We have f : R → R and f(x) = x³ – 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)

⇒ x₁³ – 3 = x₂³ – 3

⇒ x₁³ = x₂³

⇒ (x₁ – x₂)(x₁² + x₁x₂ + x₂²) = 0

⇒ x₁ – x₂ = 0 (as x₁, x₂ϵ R⁺)

∴ 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

⇒ x³ – 3 = y

⇒ x³ = 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

Hence, we have

Thus, f^-1(24) = 3 and f^-1(5) = 2.