Answer
$4x+3$, $(-\infty,\infty)$
$2x+13$, $(-\infty,\infty)$
$3x^2-7x-40$, $(-\infty,\infty)$
$\frac{3x+8}{x-5}$, $(-\infty,-5)U(-5, \infty)$
Work Step by Step
Step 1. Given $f(x)=3x+8$ and $g(x)=x-5$, we have $f+g=3x+8+x-5=4x+3$ with domain $(-\infty,\infty)$
Step 2. We have $f-g=3x+8-(x-5)=2x+13$ with domain $(-\infty,\infty)$
Step 3. We have $f\cdot g=(3x+8)(x-5)=3x^2-7x-40$ with domain $(-\infty,\infty)$
Step 4. We have $\frac{f}{g}=\frac{3x+8}{x-5}$ with domain $(-\infty,-5)U(-5, \infty)$
