It $f(x)=x+\sin x$, then can we find $f^{-1} (x)$?

Question

We have a bijective function $f(x)=x+\sin x$. So what is $f^{-1} (x)$?

Let $f^{-1}(x)$ be $g(x)$. Suppose we have to find $g\left(\dfrac{\pi}{6}+\dfrac{1}{2}\right)$ and $g'\left(\dfrac{\pi}{6}+\dfrac{1}{2}\right)$.

How to do it?

Answer 1

$x+\sin x$ is strictly increasing, since its derivative, $1+\cos x$, is strictly positive, save on isolated points of the form $x=k\pi$, when $\cos x=-1$. So it follows that the function is bijective on R. So since $x=\dfrac\pi6$ is one obvious possible solution for $x+\sin x=\dfrac\pi6+\dfrac12$ , it follows that it is the only solution. Then, as far as the derivative of its inverse is concerned, we know that $\bigg(f^{-1}\bigg)'(y)=$ $=\dfrac1{f'(x)}$ , where $y=f(x)$.