Recall that a function is invertible only when it is both one-one and onto.

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

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

⇒ 3x₁ + 5 = 3x₂ + 5

⇒ 3x₁ = 3x₂

∴ 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 Q (co-domain) such that f(x) = y

⇒ 3x + 5 = y

⇒ 3x = y – 5

Clearly, for every y ϵ Q, there exists x ϵ Q (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