Answer
Step 1. $\sqrt {x-5}+\sqrt {5-x}$, $\{5\}$.
Step 2. $\sqrt {x-5}-\sqrt {5-x}$, $\{5\}$.
Step 3. $\sqrt {x-5}\sqrt {5-x}$, $\{5\}$.
Step 4. $\frac{\sqrt {x-5}}{\sqrt {5-x}}$, $\emptyset$.
Work Step by Step
Step 1. Given $f(x)=\sqrt {x-5}$ with $x\geq5$ and $g(x)=\sqrt {5-x}$ with $x\leq5$, we have $f+g=\sqrt {x-5}+\sqrt {5-x}$ with a domain of $x=5$ or $\{5\}$, thus $f+g=0$.
Step 2. we have $f-g=\sqrt {x-5}-\sqrt {5-x}$ with a domain of $x=5$ or $\{5\}$, thus $f-g=0$.
Step 3. We have $f\cdot g=\sqrt {x-5}\sqrt {5-x}$ with a domain of $x=5$ or $\{5\}$, thus $f\cdot g=0$.
Step 4. For $\frac{f}{g}=\frac{\sqrt {x-5}}{\sqrt {5-x}}$, the domain is an empty set or $\emptyset$, thus $\frac{f}{g}$ has no possible value.
