Q:

Obtain value of variable through inverse function

I am trying to implement a controller for a strict feedback system and I am reading a paper regarding the procedure. My problem has to do only with some math involved so this is why I am posting here. Suppose I have a function defined as:
$$ S_1(ε_1) = \frac{e^{2ε_1}}{e^{2ε_1}+1} $$
My goal is to compute the value of $ε_1$ in order to be able to proceed with the design procedure. I have an equation which holds for $S_1(ε_1)$: 
$$ x_1(t) = p_1(t)S_1(ε_1)+a_0(t) $$
where $p_1(t)$ is a known performance function, $x_1(t)$ is the state of the system which is known and $a_0(t)$ is some reference trajectory also known. By working out the math of this equation, we come up with:
$$ S_1(ε_1) = \frac{x_1(t)-a_0(t)}{p_1(t)} \Rightarrow ε_1(t) = S_1^{-1}(\frac{x_1(t)-a_0(t)}{p_1(t)}) $$
So now, do I have to find the inverse function of 
$$ S_1(ε_1) = \frac{e^{2ε_1}}{e^{2ε_1}+1} $$
and replace $ε_1$ at the inverse function with the expression $ \frac{x_1(t)-a_0(t)}{p_1(t)} $ or do I misunderstand something ?

A:

Let $S(x):= \frac{e^{2x}}{e^{2x}+1}$. Then we have $0<S(x)<1$ for all $x \in \mathbb R.$
Try to show that $S : \mathbb R \to (0,1)$ is bijective. To determine $S^{-1}$, let $y \in (0,1)$ and consider the equation $S(x)=y.$
Elementary computations give
$$x= \frac{1}{2} \ln (\frac{y}{1-y}).$$
Thus
$$S^{-1}(y)=\frac{1}{2} \ln (\frac{y}{1-y}).$$

