We have f : R → R and f(x) = x³ + 4.

Recall that a function is a bijection only if it is both one-one and onto.

First, we will check if f is one-one.

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

⇒ x₁³ + 4 = x₂³ + 4

⇒ x₁³ = x₂³

⇒ (x₁ – x₂)(x₁² + x₁x₂ + x₂²) = 0

As x₁, x₂ϵ R and the second factor has no real roots,

x₁ – x₂ = 0

∴ x₁ = x₂

So, we have f(x₁) = f(x₂) ⇒ x₁ = x₂.

Thus, function f is one-one.

Now, we will check if f is onto.

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

⇒ x³ + 4 = y

⇒ x³ = y – 4

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 is a bijection and 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

Hence, we have

Thus, f^-1(3) = –1.