Given f(x) = x + sinx
⇒ f'(x) = 1 + cos x ≥ 0 for all x in R.
⇒ f is strictly increasing function
⇒ f is one-one function.
Also, the range of a function is R.
⇒ f is a onto function.
Thus, f is a bijective function.
Hence, f–1 is exists.
Therefore, f–1(x) = x – sinx
