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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?

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  • $\begingroup$ Are you familiar with calculus? $\endgroup$
    – Alex Becker
    Sep 30, 2012 at 22:29
  • $\begingroup$ No, I am currently taking Pre-Calculus. $\endgroup$
    – Kot
    Sep 30, 2012 at 22:30
  • $\begingroup$ Look at the behavior of the function at $\pm \infty$ $\endgroup$
    – PAD
    Sep 30, 2012 at 22:37
  • $\begingroup$ @PantelisDamianou the limits are both $0$, this wouldn't help much. $\endgroup$
    – user39572
    Sep 30, 2012 at 22:38

3 Answers 3

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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 $s(-2)=0$.)

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    $\begingroup$ you lost a $4$ from the discriminant $b^2-4ac$ in the next line. $\endgroup$
    – Berci
    Sep 30, 2012 at 22:50
  • $\begingroup$ Thanks, Berci. It is fixed, this did not effect the result. $\endgroup$
    – Matthew Conroy
    Sep 30, 2012 at 23:03
  • $\begingroup$ Can you see a siliar question math.stackexchange.com/questions/2836653/… $\endgroup$
    – Hydrous Caperilla
    Jul 1, 2018 at 14:27
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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.

Once you understand this, then you can apply the variety of analyses that show you how to get the range for your particular function.

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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 $x$)

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