# What is the range of the function f(x)=x/(x^2-5x+9)?

Mar 23, 2018

$- \frac{1}{11} \le f \left(x\right) \le 1$

#### Explanation:

The range is the set of $y$ values given for $f \left(x\right)$

First, we rearrange to get: $y {x}^{2} - 5 x y - x + 9 y = 0$

By using the quadratic formula we get:
$x = \frac{5 y + 1 \pm \sqrt{{\left(- 5 y - 1\right)}^{2} - 4 \left(y \cdot 9 y\right)}}{2 y} = \frac{5 y + 1 \pm \sqrt{- 11 {y}^{2} + 10 y + 1}}{2 y}$

$x = \frac{5 y + 1 + \sqrt{- 11 {y}^{2} + 10 y + 1}}{2 y}$
$x = \frac{5 y + 1 - \sqrt{- 11 {y}^{2} + 10 y + 1}}{2 y}$

Since we want the two equations to have similar values of $x$ we do:
$x - x = 0$
$\frac{5 y + 1 - \sqrt{- 11 {y}^{2} + 10 y + 1}}{2 y} - \frac{5 y + 1 + \sqrt{- 11 {y}^{2} + 10 y + 1}}{2 y} = - \frac{\sqrt{- 11 {y}^{2} + 10 y + 1}}{y}$

$- \frac{\sqrt{- 11 {y}^{2} + 10 y + 1}}{y} = 0$

$- 11 {y}^{2} + 10 y + 1 = 0$

$y = - \frac{- 10 \pm \sqrt{{10}^{2} - 4 \left(- 11\right)}}{22} = - \frac{- 10 \pm \sqrt{144}}{22} = 1 \mathmr{and} - \frac{1}{11}$

$- \frac{1}{11} \le f \left(x\right) \le 1$