Precalculus Examples

$g (x) = \frac{1}{x^{2} + 4} - \frac{1}{x^{2} - 4}$

Step 1

Set the denominator in $\frac{1}{x^{2} + 4}$ equal to $0$ to find where the expression is undefined.

$x^{2} + 4 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Subtract $4$ from both sides of the equation.

$x^{2} = - 4$

Step 2.2

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

$x = \pm \sqrt{- 4}$

Step 2.3

Simplify $\pm \sqrt{- 4}$ .

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Step 2.3.1

Rewrite $- 4$ as $- 1 (4)$ .

$x = \pm \sqrt{- 1 (4)}$

Step 2.3.2

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{4}$

Step 2.3.3

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

$x = \pm i \cdot \sqrt{4}$

Step 2.3.4

Rewrite $4$ as $2^{2}$ .

$x = \pm i \cdot \sqrt{2^{2}}$

Step 2.3.5

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

$x = \pm i \cdot 2$

Step 2.3.6

Move $2$ to the left of $i$ .

$x = \pm 2 i$

Step 2.4

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

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Step 2.4.1

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

$x = 2 i$

Step 2.4.2

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

$x = - 2 i$

Step 2.4.3

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

$x = 2 i, - 2 i$

Step 3

Set the denominator in $\frac{1}{x^{2} - 4}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 4 = 0$

Step 4

Solve for $x$ .

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Step 4.1

Add $4$ to both sides of the equation.

$x^{2} = 4$

Step 4.2

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

$x = \pm \sqrt{4}$

Step 4.3

Simplify $\pm \sqrt{4}$ .

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Step 4.3.1

Rewrite $4$ as $2^{2}$ .

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

Step 4.3.2

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

$x = \pm 2$

Step 4.4

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

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Step 4.4.1

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

$x = 2$

Step 4.4.2

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

$x = - 2$

Step 4.4.3

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

$x = 2, - 2$

Step 5

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

Interval Notation:

$(- \infty, - 2) \cup (- 2, 2) \cup (2, \infty)$

Set-Builder Notation:

${x | x \neq 2, - 2}$

Step 6