Contents

Preface
1Foundations
- Introduction
- 1.1Use the Language of Algebra
- 1.2Integers
- 1.3Fractions
- 1.4Decimals
- 1.5Properties of Real Numbers
- Chapter Review
  Key Terms
  Key Concepts
- Exercises
  Review Exercises
  Practice Test
2Solving Linear Equations
3Graphs and Functions
4Systems of Linear Equations
5Polynomials and Polynomial Functions
6Factoring
7Rational Expressions and Functions
8Roots and Radicals
9Quadratic Equations and Functions
10Exponential and Logarithmic Functions
11Conics
12Sequences, Series and Binomial Theorem
- Introduction
- 12.1Sequences
- 12.2Arithmetic Sequences
- 12.3Geometric Sequences and Series
- 12.4Binomial Theorem
- Chapter Review
  Key Terms
Answer Key
Index

Solve Rational Inequalities

Learning Objectives

By the end of this section, you will be able to:

Solve rational inequalities
Solve an inequality with rational functions

Be Prepared 7.6

Before you get started, take this readiness quiz.

Find the value of $x - 5 x - 5$ when ⓐ $x = 6 x = 6$ ⓑ $x = −3 x = −3$ ⓒ $x = 5 . x = 5 .$
If you missed this problem, review Example 1.6.
Solve: $8 - 2 x < 12 . 8 - 2 x < 12 .$
If you missed this problem, review Example 2.52.
Write in interval notation: $−3 \leq x < 5 . −3 \leq x < 5 .$
If you missed this problem, review Example 2.49.

Solve Rational Inequalities

We learned to solve linear inequalities after learning to solve linear equations. The techniques were very much the same with one major exception. When we multiplied or divided by a negative number, the inequality sign reversed.

Having just learned to solve rational equations we are now ready to solve rational inequalities. A rational inequality is an inequality that contains a rational expression.

Rational Inequality

A rational inequality is an inequality that contains a rational expression.

Inequalities such as $\frac{3}{2 x} > 1, \frac{2 x}{x - 3} < 4, \frac{2 x - 3}{x - 6} \geq x, \frac{3}{2 x} > 1,$ and $\frac{1}{4} - \frac{2}{x^{2}} \leq \frac{3}{x} \frac{1}{4} - \frac{2}{x^{2}} \leq \frac{3}{x}$ are rational inequalities as they each contain a rational expression.

When we solve a rational inequality, we will use many of the techniques we used solving linear inequalities. We especially must remember that when we multiply or divide by a negative number, the inequality sign must reverse.

Another difference is that we must carefully consider what value might make the rational expression undefined and so must be excluded.

When we solve an equation and the result is $x = 3, x = 3,$ we know there is one solution, which is 3.

When we solve an inequality and the result is $x > 3, x > 3,$ we know there are many solutions. We graph the result to better help show all the solutions, and we start with 3. Three becomes a critical point and then we decide whether to shade to the left or right of it. The numbers to the right of 3 are larger than 3, so we shade to the right.

This figure shows the solution, the interval 3 to infinity, of the inequality x is greater than 3 on a number line. The values range from negative 5 to 5 on the number line. The inequality is modeled by an open parenthesis at the critical point 3 and shading the right.

To solve a rational inequality, we first must write the inequality with only one quotient on the left and 0 on the right.

Next we determine the critical points to use to divide the number line into intervals. A critical point is a number which make the rational expression zero or undefined.

We then will evaluate the factors of the numerator and denominator, and find the quotient in each interval. This will identify the interval, or intervals, that contains all the solutions of the rational inequality.

We write the solution in interval notation being careful to determine whether the endpoints are included.

Example 7.54

Solve and write the solution in interval notation: $\frac{x - 1}{x + 3} \geq 0 . \frac{x - 1}{x + 3} \geq 0 .$

Show solution

Solution

Step 1. Write the inequality as one quotient on the left and zero on the right.

Our inequality is in this form. $\frac{x - 1}{x + 3} \geq 0$

Step 2. Determine the critical points—the points where the rational expression will be zero or undefined.

The rational expression will be zero when the numerator is zero. Since $x - 1 = 0 x - 1 = 0$ when $x = 1, x = 1,$ then $11$ is a critical point.

The rational expression will be undefined when the denominator is zero. Since $x + 3 = 0 x + 3 = 0$ when $x = −3, x = −3,$ then $−3 −3$ is a critical point.

The critical points are 1 and $−3 . −3 .$

Step 3. Use the critical points to divide the number line into intervals.

This figure shows a number line divided into three intervals by its critical points marked at negative 3 and 0.

The number line is divided into three intervals:

$(− \infty, −3) (−3, 1) (1, \infty)$

Step 4. Test a value in each interval. Above the number line show the sign of each factor of the rational expression in each interval. Below the number line show the sign of the quotient.

To find the sign of each factor in an interval, we choose any point in that interval and use it as a test point. Any point in the interval will give the expression the same sign, so we can choose any point in the interval.

Interval (− \infty, −3) Interval

The number $−4 −4$ is in the interval $(− \infty, −3) . (− \infty, −3) .$ Test $x = −4 x = −4$ in the expression in the numerator and the denominator.

This figure labels the expression, x minus 1, as the “numerator”. It shows that when negative 4 is substituted into the expression for x, the result is negative 5. It labels the result as “negative”. It also labels the expression, x plus 3, as “the denominator”. It shows that when negative 4 is substituted into the expression for x, the result is negative 1. It labels the result “negative”.

Above the number line, mark the factor $x - 1 x - 1$ negative and mark the factor $x + 3 x + 3$ negative.

Since a negative divided by a negative is positive, mark the quotient positive in the interval $(− \infty, −3) . (− \infty, −3) .$

This figure shows the quotient of the quantity x minus 1 and the quantity x plus 3, the numerator is negative and the denominator is negative, which is positive. It shows a number line divided into three intervals by its critical points marked at negative 3 and 0. The factors x minus 1 and x plus 3 are marked as negative above the number line for the interval negative infinity to negative 3. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative infinity to negative 3.

Interval (−3, 1) Interval

The number 0 is in the interval $(−3, 1) . (−3, 1) .$ Test $x = 0 . x = 0 .$

This figure labels the expression, x minus 1, as the “numerator”. It shows that when 0 is substituted into the expression for x, the result is negative 1. It labels the result as “negative”. It also labels the expression, x plus 3, as “the denominator”. It shows that when 0 is substituted into the expression for x, the result is 3. It labels the result “positive”.

Above the number line, mark the factor $x - 1 x - 1$ negative and mark $x + 3 x + 3$ positive.

Since a negative divided by a positive is negative, the quotient is marked negative in the interval $(−3, 1) . (−3, 1) .$

This figure shows a shows the quotient of the quantity x minus 1 and the quantity x plus 3, the numerator is negative and the denominator is positive, which is negative. It shows a number line divided into three intervals by its critical points marked at negative 3 and 0. The factors x minus 1 and x plus 3 are marked as negative above the number line for the interval negative infinity to negative 3. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative infinity to negative 3. The factor x minus 1 is marked as negative and the factor x plus 3 is marked as positive above the number line for the interval negative 3 to 1. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as negative below the number line for the interval negative 3 to 1.

Interval (1, \infty) Interval

The number 2 is in the interval $(1, \infty) . (1, \infty) .$ Test $x = 2 . x = 2 .$

This figure labels the expression, x minus 1, as the “numerator”. It shows that when 2 is substituted into the expression for x, the result is 1. It labels the result as “positive”. It also labels the expression, x plus 3, as “the denominator”. It shows that when 2 is substituted into the expression for x, the result is 5. It labels the result “positive”.

Above the number line, mark the factor $x - 1 x - 1$ positive and mark $x + 3 x + 3$ positive.

Since a positive divided by a positive is positive, mark the quotient positive in the interval $(1, \infty) . (1, \infty) .$

The figure shows that in the quotient of the quantity x minus 1 and the quantity x plus 3, the numerator is negative and the denominator is positive, which is negative. It shows a number line is divided into intervals by critical points at negative 3 and 1. The factors x minus 1 and x plus 3 are marked as negative above the number line for the interval negative infinity to negative 3. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative infinity to negative 3. The factor x minus 1 is marked as negative and the factor x plus 3 is marked as positive above the number line for the interval negative 3 to 1. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as negative below the number line for the interval negative 3 to 1. The factors x minus 1 and x plus 3 are marked as positive above the number line for the interval 1 to infinity. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative 1 to infinity.

Step 5. Determine the intervals where the inequality is correct. Write the solution in interval notation.

We want the quotient to be greater than or equal to zero, so the numbers in the intervals $(− \infty, −3) (− \infty, −3)$ and $(1, \infty) (1, \infty)$ are solutions.

But what about the critical points?

The critical point $x = −3 x = −3$ makes the denominator 0, so it must be excluded from the solution and we mark it with a parenthesis.

The critical point $x = 1 x = 1$ makes the whole rational expression 0. The inequality requires that the rational expression be greater than or equal to. So, 1 is part of the solution and we will mark it with a bracket.

The number line is divided into intervals by critical points at negative 3 and 1. A closed parenthesis is used at 3 and an open bracket is used at 1. The number is shaded to the left of 3 and to the right of 1. The factors x minus 1 and x plus 3 are marked as negative above the number line for the interval negative infinity to negative 3. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative infinity to negative 3. The factor x minus 1 is marked as negative and the factor x plus 3 is marked as positive above the number line for the interval negative 3 to 1. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as negative below the number line for the interval negative 3 to 1. The factors x minus 1 and x plus 3 are marked as positive above the number line for the interval 1 to infinity. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative 1 to infinity.

Recall that when we have a solution made up of more than one interval we use the union symbol, $\cup, \cup,$ to connect the two intervals. The solution in interval notation is $(− \infty, −3) \cup [1, \infty) . (− \infty, −3) \cup [1, \infty) .$

Try It 7.107

Solve and write the solution in interval notation: $\frac{x - 2}{x + 4} \geq 0 . \frac{x - 2}{x + 4} \geq 0 .$

Try It 7.108

Solve and write the solution in interval notation: $\frac{x + 2}{x - 4} \geq 0 . \frac{x + 2}{x - 4} \geq 0 .$

We summarize the steps for easy reference.

How To

Solve a rational inequality.

Step 1. Write the inequality as one quotient on the left and zero on the right.
Step 2. Determine the critical points–the points where the rational expression will be zero or undefined.
Step 3. Use the critical points to divide the number line into intervals.
Step 4. Test a value in each interval. Above the number line show the sign of each factor of the numerator and denominator in each interval. Below the number line show the sign of the quotient.
Step 5. Determine the intervals where the inequality is correct. Write the solution in interval notation.

The next example requires that we first get the rational inequality into the correct form.

Example 7.55

Solve and write the solution in interval notation: $\frac{4 x}{x - 6} < 1 . \frac{4 x}{x - 6} < 1 .$

Show solution

Solution

	$\frac{4 x}{x - 6} < 1 \frac{4 x}{x - 6} < 1$
Subtract 1 to get zero on the right.	$\frac{4 x}{x - 6} - 1 < 0 \frac{4 x}{x - 6} - 1 < 0$
Rewrite 1 as a fraction using the LCD.	$\frac{4 x}{x - 6} - \frac{x - 6}{x - 6} < 0 \frac{4 x}{x - 6} - \frac{x - 6}{x - 6} < 0$
Subtract the numerators and place the difference over the common denominator.	$\frac{4 x - (x - 6)}{x - 6} < 0 \frac{4 x - (x - 6)}{x - 6} < 0$
Simplify.	$\frac{3 x + 6}{x - 6} < 0 \frac{3 x + 6}{x - 6} < 0$
Factor the numerator to show all factors.	$\frac{3 (x + 2)}{x - 6} < 0 \frac{3 (x + 2)}{x - 6} < 0$
Find the critical points.
The quotient will be zero when the numerator is zero. The quotient is undefined when the denominator is zero.	$\begin{matrix} x + 2 & = & 0 & x - 6 & = & 0 \\ x & = & − 2 & x & = & 6 \end{matrix} \begin{matrix} x + 2 & = & 0 & x - 6 = 0 \\ x & = & − 2 & x = 6 \end{matrix}$
Use the critical points to divide the number line into intervals.

Test a value in each interval.

Above the number line show the sign of each factor of the rational expression in each interval. Below the number line show the sign of the quotient.

Determine the intervals where the inequality is correct. We want the quotient to be negative, so the solution includes the points between −2 and 6. Since the inequality is strictly less than, the endpoints are not included.
We write the solution in interval notation as (−2, 6).

Try It 7.109

Solve and write the solution in interval notation: $\frac{3 x}{x - 3} < 1 . \frac{3 x}{x - 3} < 1 .$

Try It 7.110

Solve and write the solution in interval notation: $\frac{3 x}{x - 4} < 2 . \frac{3 x}{x - 4} < 2 .$

In the next example, the numerator is always positive, so the sign of the rational expression depends on the sign of the denominator.

Example 7.56

Solve and write the solution in interval notation: $\frac{5}{x^{2} - 2 x - 15} > 0 . \frac{5}{x^{2} - 2 x - 15} > 0 .$

Show solution

Solution

The inequality is in the correct form.	$\frac{5}{x^{2} - 2 x - 15} > 0 \frac{5}{x^{2} - 2 x - 15} > 0$
Factor the denominator.	$\frac{5}{(x + 3) (x - 5)} > 0 \frac{5}{(x + 3) (x - 5)} > 0$
Find the critical points. The quotient is 0 when the numerator is 0. Since the numerator is always 5, the quotient cannot be 0.
The quotient will be undefined when the denominator is zero.	$\begin{matrix} (x + 3) (x - 5) = 0 \\ x = - 3, x = 5 \end{matrix} \begin{matrix} (x + 3) (x - 5) = 0 \\ x = - 3, x = 5 \end{matrix}$
Use the critical points to divide the number line into intervals.

Test values in each interval. Above the number line show the sign of each factor of the denominator in each interval. Below the number line, show the sign of the quotient.
Write the solution in interval notation.	$(− \infty, - 3) \cup (5, \infty) (− \infty, - 3) \cup (5, \infty)$

Try It 7.111

Solve and write the solution in interval notation: $\frac{1}{x^{2} + 2 x - 8} > 0 . \frac{1}{x^{2} + 2 x - 8} > 0 .$

Try It 7.112

Solve and write the solution in interval notation: $\frac{3}{x^{2} + x - 12} > 0 . \frac{3}{x^{2} + x - 12} > 0 .$

The next example requires some work to get it into the needed form.

Example 7.57

Solve and write the solution in interval notation: $\frac{1}{3} - \frac{2}{x^{2}} < \frac{5}{3 x} . \frac{1}{3} - \frac{2}{x^{2}} < \frac{5}{3 x} .$

Show solution

Solution

	$\frac{1}{3} - \frac{2}{x^{2}} < \frac{5}{3 x} \frac{1}{3} - \frac{2}{x^{2}} < \frac{5}{3 x}$
Subtract $\frac{5}{3 x} \frac{5}{3 x}$ to get zero on the right.	$\frac{1}{3} - \frac{2}{x^{2}} - \frac{5}{3 x} < 0 \frac{1}{3} - \frac{2}{x^{2}} - \frac{5}{3 x} < 0$
Rewrite to get each fraction with the LCD $3 x^{2} . 3 x^{2} .$	$\frac{1 \cdot x^{2}}{3 \cdot x^{2}} - \frac{2 \cdot 3}{x^{2} \cdot 3} - \frac{5 \cdot x}{3 x \cdot x} < 0 \frac{1 \cdot x^{2}}{3 \cdot x^{2}} - \frac{2 \cdot 3}{x^{2} \cdot 3} - \frac{5 \cdot x}{3 x \cdot x} < 0$
Simplify.	$\frac{x^{2}}{3 x^{2}} - \frac{6}{3 x^{2}} - \frac{5 x}{3 x^{2}} < 0 \frac{x^{2}}{3 x^{2}} - \frac{6}{3 x^{2}} - \frac{5 x}{3 x^{2}} < 0$
Subtract the numerators and place the difference over the common denominator.	$\frac{x^{2} - 5 x - 6}{3 x^{2}} < 0 \frac{x^{2} - 5 x - 6}{3 x^{2}} < 0$
Factor the numerator.	$\frac{(x - 6) (x + 1)}{3 x^{2}} < 0 \frac{(x - 6) (x + 1)}{3 x^{2}} < 0$
Find the critical points.	$\begin{matrix} 3 x^{2} & = & 0 & x - 6 & = & 0 & x + 1 & = & 0 \\ x & = & 0 & x & = & 6 & x & = & − 1 \end{matrix} \begin{matrix} 3 x^{2} & = & 0 & x - 6 = 0 x + 1 = 0 & x & = & 0 & x = 6 x = − 1 \end{matrix}$
Use the critical points to divide the number line into intervals.

Above the number line show the sign of each factor in each interval. Below the number line, show the sign of the quotient.
Since, 0 is excluded, the solution is the two intervals, $(−1, 0) (−1, 0)$ and $(0, 6) . (0, 6) .$	$(- 1, 0) \cup (0, 6) (- 1, 0) \cup (0, 6)$

Try It 7.113

Solve and write the solution in interval notation: $\frac{1}{2} + \frac{4}{x^{2}} < \frac{3}{x} . \frac{1}{2} + \frac{4}{x^{2}} < \frac{3}{x} .$

Try It 7.114

Solve and write the solution in interval notation: $\frac{1}{3} + \frac{6}{x^{2}} < \frac{3}{x} . \frac{1}{3} + \frac{6}{x^{2}} < \frac{3}{x} .$

Solve an Inequality with Rational Functions

When working with rational functions, it is sometimes useful to know when the function is greater than or less than a particular value. This leads to a rational inequality.

Example 7.58

Given the function $R (x) = \frac{x + 3}{x - 5}, R (x) = \frac{x + 3}{x - 5},$ find the values of x that make the function less than or equal to 0.

Show solution

Solution

We want the function to be less than or equal to 0.

	$R (x) \leq 0 R (x) \leq 0$
Substitute the rational expression for $R (x) . R (x) .$	$\frac{x + 3}{x - 5} \leq 0 x \neq 5 \frac{x + 3}{x - 5} \leq 0$
Find the critical points.	$\begin{matrix} x + 3 & = & 0 & x - 5 & = & 0 \\ x & = & −3 & x & = & 5 \end{matrix} \begin{matrix} x + 3 & = & 0 & x - 5 = 0 \\ x & = & −3 & x = 5 \end{matrix}$
Use the critical points to divide the number line into intervals.

Test values in each interval. Above the number line, show the sign of each factor in each interval. Below the number line, show the sign of the quotient
Write the solution in interval notation. Since 5 is excluded we, do not include it in the interval.	$[−3, 5) [−3, 5)$

Try It 7.115

Given the function $R (x) = \frac{x - 2}{x + 4}, R (x) = \frac{x - 2}{x + 4},$ find the values of x that make the function less than or equal to 0.

Try It 7.116

Given the function $R (x) = \frac{x + 1}{x - 4}, R (x) = \frac{x + 1}{x - 4},$ find the values of x that make the function less than or equal to 0.

In economics, the function $C (x) C (x)$ is used to represent the cost of producing x units of a commodity. The average cost per unit can be found by dividing $C (x) C (x)$ by the number of items $x . x .$ Then, the average cost per unit is $c (x) = \frac{C (x)}{x} . c (x) = \frac{C (x)}{x} .$

Example 7.59

The function $C (x) = 10 x + 3000 C (x) = 10 x + 3000$ represents the cost to produce $x, x,$ number of items. Find ⓐ the average cost function, $c (x) c (x)$ ⓑ how many items should be produced so that the average cost is less than $40.

Show solution

Solution

ⓐ

$\begin{matrix} C (x) = 10 x + 3000 \\ The average cost function is c (x) = \frac{C (x)}{x} . \\ \begin{matrix} To find the average cost function, divide the \\ cost function by x . \end{matrix} & \begin{matrix} c (x) = \frac{C (x)}{x} \\ c (x) = \frac{10 x + 3000}{x} \end{matrix} \\ The average cost function is c (x) = \frac{10 x + 3000}{x} . \end{matrix} \begin{matrix} C (x) = 10 x + 3000 \\ The average cost function is \\ \begin{matrix} To find the average cost function, divide the \\ cost function by \end{matrix} & \begin{matrix} c (x) = \frac{C (x)}{x} \\ c (x) = \frac{10 x + 3000}{x} \end{matrix} \\ The average cost function is \end{matrix}$

ⓑ

$\begin{matrix} We want the function c (x) to be less than 40 . & c (x) < 40 \\ Substitute the rational expression for c (x) . & \frac{10 x + 3000}{x} < 40 x \neq 0 \\ Subtract 40 to get 0 on the right. & \frac{10 x + 3000}{x} - 40 < 0 \\ \begin{matrix} Rewrite the left side as one quotient by finding \\ the LCD and performing the subtraction. \end{matrix} & \frac{10 x + 3000}{x} - 40 (\frac{x}{x}) < 0 \\ \frac{10 x + 3000}{x} - \frac{40 x}{x} < 0 \\ \frac{10 x + 3000 - 40 x}{x} < 0 \\ \frac{−30 x + 3000}{x} < 0 \\ Factor the numerator to show all factors. & \frac{−30 (x - 100)}{x} < 0 \\ Find the critical points. & \begin{matrix} - 30 (x - 100) & = & 0 & x = 0 \\ - 30 \neq 0 x - 100 & = & 0 \\ x & = & 100 \end{matrix} \end{matrix} \begin{matrix} We want the function \\ Substitute the rational expression for \\ Subtract 40 to get 0 on the right. \\ \begin{matrix} Rewrite the left side as one quotient by finding \\ the LCD and performing the subtraction. \end{matrix} & \frac{10 x + 3000}{x} - 40 (\frac{x}{x}) < 0 \\ Factor the numerator to show all factors. \\ Find the critical points. & \begin{matrix} - 30 (x - 100) & = & 0 \\ - 30 \neq 0 & = & 0 \\ x & = & 100 \end{matrix} \end{matrix}$

More than 100 items must be produced to keep the average cost below $40 per item.

Try It 7.117

The function $C (x) = 20 x + 6000 C (x) = 20 x + 6000$ represents the cost to produce $x, x,$ number of items. Find ⓐ the average cost function, $c (x) c (x)$ ⓑ how many items should be produced so that the average cost is less than $60?

Try It 7.118

The function $C (x) = 5 x + 900 C (x) = 5 x + 900$ represents the cost to produce $x, x,$ number of items. Find ⓐ the average cost function, $c (x) c (x)$ ⓑ how many items should be produced so that the average cost is less than $20?

Practice Makes Perfect

Solve Rational Inequalities

In the following exercises, solve each rational inequality and write the solution in interval notation.

339.

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

340.

$\frac{x + 6}{x - 5} \geq 0 \frac{x + 6}{x - 5} \geq 0$

341.

$\frac{x + 1}{x - 3} \leq 0 \frac{x + 1}{x - 3} \leq 0$

342.

$\frac{x - 4}{x + 2} \leq 0 \frac{x - 4}{x + 2} \leq 0$

343.

$\frac{x - 7}{x - 1} > 0 \frac{x - 7}{x - 1} > 0$

344.

$\frac{x + 8}{x + 3} > 0 \frac{x + 8}{x + 3} > 0$

345.

$\frac{x - 6}{x + 5} < 0 \frac{x - 6}{x + 5} < 0$

346.

$\frac{x + 5}{x - 2} < 0 \frac{x + 5}{x - 2} < 0$

347.

$\frac{3 x}{x - 5} < 1 \frac{3 x}{x - 5} < 1$

348.

$\frac{5 x}{x - 2} < 1 \frac{5 x}{x - 2} < 1$

349.

$\frac{6 x}{x - 6} > 2 \frac{6 x}{x - 6} > 2$

350.

$\frac{3 x}{x - 4} > 2 \frac{3 x}{x - 4} > 2$

351.

$\frac{2 x + 3}{x - 6} \leq 1 \frac{2 x + 3}{x - 6} \leq 1$

352.

$\frac{4 x - 1}{x - 4} \leq 1 \frac{4 x - 1}{x - 4} \leq 1$

353.

$\frac{3 x - 2}{x - 4} \geq 2 \frac{3 x - 2}{x - 4} \geq 2$

354.

$\frac{4 x - 3}{x - 3} \geq 2 \frac{4 x - 3}{x - 3} \geq 2$

355.

$\frac{1}{x^{2} + 7 x + 12} > 0 \frac{1}{x^{2} + 7 x + 12} > 0$

356.

$\frac{1}{x^{2} - 4 x - 12} > 0 \frac{1}{x^{2} - 4 x - 12} > 0$

357.

$\frac{3}{x^{2} - 5 x + 4} < 0 \frac{3}{x^{2} - 5 x + 4} < 0$

358.

$\frac{4}{x^{2} + 7 x + 12} < 0 \frac{4}{x^{2} + 7 x + 12} < 0$

359.

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

360.

$\frac{6}{3 x^{2} - 2 x - 5} \geq 0 \frac{6}{3 x^{2} - 2 x - 5} \geq 0$

361.

$\frac{−2}{6 x^{2} - 13 x + 6} \leq 0 \frac{−2}{6 x^{2} - 13 x + 6} \leq 0$

362.

$\frac{−1}{10 x^{2} + 11 x - 6} \leq 0 \frac{−1}{10 x^{2} + 11 x - 6} \leq 0$

363.

$\frac{1}{2} + \frac{12}{x^{2}} > \frac{5}{x} \frac{1}{2} + \frac{12}{x^{2}} > \frac{5}{x}$

364.

$\frac{1}{3} + \frac{1}{x^{2}} > \frac{4}{3 x} \frac{1}{3} + \frac{1}{x^{2}} > \frac{4}{3 x}$

365.

$\frac{1}{2} - \frac{4}{x^{2}} \leq \frac{1}{x} \frac{1}{2} - \frac{4}{x^{2}} \leq \frac{1}{x}$

366.

$\frac{1}{2} - \frac{3}{2 x^{2}} \geq \frac{1}{x} \frac{1}{2} - \frac{3}{2 x^{2}} \geq \frac{1}{x}$

367.

$\frac{1}{x^{2} - 16} < 0 \frac{1}{x^{2} - 16} < 0$

368.

$\frac{4}{x^{2} - 25} > 0 \frac{4}{x^{2} - 25} > 0$

369.

$\frac{4}{x - 2} \geq \frac{3}{x + 1} \frac{4}{x - 2} \geq \frac{3}{x + 1}$

370.

$\frac{5}{x - 1} \leq \frac{4}{x + 2} \frac{5}{x - 1} \leq \frac{4}{x + 2}$

Solve an Inequality with Rational Functions

In the following exercises, solve each rational function inequality and write the solution in interval notation.

371.

Given the function $R (x) = \frac{x - 5}{x - 2}, R (x) = \frac{x - 5}{x - 2},$ find the values of $x x$ that make the function less than or equal to 0.

372.

Given the function $R (x) = \frac{x + 1}{x + 3}, R (x) = \frac{x + 1}{x + 3},$ find the values of $x x$ that make the function less than or equal to 0.

373.

Given the function $R (x) = \frac{x - 6}{x + 2} R (x) = \frac{x - 6}{x + 2}$ , find the values of x that make the function less than or equal to 0.

374.

Given the function $R (x) = \frac{x + 1}{x - 4}, R (x) = \frac{x + 1}{x - 4},$ find the values of x that make the function less than or equal to 0.

Writing Exercises

375.

Write the steps you would use to explain solving rational inequalities to your little brother.

376.

Create a rational inequality whose solution is $(− \infty, −2] \cup [4, \infty) . (− \infty, −2] \cup [4, \infty) .$

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four columns and three rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was solve rational inequalities. In row 3, the I can was solve an inequality with rational functions.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?