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

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

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

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

Solve for $x$ .

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

Factor $x^{2}$ out of $x^{3}$ .

$x^{2} x + x^{2} = 0$

Multiply by $1$ .

$x^{2} x + x^{2} \cdot 1 = 0$

Factor $x^{2}$ out of $x^{2} x + x^{2} \cdot 1$ .

$x^{2} (x + 1) = 0$

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

$x^{2} = 0$

$x + 1 = 0$

Set $x^{2}$ equal to $0$ and solve for $x$ .

Set $x^{2}$ equal to $0$ .

$x^{2} = 0$

Solve $x^{2} = 0$ for $x$ .

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

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 $x^{2} (x + 1) = 0$ true.

$x = 0, - 1$

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:

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

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

Interval Notation:

$(0, \infty)$

Set-Builder Notation:

${y | y > 0}$

Determine the domain and range.

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

Range: $(0, \infty), {y | y > 0}$

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

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