Find the Domain and Range y=(x-1)/(x^2-5x+4)

$y = \frac{x - 1}{x^{2} - 5 x + 4}$

Set the denominator in $\frac{x - 1}{x^{2} - 5 x + 4}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 5 x + 4 = 0$

Solve for $x$ .

Factor $x^{2} - 5 x + 4$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $4$ and whose sum is $- 5$ .

$- 4, - 1$

Write the factored form using these integers.

$(x - 4) (x - 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 4 = 0$

$x - 1 = 0$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Set $x - 1$ equal to $0$ and solve for $x$ .

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $(x - 4) (x - 1) = 0$ true.

$x = 4, 1$

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

Interval Notation:

$(- \infty, 1) \cup (1, 4) \cup (4, \infty)$

Set-Builder Notation:

${x | x \neq 1, 4}$

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

Interval Notation:

$(- \infty, - \frac{1}{3}) \cup (- \frac{1}{3}, 0) \cup (0, \infty)$

Set-Builder Notation:

${y | y \neq 0, - \frac{1}{3}}$

Determine the domain and range.

Domain: $(- \infty, 1) \cup (1, 4) \cup (4, \infty), {x | x \neq 1, 4}$

Range: $(- \infty, - \frac{1}{3}) \cup (- \frac{1}{3}, 0) \cup (0, \infty), {y | y \neq 0, - \frac{1}{3}}$

Find the Domain and Range y=(x-1)/(x^2-5x+4)

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