## Thomas' Calculus 13th Edition

The domain is $(-\infty,3)\cup(3,\infty)$ The range is $(-\infty,0)\cup(0,\infty)$
$f(t)=\frac{4}{3-t}$ $3-t\neq0$ - Because we can't divide by 0 $t \neq -3$ The domain is $(-\infty,3)\cup(3,\infty)$ When $t<3$, we know that $3-t>0$ and therefore $\frac{4}{3-t}>0$. When $t>3$, we know that $3-t<0$ and therefore $\frac{4}{3-t}<0$. This means $t$ can be any real number except $0$, which makes the range $(-\infty,0)\cup(0,\infty)$.