# 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?

• Are you familiar with calculus? – Alex Becker Sep 30 '12 at 22:29
• No, I am currently taking Pre-Calculus. – Kot Sep 30 '12 at 22:30
• Look at the behavior of the function at $\pm \infty$ – PAD Sep 30 '12 at 22:37
• @PantelisDamianou the limits are both $0$, this wouldn't help much. – user39572 Sep 30 '12 at 22:38

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 $s(-2)=0$.)
Going back to your original problem statement, you have to understand that the range of a function is the set of values that the function takes on for arguments in the function's domain. This avoids worrying about functions that are not 1-1. (I don't see the need for the "domain of $s^{-1}$" statement.) I assume that the domain in your case is the real numbers.
Hint: one way would be to sketch it, notice the minimum and maximum (call them $m$ and $M$), and show that $y>M$ and $y<m$ lead to contradictions (show also that $y=M$ and $y=m$ for certain values of $x$)