Answer
$f+g,f-g,fg$: for all Domain=$\{ x:x\geq3\}$.
$\frac{f}{g}$: Domain=$\{ x:x\gt3\}$.
Work Step by Step
$f(x)=\sqrt {x+6}$ and $g(x)=\sqrt {x-3}$.
$f(x)+g(x)=\sqrt {x+6} + \sqrt {x-3}$. To find the domain of the sum function, $x+6\geq0$, $x\geq-6$ and $x-3\geq0$, $x\geq3$. Therefore,
Domain=$\{ x:x\geq3\}$.
$f(x)-g(x)=\sqrt {x+6} - \sqrt {x-3}$. To find the domain of the difference function, $x+6\geq0$, $x\geq-6$ and $x-3\geq0$, $x\geq3$. Therefore,
Domain=$\{ x:x\geq3\}$.
$f(x) \times g(x)=\sqrt {x+6} \times \sqrt {x-3}=\sqrt {(x+6)(x-3)}$. To find the domain
of the product function, $x+6\geq0$, $x\geq-6$ and $x-3\geq0$, $x\geq3$. Therefore,
Domain=$\{ x:x\geq3\}$.
$\frac{f(x)}{g(x)}=\frac{\sqrt {x+6}}{\sqrt {x-3}}=\sqrt {\frac{x+6}{x-3}}$.To find the domain of the quotient function, $x+6\geq0$, $x\geq-6$ and $x-3\gt0$, $x\gt3$ .Therefore,
Domain=$\{ x:x\gt3\}$.
