Find the Domain and Range f(x)=1/(x^2-2x+1)

$f (x) = \frac{1}{x^{2} - 2 x + 1}$

Step 1

Set the denominator in $\frac{1}{x^{2} - 2 x + 1}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 2 x + 1 = 0$

Step 2

Solve for $x$ .

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

$x^{2} - 2 x + 1^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Rewrite the polynomial.

$x^{2} - 2 \cdot x \cdot 1 + 1^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 1$ .

${(x - 1)}^{2} = 0$

Set the $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Step 3

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, 1) \cup (1, \infty)$

Set-Builder Notation:

${x | x \neq 1}$

Step 4

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$(0, \infty)$

Set-Builder Notation:

${y | y > 0}$

Step 5

Determine the domain and range.

Domain: $(- \infty, 1) \cup (1, \infty), {x | x \neq 1}$

Range: $(0, \infty), {y | y > 0}$

Step 6