Precalculus Examples

$\frac{(\sqrt{x^{2} - 16})}{(\frac{x^{2}}{x^{2} + 1})}$

Step 1

Set the radicand in $\sqrt{x^{2} - 16}$ greater than or equal to $0$ to find where the expression is defined.

$x^{2} - 16 \geq 0$

Step 2

Solve for $x$ .

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Step 2.1

Add $16$ to both sides of the inequality.

$x^{2} \geq 16$

Step 2.2

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

$\sqrt{x^{2}} \geq \sqrt{16}$

Step 2.3

Simplify the equation.

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Step 2.3.1

Simplify the left side.

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Step 2.3.1.1

Pull terms out from under the radical.

$| x | \geq \sqrt{16}$

Step 2.3.2

Simplify the right side.

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Step 2.3.2.1

Simplify $\sqrt{16}$ .

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Step 2.3.2.1.1

Rewrite $16$ as $4^{2}$ .

$| x | \geq \sqrt{4^{2}}$

Step 2.3.2.1.2

Pull terms out from under the radical.

$| x | \geq | 4 |$

Step 2.3.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $4$ is $4$ .

$| x | \geq 4$

Step 2.4

Write $| x | \geq 4$ as a piecewise.

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Step 2.4.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

Step 2.4.2

In the piece where $x$ is non-negative, remove the absolute value.

$x \geq 4$

Step 2.4.3

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

Step 2.4.4

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \geq 4$

Step 2.4.5

Write as a piecewise.

${\begin{cases} x \geq 4 & x \geq 0 \\ - x \geq 4 & x < 0 \end{cases}$

Step 2.5

Find the intersection of $x \geq 4$ and $x \geq 0$ .

$x \geq 4$

Step 2.6

Divide each term in $- x \geq 4$ by $- 1$ and simplify.

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Step 2.6.1

Divide each term in $- x \geq 4$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{4}{- 1}$

Step 2.6.2

Simplify the left side.

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Step 2.6.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{4}{- 1}$

Step 2.6.2.2

Divide $x$ by $1$ .

$x \leq \frac{4}{- 1}$

Step 2.6.3

Simplify the right side.

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Step 2.6.3.1

Divide $4$ by $- 1$ .

$x \leq - 4$

Step 2.7

Find the union of the solutions.

$x \leq - 4$ or $x \geq 4$

Step 3

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

$x^{2} + 1 = 0$

Step 4

Solve for $x$ .

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Step 4.1

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

Step 4.2

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

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

Step 4.3

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

$x = \pm i$

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.

$x = i$

Step 4.4.2

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

$x = - i$

Step 4.4.3

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

$x = i, - i$

Step 5

Set the denominator in $\frac{\sqrt{x^{2} - 16}}{\frac{x^{2}}{x^{2} + 1}}$ equal to $0$ to find where the expression is undefined.

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

Step 6

Solve for $x$ .

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Step 6.1

Set the numerator equal to zero.

$x^{2} = 0$

Step 6.2

Solve the equation for $x$ .

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Step 6.2.1

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

$x = \pm \sqrt{0}$

Step 6.2.2

Simplify $\pm \sqrt{0}$ .

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Step 6.2.2.1

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

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

Step 6.2.2.2

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

$x = \pm 0$

Step 6.2.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 7

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

Interval Notation:

$(- \infty, - 4] \cup [4, \infty)$

Set-Builder Notation:

${x | x \leq - 4, x \geq 4}$

Step 8