Precalculus Examples

$f (x) = \frac{2 x + 1}{\sqrt{- x^{2} - 2 x + 15}}$

Step 1

Set the radicand in $\sqrt{- x^{2} - 2 x + 15}$ greater than or equal to $0$ to find where the expression is defined.

$- x^{2} - 2 x + 15 \geq 0$

Step 2

Solve for $x$ .

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Step 2.1

Convert the inequality to an equation.

$- x^{2} - 2 x + 15 = 0$

Step 2.2

Factor the left side of the equation.

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Step 2.2.1

Factor $- 1$ out of $- x^{2} - 2 x + 15$ .

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Step 2.2.1.1

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) - 2 x + 15 = 0$

Step 2.2.1.2

Factor $- 1$ out of $- 2 x$ .

$- (x^{2}) - (2 x) + 15 = 0$

Step 2.2.1.3

Rewrite $15$ as $- 1 (- 15)$ .

$- (x^{2}) - (2 x) - 1 \cdot - 15 = 0$

Step 2.2.1.4

Factor $- 1$ out of $- (x^{2}) - (2 x)$ .

$- (x^{2} + 2 x) - 1 \cdot - 15 = 0$

Step 2.2.1.5

Factor $- 1$ out of $- (x^{2} + 2 x) - 1 (- 15)$ .

$- (x^{2} + 2 x - 15) = 0$

Step 2.2.2

Factor.

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Step 2.2.2.1

Factor $x^{2} + 2 x - 15$ using the AC method.

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Step 2.2.2.1.1

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 $- 15$ and whose sum is $2$ .

$- 3, 5$

Step 2.2.2.1.2

Write the factored form using these integers.

$- ((x - 3) (x + 5)) = 0$

Step 2.2.2.2

Remove unnecessary parentheses.

$- (x - 3) (x + 5) = 0$

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 $0$ .

$x - 3 = 0$

$x + 5 = 0$

Step 2.4

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 2.4.2

Add $3$ to both sides of the equation.

$x = 3$

Step 2.5

Set $x + 5$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Step 2.5.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 2.6

The final solution is all the values that make $- (x - 3) (x + 5) = 0$ true.

$x = 3, - 5$

Step 2.7

Use each root to create test intervals.

$x < - 5$

$- 5 < x < 3$

$x > 3$

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 $x < - 5$ to see if it makes the inequality true.

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Step 2.8.1.1

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

Step 2.8.1.2

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

$- {(- 8)}^{2} - 2 \cdot - 8 + 15 \geq 0$

Step 2.8.1.3

The left side $- 33$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.8.2

Test a value on the interval $- 5 < x < 3$ to see if it makes the inequality true.

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Step 2.8.2.1

Choose a value on the interval $- 5 < x < 3$ 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.

$- {(0)}^{2} - 2 \cdot 0 + 15 \geq 0$

Step 2.8.2.3

The left side $15$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.8.3

Test a value on the interval $x > 3$ to see if it makes the inequality true.

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Step 2.8.3.1

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Step 2.8.3.2

Replace $x$ with $6$ in the original inequality.

$- {(6)}^{2} - 2 \cdot 6 + 15 \geq 0$

Step 2.8.3.3

The left side $- 33$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.8.4

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

$x < - 5$ False

$- 5 < x < 3$ True

$x > 3$ False

$x < - 5$ False

$- 5 < x < 3$ True

$x > 3$ False

Step 2.9

The solution consists of all of the true intervals.

$- 5 \leq x \leq 3$

Step 3

Set the denominator in $\frac{2 x + 1}{\sqrt{- x^{2} - 2 x + 15}}$ equal to $0$ to find where the expression is undefined.

$\sqrt{- x^{2} - 2 x + 15} = 0$

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.

${\sqrt{- x^{2} - 2 x + 15}}^{2} = 0^{2}$

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 $\sqrt{- x^{2} - 2 x + 15}$ as ${(- x^{2} - 2 x + 15)}^{\frac{1}{2}}$ .

${({(- x^{2} - 2 x + 15)}^{\frac{1}{2}})}^{2} = 0^{2}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Simplify ${({(- x^{2} - 2 x + 15)}^{\frac{1}{2}})}^{2}$ .

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Step 4.2.2.1.1

Multiply the exponents in ${({(- x^{2} - 2 x + 15)}^{\frac{1}{2}})}^{2}$ .

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Step 4.2.2.1.1.1

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

${(- x^{2} - 2 x + 15)}^{\frac{1}{2} \cdot 2} = 0^{2}$

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.

${(- x^{2} - 2 x + 15)}^{\frac{1}{2} \cdot 2} = 0^{2}$

Step 4.2.2.1.1.2.2

Rewrite the expression.

${(- x^{2} - 2 x + 15)}^{1} = 0^{2}$

Step 4.2.2.1.2

Simplify.

$- x^{2} - 2 x + 15 = 0^{2}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

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

$- x^{2} - 2 x + 15 = 0$

Step 4.3

Solve for $x$ .

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Step 4.3.1

Factor the left side of the equation.

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Step 4.3.1.1

Factor $- 1$ out of $- x^{2} - 2 x + 15$ .

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Step 4.3.1.1.1

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) - 2 x + 15 = 0$

Step 4.3.1.1.2

Factor $- 1$ out of $- 2 x$ .

$- (x^{2}) - (2 x) + 15 = 0$

Step 4.3.1.1.3

Rewrite $15$ as $- 1 (- 15)$ .

$- (x^{2}) - (2 x) - 1 \cdot - 15 = 0$

Step 4.3.1.1.4

Factor $- 1$ out of $- (x^{2}) - (2 x)$ .

$- (x^{2} + 2 x) - 1 \cdot - 15 = 0$

Step 4.3.1.1.5

Factor $- 1$ out of $- (x^{2} + 2 x) - 1 (- 15)$ .

$- (x^{2} + 2 x - 15) = 0$

Step 4.3.1.2

Factor.

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Step 4.3.1.2.1

Factor $x^{2} + 2 x - 15$ using the AC method.

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Step 4.3.1.2.1.1

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 $- 15$ and whose sum is $2$ .

$- 3, 5$

Step 4.3.1.2.1.2

Write the factored form using these integers.

$- ((x - 3) (x + 5)) = 0$

Step 4.3.1.2.2

Remove unnecessary parentheses.

$- (x - 3) (x + 5) = 0$

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 $0$ .

$x - 3 = 0$

$x + 5 = 0$

Step 4.3.3

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 4.3.3.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 4.3.3.2

Add $3$ to both sides of the equation.

$x = 3$

Step 4.3.4

Set $x + 5$ equal to $0$ and solve for $x$ .

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Step 4.3.4.1

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Step 4.3.4.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 4.3.5

The final solution is all the values that make $- (x - 3) (x + 5) = 0$ true.

$x = 3, - 5$

Step 5

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

Interval Notation:

$(- 5, 3)$

Set-Builder Notation:

${x | - 5 < x < 3}$

Step 6