#### Answer

The domain is $(-\infty,3)\cup(3,\infty)$
The range is $(-\infty,0)\cup(0,\infty)$

#### Work Step by Step

$f(t)=\frac{4}{3-t}$
$3-t\neq0$ - 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 $(-\infty,0)\cup(0,\infty)$.