Consider a function $g: (-1, +1) \rightarrow \mathbb{R}$ Consider a function $g: (-1, +1) \rightarrow \mathbb{R}$ given by
$g(x) = \frac{x}{1-|x|}$
Show that $g$ is 1-1 and find $g((-1,1))$
Find $g^{-1}$
Are $g$ and $g^-1$ continuous

I found that $g$ is 1-1 by contradiction. I assumed that $g(x_0)=g(x_1)$ where $x_0 \not= x_1$. I showed through trichotomy (moving the zero around) that it is impossible for the first condition to hold if the latter is true. 
To find $g((-1,1))$ do I take limits?  If not, any hints?
For part II, How does one take the inverse of an absolute value? I'd imagine that it is not continuous.
Obviously, without an answer to part II, I cannot have a justified answer for the final part.
 A: first observe that $y = g(x) = \dfrac{x}{1-|x|}$ is odd. we only need to worry about $x \ge 0.$  in this domain $y = \dfrac{x}{1-x} = -1+\dfrac{1}{1-x}.$ that is $g$ is increasing on $[0, 1)$ and has a vertical asymptote at $x = 1.$ 
you can find the inverse by solving for $y = -1 + \dfrac{1}{1-x}$ which gives $x = 1 - \dfrac{1}{1+y} = \dfrac{y}{1+y}$
therefore $g^{-1}(x) =\dfrac{x}{1+x} \text{ for }x\ge 0.$ you can extend it as an odd function to get $$g^{-1}(x) = \dfrac{x}{1+|x|}.  $$
A: $\;g\;$ is clearly continuous on $\;(-1\,,\,1)\;$ as a quotient of continuous functions here without the denominator's function vanishing in the domain.
For the inverse function:
$$y=\frac x{1-|x|}\implies y-y|x|-x=0=\begin{cases}y-(y+1)x=0\implies x=\frac y{1+y}&,\;0\le x\\{}\\y+(y-1)x=0\implies x=\frac y{1-y}&,\;x<0\end{cases}$$
So we get that
$$g^{-1}(x)=\begin{cases}\frac x{1+x}&,\;\;0\le x\\{}\\\frac x{1-x}&,\;\;x<0\end{cases}$$
So...is the inverse continuous?
A: To find the inverse function divide the given function in two parts:
$$
y=\begin{cases}
\dfrac{x}{1-x} \quad  0\le x<1\\
\dfrac{x}{1+x} \quad  -1< x < 0
\end{cases}
$$
then invert each part being careful to the domain of $y$.
