Precalculus Examples

$2 x > \frac{4}{x + 1}$

Step 1

Subtract $\frac{4}{x + 1}$ from both sides of the inequality.

$2 x - \frac{4}{x + 1} > 0$

Step 2

Simplify $2 x - \frac{4}{x + 1}$ .

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Step 2.1

To write $2 x$ as a fraction with a common denominator, multiply by $\frac{x + 1}{x + 1}$ .

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

Step 2.2

Combine the numerators over the common denominator.

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

Step 2.3

Simplify the numerator.

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Step 2.3.1

Factor $2$ out of $2 x (x + 1) - 4$ .

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Step 2.3.1.1

Factor $2$ out of $2 x (x + 1)$ .

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

Step 2.3.1.2

Factor $2$ out of $- 4$ .

$\frac{2 (x (x + 1)) + 2 \cdot - 2}{x + 1} > 0$

Step 2.3.1.3

Factor $2$ out of $2 (x (x + 1)) + 2 \cdot - 2$ .

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

Step 2.3.2

Apply the distributive property.

$\frac{2 (x \cdot x + x \cdot 1 - 2)}{x + 1} > 0$

Step 2.3.3

Multiply $x$ by $x$ .

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

Step 2.3.4

Multiply $x$ by $1$ .

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

Step 2.3.5

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

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Step 2.3.5.1

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 $- 2$ and whose sum is $1$ .

$- 1, 2$

Step 2.3.5.2

Write the factored form using these integers.

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

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

Step 3

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

$x - 1 = 0$

$x + 2 = 0$

$x + 1 = 0$

Step 4

Add $1$ to both sides of the equation.

$x = 1$

Step 5

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 6

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 7

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

$x = 1$

$x = - 2$

$x = - 1$

Step 8

Consolidate the solutions.

$x = 1, - 2, - 1$

Step 9

Find the domain of $\frac{2 (x - 1) (x + 2)}{x + 1}$ .

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Step 9.1

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

$x + 1 = 0$

Step 9.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 9.3

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

$(- \infty, - 1) \cup (- 1, \infty)$

Step 10

Use each root to create test intervals.

$x < - 2$

$- 2 < x < - 1$

$- 1 < x < 1$

$x > 1$

Step 11

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 11.1

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

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Step 11.1.1

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

$x = - 4$

Step 11.1.2

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

$2 (- 4) > \frac{4}{(- 4) + 1}$

Step 11.1.3

The left side $- 8$ is not greater than the right side $- 1.\overline{3}$ , which means that the given statement is false.

False

Step 11.2

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

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Step 11.2.1

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

$x = - 1.5$

Step 11.2.2

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

$2 (- 1.5) > \frac{4}{(- 1.5) + 1}$

Step 11.2.3

The left side $- 3$ is greater than the right side $- 8$ , which means that the given statement is always true.

True

Step 11.3

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

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Step 11.3.1

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

$x = 0$

Step 11.3.2

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

$2 (0) > \frac{4}{(0) + 1}$

Step 11.3.3

The left side $0$ is not greater than the right side $4$ , which means that the given statement is false.

False

Step 11.4

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

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Step 11.4.1

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

$x = 4$

Step 11.4.2

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

$2 (4) > \frac{4}{(4) + 1}$

Step 11.4.3

The left side $8$ is greater than the right side $0.8$ , which means that the given statement is always true.

True

Step 11.5

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

$x < - 2$ False

$- 2 < x < - 1$ True

$- 1 < x < 1$ False

$x > 1$ True

$x < - 2$ False

$- 2 < x < - 1$ True

$- 1 < x < 1$ False

$x > 1$ True

Step 12

The solution consists of all of the true intervals.

$- 2 < x < - 1$ or $x > 1$

Step 13

Convert the inequality to interval notation.

$(- 2, - 1) \cup (1, \infty)$

Step 14