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