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

$a.\quad f(x)=\sqrt {x+1}$ $\mathrm{Domain:}\quad [-1,\infty)$ $\mathrm{Range:}\quad [0,\infty)$ $b.\quad g(x)=\sqrt {x-1}$ $\mathrm{Domain:}\quad [1,\infty)$ $\mathrm{Range:}\quad [0,\infty)$ $c.\quad f(x)+g(x)=\sqrt{x+1}+\sqrt{x-1}$ $\mathrm{Domain:}\quad [1,\infty)$ $\mathrm{Range:}\quad [\sqrt{2},\infty)$ $d.\quad f(x)\cdot g(x)=\sqrt {x^2-1}$ $\mathrm{Domain:}\quad (-\infty,-1]\cup [1,\infty)$ $\mathrm{Range:}\quad [0,\infty)$
$a.\quad f(x)=\sqrt{x+1}$ We can take square root only of positve numbers, so $\ x+1\ge0\ \rightarrow x\ge-1$. Range will simply be greater than or equal to zero. $b.\quad g(x)=\sqrt{x-1}$ We can take square root only of positve numbers, so $\ x-1\ge0\ \rightarrow x\ge 1$. Range will simply be greater than or equal to zero. $c.\quad f(x)+g(x)=\sqrt{x+1}+\sqrt{x-1}$ The domain of the combined function will be obtained by the intersection of domain of each individual function. $[-1,\infty)\$ seems to include the $\ [1,\infty)\$ in itself. So the domain is latter one. Range would be $\ [\sqrt{2},\infty)\$ as we will get $\ \sqrt{2}+\sqrt{0}=\sqrt{2}\$ while puting $\ x=1\$ from the domain. $d.\quad f(x)\cdot g(x)=\sqrt{x+1}\cdot\sqrt{x-1}=\sqrt{x^2-1}$ We can take square root only of positive numbers. So, $x^2-1\ge 0$ Zeros of the inequality are $\ \pm 1\$, so the domain is $\ (-\infty,-1]\cup [1,\infty).$ Range is simply $\ [0,\infty).$