Find the Domain and Range r(x)=(x^2+5x-14)/((x+1)(x-3)(x+7))

$r (x) = \frac{x^{2} + 5 x - 14}{(x + 1) (x - 3) (x + 7)}$

Step 1

Set the denominator in $\frac{x^{2} + 5 x - 14}{(x + 1) (x - 3) (x + 7)}$ equal to $0$ to find where the expression is undefined.

$(x + 1) (x - 3) (x + 7) = 0$

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 $0$ .

$x + 1 = 0$

$x - 3 = 0$

$x + 7 = 0$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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 $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

The final solution is all the values that make $(x + 1) (x - 3) (x + 7) = 0$ true.

$x = - 1, 3, - 7$

Step 3

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

Interval Notation:

$(- \infty, - 7) \cup (- 7, - 1) \cup (- 1, 3) \cup (3, \infty)$

Set-Builder Notation:

${x | x \neq - 1, 3, - 7}$

Step 4

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

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${y | y \in ℝ}$

Step 5

Determine the domain and range.

Domain: $(- \infty, - 7) \cup (- 7, - 1) \cup (- 1, 3) \cup (3, \infty), {x | x \neq - 1, 3, - 7}$

Range: $(- \infty, \infty), {y | y \in ℝ}$

Step 6