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]}

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