We have f : R → R and f(x) = 4x + 3.

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 R (domain) such that f(x₁) = f(x₂)

⇒ 4x₁ + 3 = 4x₂ + 3

⇒ 4x₁ = 4x₂

∴ 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 R (co-domain) such that f(x) = y

⇒ 4x + 3 = y

⇒ 4x = y – 3

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