Answer
Step 1. $\sqrt {x+4}+\sqrt {x-1}$, $[1,\infty)$.
Step 2. $\sqrt {x+4}-\sqrt {x-1}$, $[1,\infty)$.
Step 3. $\sqrt {x^2+3x-4}$, $[1,\infty)$.
Step 4. $\sqrt {\frac{x+4}{x-1}}$, $(1,\infty)$.
Work Step by Step
Step 1. Given $f(x)=\sqrt {x+4}$ with $x\geq-4$ and $g(x)=\sqrt {x-1}$ with $x\geq1$, we have $f+g=\sqrt {x+4}+\sqrt {x-1}$ with a domain of $x\geq1$ or $[1,\infty)$.
Step 2. We have $f-g=\sqrt {x+4}-\sqrt {x-1}$ with a domain of $x\geq1$ or $[1,\infty)$.
Step 3. We have $f\cdot g=\sqrt {x+4}\sqrt {x-1}=\sqrt {x^2+3x-4}$ with a domain of $x\geq1$ or $[1,\infty)$.
Step 4. We have $\frac{f}{g}=\frac{\sqrt {x+4}}{\sqrt {x-1}}=\sqrt {\frac{x+4}{x-1}}$ with a domain of $x\gt1$ or $(1,\infty)$. (remove $x=1$ from above)
