Answer :

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 x_{1}, x_{2}ϵ R (domain) such that f(x_{1}) = f(x_{2})

⇒ cos (x_{1} + 2) = cos (x_{2} + 2)

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

x_{1} + 2 = x_{2} + 2 or x_{1} + 2 = 2π + (x_{2} + 2)

∴ x_{1} = x_{2} or x_{1} = 2π + x_{2}

So, we have f(x_{1}) = f(x_{2}) ⇒ x_{1} = x_{2} or 2π + x_{2}

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.

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