Precalculus Examples

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

Step 1

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

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

Step 2

Solve for $x$ .

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Step 2.1

Simplify $\frac{x^{2} + 2}{x - 2} + 3$ .

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Step 2.1.1

To write $3$ as a fraction with a common denominator, multiply by $\frac{x - 2}{x - 2}$ .

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

Step 2.1.2

Combine the numerators over the common denominator.

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

Step 2.1.3

Simplify the numerator.

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Step 2.1.3.1

Apply the distributive property.

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

Step 2.1.3.2

Multiply $3$ by $- 2$ .

$\frac{x^{2} + 2 + 3 x - 6}{x - 2} \geq 0$

Step 2.1.3.3

Subtract $6$ from $2$ .

$\frac{x^{2} + 3 x - 4}{x - 2} \geq 0$

Step 2.1.3.4

Factor $x^{2} + 3 x - 4$ using the AC method.

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Step 2.1.3.4.1

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 $- 4$ and whose sum is $3$ .

$- 1, 4$

Step 2.1.3.4.2

Write the factored form using these integers.

$\frac{(x - 1) (x + 4)}{x - 2} \geq 0$

Step 2.2

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x - 1 = 0$

$x + 4 = 0$

$x - 2 = 0$

Step 2.3

Add $1$ to both sides of the equation.

$x = 1$

Step 2.4

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 2.5

Add $2$ to both sides of the equation.

$x = 2$

Step 2.6

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 1$

$x = - 4$

$x = 2$

Step 2.7

Consolidate the solutions.

$x = 1, - 4, 2$

Step 2.8

Find the domain of $\frac{x^{2} + 2}{x - 2} + 3$ .

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Step 2.8.1

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

$x - 2 = 0$

Step 2.8.2

Add $2$ to both sides of the equation.

$x = 2$

Step 2.8.3

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

$(- \infty, 2) \cup (2, \infty)$

Step 2.9

Use each root to create test intervals.

$x < - 4$

$- 4 < x < 1$

$1 < x < 2$

$x > 2$

Step 2.10

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 2.10.1

Test a value on the interval $x < - 4$ to see if it makes the inequality true.

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Step 2.10.1.1

Choose a value on the interval $x < - 4$ and see if this value makes the original inequality true.

$x = - 6$

Step 2.10.1.2

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

$\frac{{(- 6)}^{2} + 2}{(- 6) - 2} + 3 \geq 0$

Step 2.10.1.3

The left side $- 1.75$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.10.2

Test a value on the interval $- 4 < x < 1$ to see if it makes the inequality true.

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Step 2.10.2.1

Choose a value on the interval $- 4 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Step 2.10.2.2

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

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

Step 2.10.2.3

The left side $2$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.10.3

Test a value on the interval $1 < x < 2$ to see if it makes the inequality true.

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Step 2.10.3.1

Choose a value on the interval $1 < x < 2$ and see if this value makes the original inequality true.

$x = 1.5$

Step 2.10.3.2

Replace $x$ with $1.5$ in the original inequality.

$\frac{{(1.5)}^{2} + 2}{(1.5) - 2} + 3 \geq 0$

Step 2.10.3.3

The left side $- 5.5$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.10.4

Test a value on the interval $x > 2$ to see if it makes the inequality true.

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Step 2.10.4.1

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

$x = 4$

Step 2.10.4.2

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

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

Step 2.10.4.3

The left side $12$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.10.5

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

$x < - 4$ False

$- 4 < x < 1$ True

$1 < x < 2$ False

$x > 2$ True

$x < - 4$ False

$- 4 < x < 1$ True

$1 < x < 2$ False

$x > 2$ True

Step 2.11

The solution consists of all of the true intervals.

$- 4 \leq x \leq 1$ or $x > 2$

Step 3

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

$x - 2 = 0$

Step 4

Add $2$ to both sides of the equation.

$x = 2$

Step 5

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

Interval Notation:

$[- 4, 1] \cup (2, \infty)$

Set-Builder Notation:

${x | - 4 \leq x \leq 1, x > 2}$

Step 6