## University Calculus: Early Transcendentals (3rd Edition)

$a.\quad f(x)=1$ $\mathrm{Domain:}\quad (-\infty,\infty)$ $\mathrm{Range:}\quad \{1\}$ $b.\quad g(x)=1+\sqrt{x}$ $\mathrm{Domain:}\quad [0,\infty)$ $\mathrm{Range:}\quad [1,\infty)$ $c.\quad \frac{f}{g}=\frac{1}{1+\sqrt{x}}$ $\mathrm{Domain:}\quad [0,\infty)$ $\mathrm{Range:}\quad (0,1]$ $d.\frac{g}{f}=\frac{1+\sqrt{x}}{1}$ $\mathrm{Domain:}\quad [0,\infty)$ $\mathrm{Range:}\quad [1,\infty)$
$a.\quad f(x)=1$ 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 $\ b.$