Precalculus Examples

$f (x) = \frac{x + 7}{\sqrt{x^{2} - 64}}$

Step 1

Set the radicand in $\sqrt{x^{2} - 64}$ greater than or equal to $0$ to find where the expression is defined.

$x^{2} - 64 \geq 0$

Step 2

Solve for $x$ .

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Step 2.1

Add $64$ to both sides of the inequality.

$x^{2} \geq 64$

Step 2.2

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

$\sqrt{x^{2}} \geq \sqrt{64}$

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.

$| x | \geq \sqrt{64}$

Step 2.3.2

Simplify the right side.

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Step 2.3.2.1

Simplify $\sqrt{64}$ .

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Step 2.3.2.1.1

Rewrite $64$ as $8^{2}$ .

$| x | \geq \sqrt{8^{2}}$

Step 2.3.2.1.2

Pull terms out from under the radical.

$| x | \geq | 8 |$

Step 2.3.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $8$ is $8$ .

$| x | \geq 8$

Step 2.4

Write $| x | \geq 8$ 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.

$x \geq 8$

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 $- 1$ .

$- x \geq 8$

Step 2.4.5

Write as a piecewise.

${\begin{cases} x \geq 8 & x \geq 0 \\ - x \geq 8 & x < 0 \end{cases}$

Step 2.5

Find the intersection of $x \geq 8$ and $x \geq 0$ .

$x \geq 8$

Step 2.6

Divide each term in $- x \geq 8$ by $- 1$ and simplify.

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Step 2.6.1

Divide each term in $- x \geq 8$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{8}{- 1}$

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.

$\frac{x}{1} \leq \frac{8}{- 1}$

Step 2.6.2.2

Divide $x$ by $1$ .

$x \leq \frac{8}{- 1}$

Step 2.6.3

Simplify the right side.

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Step 2.6.3.1

Divide $8$ by $- 1$ .

$x \leq - 8$

Step 2.7

Find the union of the solutions.

$x \leq - 8$ or $x \geq 8$

Step 3

Set the denominator in $\frac{x + 7}{\sqrt{x^{2} - 64}}$ equal to $0$ to find where the expression is undefined.

$\sqrt{x^{2} - 64} = 0$

Step 4

Solve for $x$ .

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Step 4.1

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{x^{2} - 64}}^{2} = 0^{2}$

Step 4.2

Simplify each side of the equation.

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Step 4.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x^{2} - 64}$ as ${(x^{2} - 64)}^{\frac{1}{2}}$ .

${({(x^{2} - 64)}^{\frac{1}{2}})}^{2} = 0^{2}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Simplify ${({(x^{2} - 64)}^{\frac{1}{2}})}^{2}$ .

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Step 4.2.2.1.1

Multiply the exponents in ${({(x^{2} - 64)}^{\frac{1}{2}})}^{2}$ .

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Step 4.2.2.1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(x^{2} - 64)}^{\frac{1}{2} \cdot 2} = 0^{2}$

Step 4.2.2.1.1.2

Cancel the common factor of $2$ .

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Step 4.2.2.1.1.2.1

Cancel the common factor.

${(x^{2} - 64)}^{\frac{1}{2} \cdot 2} = 0^{2}$

Step 4.2.2.1.1.2.2

Rewrite the expression.

${(x^{2} - 64)}^{1} = 0^{2}$

Step 4.2.2.1.2

Simplify.

$x^{2} - 64 = 0^{2}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Raising $0$ to any positive power yields $0$ .

$x^{2} - 64 = 0$

Step 4.3

Solve for $x$ .

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Step 4.3.1

Add $64$ to both sides of the equation.

$x^{2} = 64$

Step 4.3.2

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

$x = \pm \sqrt{64}$

Step 4.3.3

Simplify $\pm \sqrt{64}$ .

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Step 4.3.3.1

Rewrite $64$ as $8^{2}$ .

$x = \pm \sqrt{8^{2}}$

Step 4.3.3.2

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

$x = \pm 8$

Step 4.3.4

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

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Step 4.3.4.1

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

$x = 8$

Step 4.3.4.2

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

$x = - 8$

Step 4.3.4.3

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

$x = 8, - 8$

Step 5

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

Interval Notation:

$(- \infty, - 8) \cup (8, \infty)$

Set-Builder Notation:

${x | x < - 8, x > 8}$

Step 6