Range of $f(x) = \frac{1+e^x}{1-e^x}$ 
Range of $f(x) = \frac{1+e^x}{1-e^x}$

To understand more about the range, I calculated the domain of $f(x)$ which is all real numbers except $x=0$
From the domain, I understand that there is no values of $f(x)$ when $x=0$
I had a hint from someone else that $0< e^x <1$ or $1<e^x$
I am not sure what this hint suppose to mean in finding the range.
 A: We have $$f(x)=\frac{1+e^x}{1-e^x}$$ with domain $$\mathbb{D}_{f(x)}=(-\infty, 0)\cup(0,\infty)$$
If you take the derivative to find extrema, $$f'(x)=\frac{2e^x}{\left(1-e^x\right)^2}=0$$
this has no solutions. There are not extrema so $f$ should smoothly transition from an upper bound to lower bound along the endpoints of both parts of the domain.
Taking limits at the origin, we see that
$$\lim_{x\to0^+}f(x)=-\infty$$
$$\lim_{x\to0^-}f(x)=\infty$$
Now, take the limits out to infinity, we see that
$$\lim_{x\to\infty}f(x)=-1$$
$$\lim_{x\to-\infty}f(x)=1$$
So the range is just
$$\mathbb{R}_{f(x)}=(-\infty, -1)\cup(1, \infty)$$
A: I know this is not the correct thing to do; but I don't have the reputation to comment so I'll post it as an answer.
You can take the inverse of the function and the domain of the inverse will be the range of the original function.
$$f(x)=\frac{1+e^x}{1-e^x}$$
$$f^{-1}(x)=\frac{1+e^y}{1-e^y}$$
$$x=\frac{1+e^y}{1-e^y}$$
$$x(1-e^y)={1+e^y}$$
$$x-xe^y={1+e^y}$$
$$-xe^y-e^y={1-x}$$
$$e^y(-x-1)={1-x}$$
$$e^y=\frac{1-x}{-x-1}$$
$$y=\ln\left(\frac{1-x}{-x-1}\right)$$
Hope this helps.
