Answer
Domain: all real numbers $<3$
Range: all real numbers $>0$
Work Step by Step
To evaluate $p(x)$, we must have $3-x>0$. So the domain of $p(x)$ is all real numbers $<3$. To find the range of $p(x)$, we find the inverse function of $p(x)$. Let $y=p(x)$. Solving for $x$, we get
$$
\begin{aligned}
y & =\frac{1}{\sqrt{3-x}} \\
\sqrt{3-x} & =\frac{1}{y} \\
3-x & =\frac{1}{y^2} \\
x & =3-\frac{1}{y^2} \\
p^{-1}(y) & =3-\frac{1}{y^2}
\end{aligned}
$$
The formula works for any $y$ except $y=0$. We know that $y$ must be positive, since $\sqrt{3-x}$ is positive, so the range of $p(x)$ is all real numbers $>0$.
