Answer
$a.\quad f(x)=\sqrt {x+1}$
$\mathrm{Domain:}\quad [-1,\infty)$
$\mathrm{Range:}\quad [0,\infty)$
$b.\quad g(x)=\sqrt {x-1}$
$\mathrm{Domain:}\quad [1,\infty)$
$\mathrm{Range:}\quad [0,\infty)$
$c.\quad f(x)+g(x)=\sqrt{x+1}+\sqrt{x-1}$
$\mathrm{Domain:}\quad [1,\infty)$
$\mathrm{Range:}\quad [\sqrt{2},\infty)$
$d.\quad f(x)\cdot g(x)=\sqrt {x^2-1}$
$\mathrm{Domain:}\quad (-\infty,-1]\cup [1,\infty)$
$\mathrm{Range:}\quad [0,\infty)$
Work Step by Step
$a.\quad f(x)=\sqrt{x+1}$
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 $\ [0,\infty).$
