Let $f:\Bbb R \to \Bbb R$ s.t. $f(x) = 2x + \sin x -1$ then Find $f^{-1'}(-1)$
solution
First, given function $f$ is differentiable at given domain of $x$ since $2x$, $\sin x$ and $-1$ is are all differentiable and also linear combination of these also differentiable.
In addition, $f'(x) = 2 + \cos x$ is always bigger than $1$ which is not $0$. Thus,
By inverse function theorem, there would exist only one $x$ which makes $f(x) = -1$ which is $0$ since this function is strictly increasing and this characteristic guarantees unique mapping from domain to range $-1$.
Thus, $f^{-1 '}(-1)=1 / f'(0) = 1/3$
