Find the Domain and Range f(x)=(2x^3+10x^2+8x)/(x^3+7x^2+12x)

$f (x) = \frac{2 x^{3} + 10 x^{2} + 8 x}{x^{3} + 7 x^{2} + 12 x}$

Step 1

Set the denominator in $\frac{2 x^{3} + 10 x^{2} + 8 x}{x^{3} + 7 x^{2} + 12 x}$ equal to $0$ to find where the expression is undefined.

$x^{3} + 7 x^{2} + 12 x = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $x$ out of $x^{3} + 7 x^{2} + 12 x$ .

Factor $x$ out of $x^{3}$ .

$x \cdot x^{2} + 7 x^{2} + 12 x = 0$

Factor $x$ out of $7 x^{2}$ .

$x \cdot x^{2} + x (7 x) + 12 x = 0$

Factor $x$ out of $12 x$ .

$x \cdot x^{2} + x (7 x) + x \cdot 12 = 0$

Factor $x$ out of $x \cdot x^{2} + x (7 x)$ .

$x (x^{2} + 7 x) + x \cdot 12 = 0$

Factor $x$ out of $x (x^{2} + 7 x) + x \cdot 12$ .

$x (x^{2} + 7 x + 12) = 0$

Factor.

Factor $x^{2} + 7 x + 12$ 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 $12$ and whose sum is $7$ .

$3, 4$

Write the factored form using these integers.

$x ((x + 3) (x + 4)) = 0$

Remove unnecessary parentheses.

$x (x + 3) (x + 4) = 0$

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

$x = 0$

$x + 3 = 0$

$x + 4 = 0$

Set $x$ equal to $0$ .

$x = 0$

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $x (x + 3) (x + 4) = 0$ true.

$x = 0, - 3, - 4$

Step 3

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

Interval Notation:

$(- \infty, - 4) \cup (- 4, - 3) \cup (- 3, 0) \cup (0, \infty)$

Set-Builder Notation:

${x | x \neq 0, - 3, - 4}$

Step 4

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

Interval Notation:

$(- \infty, \frac{2}{3}) \cup (\frac{2}{3}, 2) \cup (2, 6) \cup (6, \infty)$

Set-Builder Notation:

${y | y \neq \frac{2}{3}, 2, 6}$

Step 5

Determine the domain and range.

Domain: $(- \infty, - 4) \cup (- 4, - 3) \cup (- 3, 0) \cup (0, \infty), {x | x \neq 0, - 3, - 4}$

Range: $(- \infty, \frac{2}{3}) \cup (\frac{2}{3}, 2) \cup (2, 6) \cup (6, \infty), {y | y \neq \frac{2}{3}, 2, 6}$

Step 6

Find the Domain and Range f(x)=(2x^3+10x^2+8x)/(x^3+7x^2+12x)

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