I must invert the following function, to obtain $x$ as a function of $y$:
$y(x) = f(x)g(x)$
where:
$f(x) = (1 - e^{-ax})$
$g(x) = (1 - e^{-bx})(1 - e^{-b(\alpha - x)})$
$a<b$, $a \in \mathbb {R}_{+}$ $b \in \mathbb {R}_{+}$
However, I have found this expression not to be invertible, or, at least, I do not know how to invert it with what I know.
Is there a method I could use to approximately or analytically find the inverse of this function?
Attempt at a solution:
Excluding the first term, it is trivial to invert $g(x) = (1 - e^{-bx})(1 - e^{-b(\alpha - x)})$, since it may be rewritten purely in terms of constants and a $\cosh$ function:
$f(x) = 1+e^{-b\alpha}-2 e^{-\alpha b /2} \cosh \left(b(\alpha / 2-x)\right)$.
However, I am now stuck since I do not know the correct way to proceed.
I have considered Taylor expanding the $f(x)$ to first order, $f(x) \simeq 1 - (1 + ax) = -ax$, but I do not think this helps me at all.
It is not invertible near $x=0$ if $a,b,\alpha \ne 0$, as $$ f(x) g(x) \sim a b (1 - e^{-b\alpha}) x^2 + O(x^3)$$ Note also that $f(x) g(x) = 0$ at both $x=0$ and $x=\alpha$.