I haven't seen this done before, but I'm sure it's possible.
Suppose $f$ is analytic on $[x_1,x_2]$ with range $[y_1,y_2],$ then its Taylor series converges on this interval and it is branch-invertible. Let $f_2$ be the second-order Taylor approximation of the $n$th-order Taylor approximation $f_n,$ such that $f_2(x) = ax^2 + bx + c.$
The branch-inverses of $ax^2 + bx + c$ are the two functions $\frac{-b \pm \sqrt{b^2 + 4ax - 4ac}}{2a}.$
Now, the inverse function of $f,$ $f^{-1},$ is also bounded on the domain $[y_1,y_2].$
My more general question is: as $n \rightarrow \infty,$ does $f_n^{-1}$ converge to $f^{-1}$? And thus can I claim that $\frac{-b \pm \sqrt{b^2 + 4ax - 4ac}}{2a}$ is a reasonable local approximation (heuristically speaking) of $f^{-1},$ such that for some upper bound M where $|f^{-1}(x)| < M,$ $|f_2^{-1}|<M$ over some subset of $[y_1,y_2]$?
If you feel inclined to adjust this approach to "suppose $f$ is monotone..." I am open to that as well.