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.
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