Answer
$f+g=\sqrt {x+7}+\sqrt {x-2}$ with domain $[2, \infty)$
$f-g=\sqrt {x+7}-\sqrt {x-2}$ with domain $[2, \infty)$
$fg= \sqrt {x^2+5x-14}$ with domain $[2, \infty)$
$\frac{f}{g}=\frac{\sqrt {x+7}}{\sqrt {x-2}}$ with domain $(2, \infty)$
Work Step by Step
Step 1. Given $f(x)=\sqrt {x+7}, x\geq-7$ and $g(x)=\sqrt {x-2}, x\geq2$, we have $f+g=\sqrt {x+7}+\sqrt {x-2}$ with domain $[2, \infty)$
Step 2. We have $f-g=\sqrt {x+7}-\sqrt {x-2}$ with domain $[2, \infty)$
Step 3. We have $fg=\sqrt {x+7}\sqrt {x-2}=\sqrt {x^2+5x-14}$ with domain $[2, \infty)$
Step 4. We have $\frac{f}{g}=\frac{\sqrt {x+7}}{\sqrt {x-2}}$ with domain $(2, \infty)$
