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).
 A: You are making things more difficult than necessary in your effort to find the range. It is not really necessary to yield an inverse (as you seem to do). You could do it in simple steps:


*

*range of $\sqrt{1+x}$ is $[0,\infty)$

*range of $3+\sqrt{1+x}$  is $[3,\infty)$

*range of $\frac{1}{3+\sqrt{1+x}}$ is $(0,\frac13]$

A: 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):
$$0 \le 16-(x-3)^2 \le 16$$
$$0 \le \sqrt{16-(x-3)^2} \le 4$$
$$3 \le 3+ \sqrt{16-(x-3)^2} \le 7$$
