You are given function f(x) = x + sin(x).

They want you find g(1), where g(x) is the inverse function of f(x).


First of all, in order you understand better everything from the very beginning, 
f(x) = x + sin(x) is a MONOTONIC function on the entire domain, which is the set of all real numbers.

The range of f(x) is the entire set of real numbers. 

It is monotonic function, since its derivative f'(x) = 1 + cos(x) is non-negative everywhere
excluding discrete points x = , k = 0, +/-1, +/-2, . . . 


g(1) is the real number x such that f(x) = x + sin(x) = 1.     (1)

Since f(x) is monotonic, such number x (the root of the equation (1), 

     (a)  exists, and 

     (b)  is UNIQUE.


To find this root x, the algebraic methods do not work. We should use numerical methods
and special computer solvers.


One such solver is Wolfram Mathworld / Transcendent solver
https://mathworld.wolfram.com/TranscendentalEquation.html


It is free of charge.


Using it, we find the root of equation (1).  It is  x = 0.510973.

So, g(1) = 0.510973  (approximate value).


Now we should find g'(1).


Use the identity for inverse function  g'(b) = 1/f'(x).  It gives

g'(1) = 1/f'(0.510973) =  =  = 0.534111  (approximate value).


ANSWER.  g(1) = 0.510973  (approximate value).  g'(1) = 0.534111  (approximate value).