Answer
The domain of $f(x)$ is $(-\infty,\infty)$
The range of $f(x)$ is $(0,\frac{1}{2})$.
Work Step by Step
$$f(x)=\frac{1}{2+e^x}$$
1) Domain:
$f(x)$ is defined on $(-\infty,\infty)$ except where $2+e^x=0$
However, since $e^x\gt0$ for all $x\in R$, $2+e^x\gt0$ for all $x\in R$.
Therefore, the domain of $f(x)$ is $(-\infty,\infty)$
2) Range:
As stated above, $e^x\gt0$ for all $x\in(-\infty,\infty)$
So $2+e^{x}\gt2$ for all $x\in(-\infty,\infty)$
That means $\frac{1}{2+e^x}\lt\frac{1}{2}$ for $x\in(-\infty,\infty)$
Also, since both $2+e^x\gt0$ and $1\gt0$, we have $\frac{1}{2+e^x}\gt0$
Therefore, overall, the range of $f(x)$ is $(0,\frac{1}{2})$.
