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

$a.\quad f(x)=x$ $\mathrm{Domain:}\quad (-\infty,\infty)$ $\mathrm{Range:}\quad (-\infty,\infty)$ $b.\quad g(x)=\sqrt {x-1}$ $\mathrm{Domain:}\quad [1,\infty)$ $\mathrm{Range:}\quad [0,\infty)$ $c.\quad f(x)+g(x)=x+\sqrt {x-1}$ $\mathrm{Domain:}\quad [1,\infty)$ $\mathrm{Range:}\quad [1,\infty)$ $d.\quad f(x)\cdot g(x)=x\sqrt {x-1}$ $\mathrm{Domain:}\quad [1,\infty)$ $\mathrm{Range:}\quad [0,\infty)$
$a.\quad f(x)=x$ The function is defined for all real numbers. $b.\quad g(x)=\sqrt{x-1}$ We take square root only of positive numbers. $x-1\ge0$ $x\ge1$ Range will be above zero and will reach infinity. $c.\quad f(x)+g(x)=x+\sqrt{x-1}$ The square root part of the function will control the domain and range. The domain will be the same as in part $\ b.$ But if we input $\ x=1\$ (minimum value in the domain), we will get output $\ 1.\$ Thus the range would be $\ [1,\infty).$ $d.\quad f(x)\cdot g(x)=x\sqrt{x-1}$ The domain and range of the combined function will depend on the square root part of the function. It will yield the same results as in part $b.$