Precalculus (6th Edition) Blitzer

Step 1. $\sqrt {x+6}+\sqrt {x-3}$, $[3,\infty)$. Step 2. $\sqrt {x+6}-\sqrt {x-3}$, $[3,\infty)$. Step 3. $\sqrt {x^2+3x-18}$, $[3,\infty)$. Step 4. $\sqrt {\frac{x+6}{x-3}}$, $(3,\infty)$
Step 1. Given $f(x)=\sqrt {x+6}$ with $x\geq-6$ and $g(x)=\sqrt {x-3}$ with $x\geq3$, we have $f+g=\sqrt {x+6}+\sqrt {x-3}$ with a domain of $x\geq3$ or $[3,\infty)$. Step 2. We have $f-g=\sqrt {x+6}-\sqrt {x-3}$ with a domain of $x\geq3$ or $[3,\infty)$. Step 3. We have $f\cdot g=\sqrt {x+6}\sqrt {x-3}=\sqrt {x^2+3x-18}$ with a domain of $x\geq3$ or $[3,\infty)$. Step 4. We have $\frac{f}{g}=\frac{\sqrt {x+6}}{\sqrt {x-3}}=\sqrt {\frac{x+6}{x-3}}$ with a domain of $x\gt3$ or $(3,\infty)$. (remove $x=3$ from above)