Precalculus Examples

$f (x) = \frac{2 x + 5}{\sqrt{x^{2} + 2 x - 48}}$

Step 1

Set the radicand in $\sqrt{x^{2} + 2 x - 48}$ greater than or equal to $0$ to find where the expression is defined.

$x^{2} + 2 x - 48 \geq 0$

Step 2

Solve for $x$ .

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Step 2.1

Convert the inequality to an equation.

$x^{2} + 2 x - 48 = 0$

Step 2.2

Factor $x^{2} + 2 x - 48$ using the AC method.

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Step 2.2.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 48$ and whose sum is $2$ .

$- 6, 8$

Step 2.2.2

Write the factored form using these integers.

$(x - 6) (x + 8) = 0$

Step 2.3

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

$x - 6 = 0$

$x + 8 = 0$

Step 2.4

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 2.4.2

Add $6$ to both sides of the equation.

$x = 6$

Step 2.5

Set $x + 8$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Step 2.5.2

Subtract $8$ from both sides of the equation.

$x = - 8$

Step 2.6

The final solution is all the values that make $(x - 6) (x + 8) = 0$ true.

$x = 6, - 8$

Step 2.7

Use each root to create test intervals.

$x < - 8$

$- 8 < x < 6$

$x > 6$

Step 2.8

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 2.8.1

Test a value on the interval $x < - 8$ to see if it makes the inequality true.

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Step 2.8.1.1

Choose a value on the interval $x < - 8$ and see if this value makes the original inequality true.

$x = - 10$

Step 2.8.1.2

Replace $x$ with $- 10$ in the original inequality.

${(- 10)}^{2} + 2 (- 10) - 48 \geq 0$

Step 2.8.1.3

The left side $32$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.8.2

Test a value on the interval $- 8 < x < 6$ to see if it makes the inequality true.

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Step 2.8.2.1

Choose a value on the interval $- 8 < x < 6$ and see if this value makes the original inequality true.

$x = 0$

Step 2.8.2.2

Replace $x$ with $0$ in the original inequality.

${(0)}^{2} + 2 (0) - 48 \geq 0$

Step 2.8.2.3

The left side $- 48$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.8.3

Test a value on the interval $x > 6$ to see if it makes the inequality true.

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Step 2.8.3.1

Choose a value on the interval $x > 6$ and see if this value makes the original inequality true.

$x = 8$

Step 2.8.3.2

Replace $x$ with $8$ in the original inequality.

${(8)}^{2} + 2 (8) - 48 \geq 0$

Step 2.8.3.3

The left side $32$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.8.4

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 8$ True

$- 8 < x < 6$ False

$x > 6$ True

$x < - 8$ True

$- 8 < x < 6$ False

$x > 6$ True

Step 2.9

The solution consists of all of the true intervals.

$x \leq - 8$ or $x \geq 6$

Step 3

Set the denominator in $\frac{2 x + 5}{\sqrt{x^{2} + 2 x - 48}}$ equal to $0$ to find where the expression is undefined.

$\sqrt{x^{2} + 2 x - 48} = 0$

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.

${\sqrt{x^{2} + 2 x - 48}}^{2} = 0^{2}$

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 $\sqrt{x^{2} + 2 x - 48}$ as ${(x^{2} + 2 x - 48)}^{\frac{1}{2}}$ .

${({(x^{2} + 2 x - 48)}^{\frac{1}{2}})}^{2} = 0^{2}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Simplify ${({(x^{2} + 2 x - 48)}^{\frac{1}{2}})}^{2}$ .

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Step 4.2.2.1.1

Multiply the exponents in ${({(x^{2} + 2 x - 48)}^{\frac{1}{2}})}^{2}$ .

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Step 4.2.2.1.1.1

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

${(x^{2} + 2 x - 48)}^{\frac{1}{2} \cdot 2} = 0^{2}$

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.

${(x^{2} + 2 x - 48)}^{\frac{1}{2} \cdot 2} = 0^{2}$

Step 4.2.2.1.1.2.2

Rewrite the expression.

${(x^{2} + 2 x - 48)}^{1} = 0^{2}$

Step 4.2.2.1.2

Simplify.

$x^{2} + 2 x - 48 = 0^{2}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

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

$x^{2} + 2 x - 48 = 0$

Step 4.3

Solve for $x$ .

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Step 4.3.1

Factor $x^{2} + 2 x - 48$ using the AC method.

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Step 4.3.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 48$ and whose sum is $2$ .

$- 6, 8$

Step 4.3.1.2

Write the factored form using these integers.

$(x - 6) (x + 8) = 0$

Step 4.3.2

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

$x - 6 = 0$

$x + 8 = 0$

Step 4.3.3

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 4.3.3.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 4.3.3.2

Add $6$ to both sides of the equation.

$x = 6$

Step 4.3.4

Set $x + 8$ equal to $0$ and solve for $x$ .

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Step 4.3.4.1

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Step 4.3.4.2

Subtract $8$ from both sides of the equation.

$x = - 8$

Step 4.3.5

The final solution is all the values that make $(x - 6) (x + 8) = 0$ true.

$x = 6, - 8$

Step 5

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

Interval Notation:

$(- \infty, - 8) \cup (6, \infty)$

Set-Builder Notation:

${x | x < - 8, x > 6}$

Step 6