Question

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

Answer 1

`-1/11<=f(x)<=1`

Explanation:

The range is the set of `y` values given for `f(x)`

First, we rearrange to get: `yx^2-5xy-x+9y=0`

By using the quadratic formula we get:
`x=(5y+1+-sqrt((-5y-1)^2-4(y*9y)))/(2y)=(5y+1+-sqrt(-11y^2+10y+1))/(2y)`

`x=(5y+1+sqrt(-11y^2+10y+1))/(2y)`
`x=(5y+1-sqrt(-11y^2+10y+1))/(2y)`

Since we want the two equations to have similar values of `x` we do:
`x-x=0`
`(5y+1-sqrt(-11y^2+10y+1))/(2y)-(5y+1+sqrt(-11y^2+10y+1))/(2y)=-sqrt(-11y^2+10y+1)/y`

`-sqrt(-11y^2+10y+1)/y=0`

`-11y^2+10y+1=0`

`y=-(-10+-sqrt(10^2-4(-11)))/22=-(-10+-sqrt144)/22=1or-1/11`

`-1/11<=f(x)<=1`