Precalculus Examples

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

Step 1

Set the denominator in $\frac{x - 2}{x^{3} + x}$ 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 $x$ out of $x^{3} + x$ .

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Step 2.1.1

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

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

Step 2.1.2

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

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

Step 2.1.3

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

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

Step 2.1.4

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

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

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 $0$ .

$x = 0$

$x^{2} + 1 = 0$

Step 2.3

Set $x$ equal to $0$ .

$x = 0$

Step 2.4

Set $x^{2} + 1$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x^{2} + 1$ equal to $0$ .

$x^{2} + 1 = 0$

Step 2.4.2

Solve $x^{2} + 1 = 0$ for $x$ .

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Step 2.4.2.1

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

Step 2.4.2.2

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

$x = \pm \sqrt{- 1}$

Step 2.4.2.3

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i$

Step 2.4.2.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 2.4.2.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x = i$

Step 2.4.2.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - i$

Step 2.4.2.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = i, - i$

Step 2.5

The final solution is all the values that make $x (x^{2} + 1) = 0$ true.

$x = 0, i, - i$

Step 3

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

Interval Notation:

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

Set-Builder Notation:

${x | x \neq 0}$

Step 4