Precalculus Examples

$y = \frac{x^{2} + 2}{x^{3} - x}$

Step 1

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

$x^{3} - x = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor the left side of the equation.

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Step 2.1.1

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

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Step 2.1.1.1

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

$x \cdot x^{2} - x = 0$

Step 2.1.1.2

Factor $x$ out of $- x$ .

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

Step 2.1.1.3

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

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

Step 2.1.2

Rewrite $1$ as $1^{2}$ .

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

Step 2.1.3

Factor.

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Step 2.1.3.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

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

Step 2.1.3.2

Remove unnecessary parentheses.

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

Step 2.2

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

$x = 0$

$x + 1 = 0$

$x - 1 = 0$

Step 2.3

Set $x$ equal to $0$ .

$x = 0$

Step 2.4

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

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Step 2.4.1

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

$x + 1 = 0$

Step 2.4.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.5

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

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Step 2.5.1

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

$x - 1 = 0$

Step 2.5.2

Add $1$ to both sides of the equation.

$x = 1$

Step 2.6

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

$x = 0, - 1, 1$

Step 3

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

Interval Notation:

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

Set-Builder Notation:

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

Step 4