Answer
$a.\quad f(x)=1$
$\mathrm{Domain:}\quad (-\infty,\infty)$
$\mathrm{Range:}\quad \{1\}$
$b.\quad g(x)=1+\sqrt{x}$
$\mathrm{Domain:}\quad [0,\infty)$
$\mathrm{Range:}\quad [1,\infty)$
$c.\quad \frac{f}{g}=\frac{1}{1+\sqrt{x}}$
$\mathrm{Domain:}\quad [0,\infty)$
$\mathrm{Range:}\quad (0,1]$
$d.\frac{g}{f}=\frac{1+\sqrt{x}}{1}$
$\mathrm{Domain:}\quad [0,\infty)$
$\mathrm{Range:}\quad [1,\infty)$
Work Step by Step
$a.\quad f(x)=1$
There is no restriction for $\ x\ $ , so the domain consists of real numbers.
For any value of $\ x\ $, the result will always be $\ 1.\ $ So the range would be $\ {1}.$
$b.\quad g(x)=1+\sqrt{x}$
$x\ge0\ $ will give us the domain $\ [0,\infty).$
For the minimum value of $\ x\ $ the function will yield $\ 1.\ $ So the range is $\ [1,\infty).$
$c.\quad \frac{f}{g}=\frac{1}{1+\sqrt{x}}$
The domain would be $\ [0,\infty)\ $ since, $\ x\ge0\ $ must be true.
Since the value of the denominator is always greater than $\ 1\ $, the value of fraction would be between $\ 0\ $ and $\ 1\ $.
$d.\quad \frac{g}{f}=\frac{1+\sqrt{x}}{1}$
The domain and range would be the same as in part $\ b.$
