<s>[INST] How do I find the domain/range of functions algebraically? 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).
 [/INST] I don't fully follow what you are doing to determine the range. In any case, when you have:

$(1/y -3)^2\ge0$

The LHS is a square and thus always positive, this inequality is satisfied for all $y$...
In the formula:
$$y = \frac{1}{3+\sqrt{x+1}}$$
the range of the monotonically increasing part $\sqrt{x+1}$ is (clearly) $[0,+\infty)$, which means the denominator is monotonically decreasing with a maximum in $x=-1$, namely $y = 1/3$. For $x \to \infty$, $y \to 0$ but since $y \ne 0$ for all $x$, the range is: $0 < y \le \tfrac{1}{3}$.

Alternatively:
$$0 \le \sqrt{x+1} < +\infty$$
$$3 \le 3+\sqrt{x+1} < +\infty$$
$$ \frac{1}{3} \ge \frac{1}{3+\sqrt{x+1}} > \frac{1}{+\infty}$$
So:
$$ 0 <  \frac{1}{3+\sqrt{x+1}} \le \frac{1}{3}$$

For the domain of:
$$g(x)=3+\sqrt{16-(x-3)^2}$$
You need:
$$16-(x-3)^2 \ge 0 \iff (x-3)^2 \le 16 \iff |x-3| \le 4 \iff -1 \le x \le 7 $$
For the range (given the domain as above):
$$0 \le 16-(x-3)^2 \le 16$$
$$0 \le \sqrt{16-(x-3)^2} \le 4$$
$$3 \le 3+ \sqrt{16-(x-3)^2} \le 7$$
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