I have functions $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ and $g$ that look like this: \begin{equation} \begin{aligned} f(x) &= Ax + b + g(x) \\ g(x_i) &= -c \log(1/x_i - 1), \quad x_i\in(0,1) \end{aligned} \end{equation}
where $A$ is square, $x_i$ are the components of $x\in\mathbb{R}^n$ and $g(x)$ is intended to indicate the vectorization of $g(x_i)$ (abuse of notation).
For $A$ full rank, I want to know if the equation $f(x) = 0$ has a unique solution $x^\star\in(0,1)^n$ (note that $g$ is undefined for $x_i\in(-\infty,0] \cup [1,\infty)$, so this specification is not particularly constraining). Namely, if there is a unique $x^\star$ which satisfies $x^\star = A^{-1}(-g(x^\star) - b)$.
Also, note that $\dfrac{\partial g}{\partial x_i} \geq 4c > 0$.
1) Does this even seem to be true or should I expect multiple roots?
2) How to prove it?
My approach so far has been via global versions of the Implicit Function Theorem, e.g. let $\tilde{f}(x,y) = Ax + b + y = 0$, where $y = g(x)$. That doesn't seem to be particularly useful here because it's approaching the problem from the opposite direction -- I've already defined the global implicit function for $y$ in terms of $x$. What I need to know is if the function $f$ has a (unique?) root.