Answer
a) $\sqrt{3-x}+\sqrt{x^2-1}$; $x\le-1$ and $1\le x\le 3$
b) $\sqrt{3-x}-\sqrt{x^2-1}$; $x\le-1$ and $1\le x\le 3$
c) $\sqrt{(x-3)(1+x)(1-x)}$; $x\le-1$ and $1\le x\le3$
d) $\frac{\sqrt{3-x}}{\sqrt{(x+1)(x-1)}}$; $x\lt-1$ and $1\lt x\le3$
Work Step by Step
a) Add f(x) to g(x). $\sqrt{3-x}+\sqrt{x^2-1}$. There is no way to simplify. To find the domain, we just need to find the values that make the square roots positive. Because $x=3$, $x=-1$, and $x=1$ make at least one of the roots zero, these are our critical points.
We check $x\lt -1$, $-1\lt x\lt 1$, $1\lt x \lt 3$, and $x\gt 3$. I tested $x=-2$, $x=0$, $x=2$, and $x=4$ to find that the only ranges that work are $x\lt-1$ and $1\lt x\lt 3$.
Then we check $x=3$, $x=-1$, and $x=1$. They all work, so our domain is $x\le-1$ and $1\le x\le 3$.
b) Subtract g(x) from f(x). $\sqrt{3-x}-\sqrt{x^2-1}$.
We check $x\lt -1$, $-1\lt x\lt 1$, $1\lt x \lt 3$, and $x\gt 3$. The only ranges that work are $x\lt-1$ and $1\lt x\lt 3$.
Then we check $x=3$, $x=-1$, and $x=1$. They all work, so our domain is $x\le-1$ and $1\le x\le 3$.
c) Multiply the two functions. $(\sqrt{3-x})(\sqrt{x^2-1})$ $=\sqrt{(3-x)(x^2-1)}$ $=\sqrt{(x-3)(1+x)(1-x)}$
The zeros of this function are $x=-1, 1, 3$. We check $x\lt -1$, $-1\lt x\lt 1$, $1\lt x \lt 3$, and $x\gt 3$. The only ranges that work are $x\lt-3$ and $1\lt x\lt 3$.
Then we check $x=3$, $x=-1$, and $x=1$. They all work, so our domain is $x\le-1$ and $1\le x\le 3$.
d) Divide the two functions. $\frac{\sqrt{3-x}}{\sqrt{(x+1)(x-1)}}$
The zeros of this function are $x=-1, 1, 3$. We check $x\lt -1$, $-1\lt x\lt 1$, $1\lt x \lt 3$, and $x\gt 3$. The only ranges that work are $x\lt-3$ and $1\lt x\lt 3$.
Then we check $x=3$, $x=-1$, and $x=1$. Only $x=3$ works, so our domain is $x\lt-1$ and $1\lt x\le3$.
