Find the Domain and Range y=((x-4)(x+2))/((x-3)(x-1))

$y = \frac{(x - 4) (x + 2)}{(x - 3) (x - 1)}$

Step 1

Set the denominator in $\frac{(x - 4) (x + 2)}{(x - 3) (x - 1)}$ equal to $0$ to find where the expression is undefined.

$(x - 3) (x - 1) = 0$

Step 2

Solve for $x$ .

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

$x - 3 = 0$

$x - 1 = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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 $(x - 3) (x - 1) = 0$ true.

$x = 3, 1$

Step 3

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

Interval Notation:

$(- \infty, 1) \cup (1, 3) \cup (3, \infty)$

Set-Builder Notation:

${x | x \neq 1, 3}$

Step 4

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

Interval Notation:

$(- \infty, \frac{9 - 3 \sqrt{5}}{2}] \cup [\frac{9 + 3 \sqrt{5}}{2}, \infty)$

Set-Builder Notation:

${y | y \leq \frac{9 - 3 \sqrt{5}}{2}, y \geq \frac{9 + 3 \sqrt{5}}{2}}$

Step 5

Determine the domain and range.

Domain: $(- \infty, 1) \cup (1, 3) \cup (3, \infty), {x | x \neq 1, 3}$

Range: $(- \infty, \frac{9 - 3 \sqrt{5}}{2}] \cup [\frac{9 + 3 \sqrt{5}}{2}, \infty), {y | y \leq \frac{9 - 3 \sqrt{5}}{2}, y \geq \frac{9 + 3 \sqrt{5}}{2}}$

Step 6

Find the Domain and Range y=((x-4)(x+2))/((x-3)(x-1))

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