Find the Domain and Range f(x)=(3x^3+2x^2)/(x^2+x)

$f (x) = \frac{3 x^{3} + 2 x^{2}}{x^{2} + x}$

Step 1

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

$x^{2} + x = 0$

Step 2

Solve for $x$ .

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

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

$x \cdot x + x = 0$

Raise $x$ to the power of $1$ .

$x \cdot x + x = 0$

Factor $x$ out of $x^{1}$ .

$x \cdot x + x \cdot 1 = 0$

Factor $x$ out of $x \cdot x + x \cdot 1$ .

$x (x + 1) = 0$

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

$x = 0$

$x + 1 = 0$

Set $x$ equal to $0$ .

$x = 0$

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.

$x = - 1$

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

$x = 0, - 1$

Step 3

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

Interval Notation:

$(- \infty, - 1) \cup (- 1, 0) \cup (0, \infty)$

Set-Builder Notation:

${x | x \neq 0, - 1}$

Step 4

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

Interval Notation:

$(- \infty, - 4 - 2 \sqrt{3}] \cup [- 4 + 2 \sqrt{3}, \infty)$

Set-Builder Notation:

${y | y \leq - 4 - 2 \sqrt{3}, y \geq - 4 + 2 \sqrt{3}}$

Step 5

Determine the domain and range.

Domain: $(- \infty, - 1) \cup (- 1, 0) \cup (0, \infty), {x | x \neq 0, - 1}$

Range: $(- \infty, - 4 - 2 \sqrt{3}] \cup [- 4 + 2 \sqrt{3}, \infty), {y | y \leq - 4 - 2 \sqrt{3}, y \geq - 4 + 2 \sqrt{3}}$

Step 6