We have f : R → R and f(x) = cos (x + 2).

Recall that a function is invertible only when 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)

⇒ cos (x₁ + 2) = cos (x₂ + 2)

As the cosine function repeats itself with a period 2π, we have

x₁ + 2 = x₂ + 2 or x₁ + 2 = 2π + (x₂ + 2)

∴ x₁ = x₂ or x₁ = 2π + x₂

So, we have f(x₁) = f(x₂) ⇒ x₁ = x₂ or 2π + x₂

This means that two different elements of the domain are mapped to the same element by the function f.

For example, consider f(0) and f(2π).

We have f(0) = cos (0 + 2) = cos 2 and

f(2π) = cos (2π + 2) = cos 2 = f(0)

Thus, f is not one-one.

Hence, f is not invertible and f^-1 does not exist.