Answer
The domain of $f(x)$ is $(-\infty,0)\cup(0,\infty)$.
The range of $f(x)$ is $(-\infty,0)\cup(3,\infty)$.
Work Step by Step
$$f(x)=\frac{3}{1-e^{2x}}$$
1) Domain:
$f(x)$ is defined on $(-\infty,\infty)$ except where $(1-e^{2x})=0$
$$1-e^{2x}=0$$ $$e^{2x}=1$$ $$2x=0$$ $$x=0$$
Therefore, $f(x)$ is defined on $(-\infty,\infty)$ except where $x=0$. In other words, the domain of $f(x)$ is $(-\infty,0)\cup(0,\infty)$
2) Range:
For all $x$ in the domain stated above:
$$e^{2x}\gt0$$ $$-e^{2x}\lt0$$ $$1-e^{2x}\lt1$$
- As $(1-e^{2x})\in(0,1)$, then $$\frac{1}{1-e^{2x}}\gt1$$ $$\frac{3}{1-e^{-2x}}\gt3$$
- On the other hand, as $(1-e^{2x})\in(-\infty,0)$, then
$$\frac{1}{1-e^{2x}}\lt0$$ $$\frac{3}{1-e^{-2x}}\lt0$$
Therefore, the range of $f(x)$ is $(-\infty,0)\cup(3,\infty)$.
