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

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Answer 1 · 2018-06-14T15:43:22

The domain is $x \in (- \infty, 3) \cup (3, + \infty)$ $x in (-oo,3)uu(3,+oo)$ . The range is $y \in [- \frac{1}{4}, + \infty)$ $y in [-1/4,+oo)$ .

The function is

$f (x) = \frac{x - 2}{x^{2} - 6 x + 9} = \frac{x - 2}{{(x - 3)}^{2}}$ $f(x)=(x-2)/(x^2-6x+9)=(x-2)/(x-3)^2$

The denominator must be $\neq 0$ $!=0$

Therefore,

${(x - 3)}^{2} \neq 0$ $(x-3)^2!=0$ , $\Rightarrow$ $=>$ , $x \neq 3$ $x!=3$

The domain is $x \in (- \infty, 3) \cup (3, + \infty)$ $x in (-oo,3)uu(3,+oo)$

To find the range, let

$y = \frac{x - 2}{x^{2} - 6 x + 9}$ $y=(x-2)/(x^2-6x+9)$

Cross multiply,

$y (x^{2} - 6 x + 9) = x - 2$ $y(x^2-6x+9)=x-2$

$y x^{2} - 6 y x - x + 9 y + 2 = 0$ $yx^2-6yx-x+9y+2=0$

$y x^{2} - (6 y + 1) x + 9 y + 2 = 0$ $yx^2-(6y+1)x+9y+2=0$

This is a quadratic equation in $x$ $x$ , 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−2x2−6x+9f(x)=(x-2)/(x^2-6x+9)?