Precalculus Examples

$f (x) = \frac{2 x - 2}{\sqrt{- x^{2} + 3 x + 10}}$

Step 1

Set the radicand in $\sqrt{- x^{2} + 3 x + 10}$ greater than or equal to $0$ to find where the expression is defined.

$- x^{2} + 3 x + 10 \geq 0$

Step 2

Solve for $x$ .

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Step 2.1

Convert the inequality to an equation.

$- x^{2} + 3 x + 10 = 0$

Step 2.2

Factor the left side of the equation.

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Step 2.2.1

Factor $- 1$ out of $- x^{2} + 3 x + 10$ .

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Step 2.2.1.1

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

$- (x^{2}) + 3 x + 10 = 0$

Step 2.2.1.2

Factor $- 1$ out of $3 x$ .

$- (x^{2}) - (- 3 x) + 10 = 0$

Step 2.2.1.3

Rewrite $10$ as $- 1 (- 10)$ .

$- (x^{2}) - (- 3 x) - 1 \cdot - 10 = 0$

Step 2.2.1.4

Factor $- 1$ out of $- (x^{2}) - (- 3 x)$ .

$- (x^{2} - 3 x) - 1 \cdot - 10 = 0$

Step 2.2.1.5

Factor $- 1$ out of $- (x^{2} - 3 x) - 1 (- 10)$ .

$- (x^{2} - 3 x - 10) = 0$

Step 2.2.2

Factor.

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Step 2.2.2.1

Factor $x^{2} - 3 x - 10$ 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 $- 10$ and whose sum is $- 3$ .

$- 5, 2$

Step 2.2.2.1.2

Write the factored form using these integers.

$- ((x - 5) (x + 2)) = 0$

Step 2.2.2.2

Remove unnecessary parentheses.

$- (x - 5) (x + 2) = 0$

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 $0$ .

$x - 5 = 0$

$x + 2 = 0$

Step 2.4

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

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Step 2.4.1

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

$x - 5 = 0$

Step 2.4.2

Add $5$ to both sides of the equation.

$x = 5$

Step 2.5

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 2.5.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.6

The final solution is all the values that make $- (x - 5) (x + 2) = 0$ true.

$x = 5, - 2$

Step 2.7

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 5$

$x > 5$

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 $x < - 2$ to see if it makes the inequality true.

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Step 2.8.1.1

Choose a value on the interval $x < - 2$ and see if this value makes the original inequality true.

$x = - 4$

Step 2.8.1.2

Replace $x$ with $- 4$ in the original inequality.

$- {(- 4)}^{2} + 3 (- 4) + 10 \geq 0$

Step 2.8.1.3

The left side $- 18$ 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 $- 2 < x < 5$ to see if it makes the inequality true.

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Step 2.8.2.1

Choose a value on the interval $- 2 < x < 5$ 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.

$- {(0)}^{2} + 3 (0) + 10 \geq 0$

Step 2.8.2.3

The left side $10$ 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 $x > 5$ to see if it makes the inequality true.

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Step 2.8.3.1

Choose a value on the interval $x > 5$ 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.

$- {(8)}^{2} + 3 (8) + 10 \geq 0$

Step 2.8.3.3

The left side $- 30$ 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.

$x < - 2$ False

$- 2 < x < 5$ True

$x > 5$ False

$x < - 2$ False

$- 2 < x < 5$ True

$x > 5$ False

Step 2.9

The solution consists of all of the true intervals.

$- 2 \leq x \leq 5$

Step 3

Set the denominator in $\frac{2 x - 2}{\sqrt{- x^{2} + 3 x + 10}}$ equal to $0$ to find where the expression is undefined.

$\sqrt{- x^{2} + 3 x + 10} = 0$

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.

${\sqrt{- x^{2} + 3 x + 10}}^{2} = 0^{2}$

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 $\sqrt{- x^{2} + 3 x + 10}$ as ${(- x^{2} + 3 x + 10)}^{\frac{1}{2}}$ .

${({(- x^{2} + 3 x + 10)}^{\frac{1}{2}})}^{2} = 0^{2}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Simplify ${({(- x^{2} + 3 x + 10)}^{\frac{1}{2}})}^{2}$ .

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Step 4.2.2.1.1

Multiply the exponents in ${({(- x^{2} + 3 x + 10)}^{\frac{1}{2}})}^{2}$ .

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Step 4.2.2.1.1.1

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

${(- x^{2} + 3 x + 10)}^{\frac{1}{2} \cdot 2} = 0^{2}$

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.

${(- x^{2} + 3 x + 10)}^{\frac{1}{2} \cdot 2} = 0^{2}$

Step 4.2.2.1.1.2.2

Rewrite the expression.

${(- x^{2} + 3 x + 10)}^{1} = 0^{2}$

Step 4.2.2.1.2

Simplify.

$- x^{2} + 3 x + 10 = 0^{2}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

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

$- x^{2} + 3 x + 10 = 0$

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 $- x^{2} + 3 x + 10$ .

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Step 4.3.1.1.1

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

$- (x^{2}) + 3 x + 10 = 0$

Step 4.3.1.1.2

Factor $- 1$ out of $3 x$ .

$- (x^{2}) - (- 3 x) + 10 = 0$

Step 4.3.1.1.3

Rewrite $10$ as $- 1 (- 10)$ .

$- (x^{2}) - (- 3 x) - 1 \cdot - 10 = 0$

Step 4.3.1.1.4

Factor $- 1$ out of $- (x^{2}) - (- 3 x)$ .

$- (x^{2} - 3 x) - 1 \cdot - 10 = 0$

Step 4.3.1.1.5

Factor $- 1$ out of $- (x^{2} - 3 x) - 1 (- 10)$ .

$- (x^{2} - 3 x - 10) = 0$

Step 4.3.1.2

Factor.

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Step 4.3.1.2.1

Factor $x^{2} - 3 x - 10$ 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 $- 10$ and whose sum is $- 3$ .

$- 5, 2$

Step 4.3.1.2.1.2

Write the factored form using these integers.

$- ((x - 5) (x + 2)) = 0$

Step 4.3.1.2.2

Remove unnecessary parentheses.

$- (x - 5) (x + 2) = 0$

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 $0$ .

$x - 5 = 0$

$x + 2 = 0$

Step 4.3.3

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

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Step 4.3.3.1

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

$x - 5 = 0$

Step 4.3.3.2

Add $5$ to both sides of the equation.

$x = 5$

Step 4.3.4

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 4.3.4.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 4.3.4.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 4.3.5

The final solution is all the values that make $- (x - 5) (x + 2) = 0$ true.

$x = 5, - 2$

Step 5

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

Interval Notation:

$(- 2, 5)$

Set-Builder Notation:

${x | - 2 < x < 5}$

Step 6