Inverse of $f(x) = a \left(1 + \frac{c}{(1+x^b)^{-\frac{1}{b}} - c}\right) \cdot (1+x^{-b})^\frac{1}{b}$?

Question

How can one find the inverse of $$ f(x) = \mathrm{a} \left(1 + \frac{\mathrm{c}}{(1+x^\mathrm{b})^{-\frac{1}{\mathrm{b}}} - \mathrm{c}}\right) \cdot (1+x^{-\mathrm{b}})^\frac{1}{\mathrm{b}} $$ with $a, b, c$ positive constants?

I have not any experience with inverting such functions, and I cannot find any textbooks that contain such inversion techniques. Any relevant suggestions would be appreciated very much.

The reason I want this inversion is to find $\frac{\mathrm{d}x}{\mathrm{d}a}$ for this equation: $$ a \left(1 + \frac{\mathrm{c}}{(1+x^\mathrm{b})^{-\frac{1}{\mathrm{b}}} - \mathrm{c}}\right) \cdot (1+x^{-\mathrm{b}})^\frac{1}{\mathrm{b}} = 1$$

Answer 1

Looking at the source paper (Michaillat and Saez, "Aggregate Demand, Idle Time, and Unemployment" (2014)) that you linked, I think the function that you'd want to invert is actually $$ g(x) = \left(1 + \frac{c}{(1+x^b)^{-1/b} - c}\right)^{\epsilon - 1} (1 + x^{-b})^{-1/b} $$ (where I guess you set $\epsilon = 2$ for simplicity) so that $g(x) = a$.

Anyways, I suspect that neither $f$ nor $g$ are invertible using elementary operations because you have $x$ trapped inside multiple levels of not-so-well-related expressions. In Mathematica, the related Solve and InverseFunction functions quickly give up, while Reduce had to be manually stopped because it wasn't returning anything within a reasonable amount of time.

But notice that the authors of the paper didn't actually bother to find a concrete expression for the inverse (redactions and emphasis mine):

Since [$g$ is a strictly increasing function of $x$], equation $(10)$ implicitly defines $x$ as an increasing function of [$a$]. (p. 18)

The idea here is simply that a strictly increasing function always has a strictly increasing inverse (with the domain and the codomain restricted appropriately). We may not have a way to compute the inverse directly, but it exists.

To calculate $\frac{dx}{da}$ at some $a$, find/estimate the $x$ such that $g(x) = a$ and apply the well-known formula $$ [g^{-1}]'(a) = \frac{1}{g'(x)} $$ for the derivative of inverse functions.