How do you find the domain and range of #f(x)=(x-2)/(x^2-6x+9)#?

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Answer 1 · 2018-06-14T15:43:22

The domain is #x in (-oo,3)uu(3,+oo)#. The range is #y in [-1/4,+oo)#.

The function is

#f(x)=(x-2)/(x^2-6x+9)=(x-2)/(x-3)^2#

The denominator must be #!=0#

Therefore,

#(x-3)^2!=0#, #=>#, #x!=3#

The domain is #x in (-oo,3)uu(3,+oo)#

To find the range, let

#y=(x-2)/(x^2-6x+9)#

Cross multiply,

#y(x^2-6x+9)=x-2#

#yx^2-6yx-x+9y+2=0#

#yx^2-(6y+1)x+9y+2=0#

This is a quadratic equation in #x#, and in order to have solutions,

the discriminant #Delta>=0#

#Delta=(6y+1)^2-4(y)(9y+2)>=0#

#36y^2+12y+1-36y^2-8y>=0#

#4y+1>=0#

#y>=-1/4#

The range is #y in [-1/4,+oo)#.

graph{(x-2)/(x^2-6x+9) [-5.24, 8.81, -2.67, 4.353]}