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

1 Answer
1s2s2p
Mar 23, 2018

-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

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