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).
Best Answer
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: