I've been having trouble when trying to find the domain/range of functions algebraically. Here is an example:

$P(x)=\frac{1}{3+\sqrt{x+1}}$

Finding the domain:

$x+1\ge0$

$x\ge-1$

Therefore, $x \in [-1,+\infty)$

Finding the range: Let $y=P(x)=\frac{1}{3+\sqrt{x+1}}$
From isolating x we find:

$x=(\frac{1}{y} -3)^2-1$ 

Therefore:

$(\frac{1}{y} -3)^2-1\ge-1$

$(\frac{1}{y} -3)^2\ge0$

$\frac{1}{y} -3\ge0$ or $\frac{1}{y} -3\le0$

$y\le \frac{1}{3}$ or $y\ge \frac{1}{3}$

This doesn't make any sense! Intuitively I can see that when $x=-1$ then $f(x)=\frac{1}{3}$ and as x approaches $+\infty$ then $f(x)$ approaches zero (without ever reaching it). How do I find this solution algebraically? What are the "rules" for working with inequalities w/ exponents and radicals (both positive and negative)? How do I find the range for other functions such as $g(x)=3+\sqrt{16-(x-3)^2}$ and $h(x)=\frac{12x-9}{6-9x}$ algebraically? A thorough explanation would be appreciated (also, feel free to point out errors in my work- there are obviously many).