## Precalculus (6th Edition) Blitzer

Step 1. $\sqrt {x-2}+\sqrt {2-x}$, $\{2\}$. Step 2. $\sqrt {x-2}-\sqrt {2-x}$, $\{2\}$. Step 3. $\sqrt {x-2}\sqrt {2-x}$, $\{2\}$. Step 4. $\frac{\sqrt {x-2}}{\sqrt {2-x}}$, $\emptyset$.
Step 1. Given $f(x)=\sqrt {x-2}$ with $x\geq2$ and $g(x)=\sqrt {2-x}$ with $x\leq2$, we have $f+g=\sqrt {x-2}+\sqrt {2-x}$ with a domain of $x=2$ or $\{2\}$, thus $f+g=0$. Step 2. we have $f-g=\sqrt {x-2}-\sqrt {2-x}$ with a domain of $x=2$ or $\{2\}$, thus $f-g=0$. Step 3. We have $f\cdot g=\sqrt {x-2}\sqrt {2-x}$ with a domain of $x=2$ or $\{2\}$, thus $f\cdot g=0$. Step 4. For $\frac{f}{g}=\frac{\sqrt {x-2}}{\sqrt {2-x}}$, the domain is an empty set or $\emptyset$, thus $\frac{f}{g}$ has no possible value.