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

1 Answer
Mar 23, 2018

Answer:

#-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#