Finding the range of rational functions I have a problem that I cannot figure out how to do. The problem is:
Suppose $s(x)=\frac{x+2}{x^2+5}$. What is the range of $s$?
I know that the range is equivalent to the domain of $s^{-1}(x)$ but that is only true for one-to-one functions. I have tried to find the inverse of function s but I got stuck trying to isolate y. Here is what I have done so far:
$y=\frac{x+2}{x^2+5}$
$x=\frac{y+2}{y^2+5}$
$x(y^2+5)=y+2$
$xy^2+5x=y+2$
$xy^2-y=2-5x$
$y(xy-1)=2-5x$
This is the step I got stuck on, usually I would just divide by the parenthesis to isolate y but since y is squared, I cannot do that. Is this the right approach to finding the range of the function? If not how would I approach this problem?

To find the range, we want to find all $y$ for which there exists an $x$ such that
$$ y = \frac{x+2}{x^2+5}.$$
We can solve this equation for $x$:
$$ y x^2 + 5y = x+2$$
$$ 0 = y x^2 -x + 5y-2$$
If $y \neq 0$, this is a quadratic equation in $x$, so we can solve it with the quadratic formula:
$$
x = \frac{ 1 \pm \sqrt{ 1 - 4y(5y-2)}}{2y}.$$
So, for a given $y$, $y$ is in the range if this expression yields a real number.  That is, if
$$ 1 - 4y(5y-2) = -20y^2 +8y +1 \ge 0$$
If you study this quadratic, you will find that it has roots at $y=1/2$ and $y=-1/10$, and between these roots it is positive, while outside these roots it is negative.  Hence, there exists an $x$ such that $s(x)=y$ only if 
$$
-\frac{1}{10} \le y \le \frac{1}{2}.
$$
Thus, this is the range of $s$.
(Note we excluded $y=0$ earlier, but we know $y=0$ is in our range since $s(-2)=0$.)
