Answer
$a.\quad f(x)=2$
$\mathrm{Domain:}\quad (-\infty,\infty)$
$\mathrm{Range:}\quad \{2\}$
$b.\quad g(x)=x^2+1$
$\mathrm{Domain:}\quad (-\infty,\infty)$
$\mathrm{Range:}\quad [1,\infty)$
$c.\quad \frac{f}{g}=\frac{2}{x^2+1}$
$\mathrm{Domain:}\quad (-\infty,\infty)$
$\mathrm{Range:}\quad (0,2]$
$d.\frac{g}{f}=\frac{x^2+1}{2}$
$\mathrm{Domain:}\quad (-\infty,\infty)$
$\mathrm{Range:}\quad [\frac{1}{2},\infty)$
Work Step by Step
$a.\quad f(x)=2$
There is no restriction for $\ x\ $ , so the domain consists of real numbers.
For any value of $\ x\ $, the result will always be $\ 2.\ $ So the range would be $\ {2}.$
$b.\quad g(x)=x^2+1$
The domain of polynomial function is real numbers.
Since, $\ x^2\ $ will always yield a positive result, $\ x\ge0\ $ shows that the range will be $\ [1,\infty).$
$c.\quad \frac{f}{g}=\frac{2}{x^2+1}$
The denominator of the rational function must not be equal to zero. Since the denominator of this combined function is always positive, there would be no restriction for $\ x.$
The maximum value this function can get is when $\ x=0\ $ which will give us $\ \frac{f}{g}=2.\ $ And when we increase or decrease the value of $\ x\ $, the value of fraction would become smaller.
$d.\quad \frac{g}{f}=\frac{x^2+1}{2}$
The domain would be the same as for the function in part $\ b.\ $
For the range, the output of the function in part $\ b\ $, is being divided by $\ 2\ $ so the range would be $\ [\frac{1}{2},\infty).$
