• 7.6 Solve Rational Inequalities
• Introduction
• 1.1 Use the Language of Algebra
• 1.2 Integers
• 1.3 Fractions
• 1.4 Decimals
• 1.5 Properties of Real Numbers
• Key Concepts
• Review Exercises
• Practice Test
• 2.1 Use a General Strategy to Solve Linear Equations
• 2.2 Use a Problem Solving Strategy
• 2.3 Solve a Formula for a Specific Variable
• 2.4 Solve Mixture and Uniform Motion Applications
• 2.5 Solve Linear Inequalities
• 2.6 Solve Compound Inequalities
• 2.7 Solve Absolute Value Inequalities
• 3.1 Graph Linear Equations in Two Variables
• 3.2 Slope of a Line
• 3.3 Find the Equation of a Line
• 3.4 Graph Linear Inequalities in Two Variables
• 3.5 Relations and Functions
• 3.6 Graphs of Functions
• 4.1 Solve Systems of Linear Equations with Two Variables
• 4.2 Solve Applications with Systems of Equations
• 4.3 Solve Mixture Applications with Systems of Equations
• 4.4 Solve Systems of Equations with Three Variables
• 4.5 Solve Systems of Equations Using Matrices
• 4.6 Solve Systems of Equations Using Determinants
• 4.7 Graphing Systems of Linear Inequalities
• 5.1 Add and Subtract Polynomials
• 5.2 Properties of Exponents and Scientific Notation
• 5.3 Multiply Polynomials
• 5.4 Dividing Polynomials
• Introduction to Factoring
• 6.1 Greatest Common Factor and Factor by Grouping
• 6.2 Factor Trinomials
• 6.3 Factor Special Products
• 6.4 General Strategy for Factoring Polynomials
• 6.5 Polynomial Equations
• 7.1 Multiply and Divide Rational Expressions
• 7.2 Add and Subtract Rational Expressions
• 7.3 Simplify Complex Rational Expressions
• 7.4 Solve Rational Equations
• 7.5 Solve Applications with Rational Equations
• 8.1 Simplify Expressions with Roots
• 8.3 Simplify Rational Exponents
• 8.7 Use Radicals in Functions
• 8.8 Use the Complex Number System
• 9.1 Solve Quadratic Equations Using the Square Root Property
• 9.2 Solve Quadratic Equations by Completing the Square
• 9.4 Solve Equations in Quadratic Form
• 9.5 Solve Applications of Quadratic Equations
• 9.6 Graph Quadratic Functions Using Properties
• 9.7 Graph Quadratic Functions Using Transformations
• 10.1 Finding Composite and Inverse Functions
• 10.2 Evaluate and Graph Exponential Functions
• 10.3 Evaluate and Graph Logarithmic Functions
• 10.4 Use the Properties of Logarithms
• 10.5 Solve Exponential and Logarithmic Equations
• 11.1 Distance and Midpoint Formulas; Circles
• 11.2 Parabolas
• 11.3 Ellipses
• 11.4 Hyperbolas
• 11.5 Solve Systems of Nonlinear Equations
• 12.1 Sequences
• 12.2 Arithmetic Sequences
• 12.3 Geometric Sequences and Series
• 12.4 Binomial Theorem ## Learning Objectives

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

• Solve rational inequalities
• Solve an inequality with rational functions

## Be Prepared 7.16

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 .

## Be Prepared 7.17

Solve: 8 − 2 x < 12 . 8 − 2 x < 12 . If you missed this problem, review Example 2.52 .

## Be Prepared 7.18

Write in interval notation: −3 ≤ x < 5 . −3 ≤ 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 3 2 x > 1 , 2 x x − 3 < 4 , 2 x − 3 x − 6 ≥ x , 3 2 x > 1 , 2 x x − 3 < 4 , 2 x − 3 x − 6 ≥ x , and 1 4 − 2 x 2 ≤ 3 x 1 4 − 2 x 2 ≤ 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.

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: x − 1 x + 3 ≥ 0 . x − 1 x + 3 ≥ 0 .

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

Our inequality is in this form. x − 1 x + 3 ≥ 0 x − 1 x + 3 ≥ 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 1 1 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.

The number line is divided into three intervals:

( − ∞ , −3 ) ( −3 , 1 ) ( 1 , ∞ ) ( − ∞ , −3 ) ( −3 , 1 ) ( 1 , ∞ )

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.

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

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 ( − ∞ , −3 ) . ( − ∞ , −3 ) .

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

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 ) .

The number 2 is in the interval ( 1 , ∞ ) . ( 1 , ∞ ) . Test x = 2 . x = 2 .

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 , ∞ ) . ( 1 , ∞ ) .

• 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 ( − ∞ , −3 ) ( − ∞ , −3 ) and ( 1 , ∞ ) ( 1 , ∞ ) 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 0. So, 1 is part of the solution and we will mark it with a bracket.

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

## Try It 7.107

Solve and write the solution in interval notation: x − 2 x + 4 ≥ 0 . x − 2 x + 4 ≥ 0 .

## Try It 7.108

Solve and write the solution in interval notation: x + 2 x − 4 ≥ 0 . x + 2 x − 4 ≥ 0 .

We summarize the steps for easy reference.

## Solve a rational inequality.

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

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: 4 x x − 6 < 1 . 4 x x − 6 < 1 .

## Try It 7.109

Solve and write the solution in interval notation: 3 x x − 3 < 1 . 3 x x − 3 < 1 .

## Try It 7.110

Solve and write the solution in interval notation: 3 x x − 4 < 2 . 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: 5 x 2 − 2 x − 15 > 0 . 5 x 2 − 2 x − 15 > 0 .

## Try It 7.111

Solve and write the solution in interval notation: 1 x 2 + 2 x − 8 > 0 . 1 x 2 + 2 x − 8 > 0 .

## Try It 7.112

Solve and write the solution in interval notation: 3 x 2 + x − 12 > 0 . 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: 1 3 − 2 x 2 < 5 3 x . 1 3 − 2 x 2 < 5 3 x .

## Try It 7.113

Solve and write the solution in interval notation: 1 2 + 4 x 2 < 3 x . 1 2 + 4 x 2 < 3 x .

## Try It 7.114

Solve and write the solution in interval notation: 1 3 + 6 x 2 < 3 x . 1 3 + 6 x 2 < 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 ) = x + 3 x − 5 , R ( x ) = x + 3 x − 5 , find the values of x that make the function less than or equal to 0.

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

## Try It 7.115

Given the function R ( x ) = x − 2 x + 4 , R ( x ) = 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 ) = x + 1 x − 4 , R ( x ) = 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 ) = C ( x ) x . c ( x ) = 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. More than 100 items must be produced to keep the average cost below$40 per item.

## Section 7.6 Exercises

Practice makes perfect.

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

x − 3 x + 4 ≥ 0 x − 3 x + 4 ≥ 0

x + 6 x − 5 ≥ 0 x + 6 x − 5 ≥ 0

x + 1 x − 3 ≤ 0 x + 1 x − 3 ≤ 0

x − 4 x + 2 ≤ 0 x − 4 x + 2 ≤ 0

x − 7 x − 1 > 0 x − 7 x − 1 > 0

x + 8 x + 3 > 0 x + 8 x + 3 > 0

x − 6 x + 5 < 0 x − 6 x + 5 < 0

x + 5 x − 2 < 0 x + 5 x − 2 < 0

3 x x − 5 < 1 3 x x − 5 < 1

5 x x − 2 < 1 5 x x − 2 < 1

6 x x − 6 > 2 6 x x − 6 > 2

3 x x − 4 > 2 3 x x − 4 > 2

2 x + 3 x − 6 ≤ 1 2 x + 3 x − 6 ≤ 1

4 x − 1 x − 4 ≤ 1 4 x − 1 x − 4 ≤ 1

3 x − 2 x − 4 ≥ 2 3 x − 2 x − 4 ≥ 2

4 x − 3 x − 3 ≥ 2 4 x − 3 x − 3 ≥ 2

1 x 2 + 7 x + 12 > 0 1 x 2 + 7 x + 12 > 0

1 x 2 − 4 x − 12 > 0 1 x 2 − 4 x − 12 > 0

3 x 2 − 5 x + 4 < 0 3 x 2 − 5 x + 4 < 0

4 x 2 + 7 x + 12 < 0 4 x 2 + 7 x + 12 < 0

2 2 x 2 + x − 15 ≥ 0 2 2 x 2 + x − 15 ≥ 0

6 3 x 2 − 2 x − 5 ≥ 0 6 3 x 2 − 2 x − 5 ≥ 0

−2 6 x 2 − 13 x + 6 ≤ 0 −2 6 x 2 − 13 x + 6 ≤ 0

−1 10 x 2 + 11 x − 6 ≤ 0 −1 10 x 2 + 11 x − 6 ≤ 0

1 2 + 12 x 2 > 5 x 1 2 + 12 x 2 > 5 x

1 3 + 1 x 2 > 4 3 x 1 3 + 1 x 2 > 4 3 x

1 2 − 4 x 2 ≤ 1 x 1 2 − 4 x 2 ≤ 1 x

1 2 − 3 2 x 2 ≥ 1 x 1 2 − 3 2 x 2 ≥ 1 x

1 x 2 − 16 < 0 1 x 2 − 16 < 0

4 x 2 − 25 > 0 4 x 2 − 25 > 0

4 x − 2 ≥ 3 x + 1 4 x − 2 ≥ 3 x + 1

5 x − 1 ≤ 4 x + 2 5 x − 1 ≤ 4 x + 2

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

Given the function R ( x ) = x − 5 x − 2 , R ( x ) = x − 5 x − 2 , find the values of x x that make the function less than or equal to 0.

Given the function R ( x ) = x + 1 x + 3 , R ( x ) = x + 1 x + 3 , find the values of x x that make the function greater than or equal to 0.

Given the function R ( x ) = x − 6 x + 2 R ( x ) = x − 6 x + 2 , find the values of x that make the function less than or equal to 0.

## Writing Exercises

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

Create a rational inequality whose solution is ( − ∞ , −2 ] ∪ [ 4 , ∞ ) . ( − ∞ , −2 ] ∪ [ 4 , ∞ ) .

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

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

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## Learning Objectives

• Solve rational inequalities
• Solve an inequality with rational functions

## Be Prepared

Before you get started, take this readiness quiz.

• Find the value of $$x-5$$ when ⓐ $$x=6$$ ⓑ $$x=-3$$ ⓒ $$x=5$$ If you missed this problem, review Example 1.2.16 .
• Solve: $$8-2 x<12$$ If you missed this problem, review Example 2.6.13 .
• Write in interval notation: $$-3 \leq x<5$$ If you missed this problem, review Example 2.6.4 .

## 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$$\quad \dfrac{3}{2 x}>1, \quad \dfrac{2 x}{x-3}<4, \quad \dfrac{2 x-3}{x-6} \geq x,\quad$$ and $$\quad \dfrac{1}{4}-\dfrac{2}{x^{2}} \leq \dfrac{3}{x}\quad$$ 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 $$x=3$$, we know there is one solution, which is 3.

When we solve an inequality and the result is $$x>3$$, 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. 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 $$\PageIndex{1}$$

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

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

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

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 $$x-1=0$$ when $$x=1$$, then 1 is a critical point.

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

The critical points are 1 and -3.

Step 3 . Use the critical points to divide the number line into intervals. The number line is divided into three intervals:

$(-\infty,-3) \quad (-3,1) \quad (1,\infty) \nonumber$

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.

$\text { Interval }(-\infty,-3) \nonumber$

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

The numerator:

$\begin{array}{l} {x-1} \\ {-4-1} \\ {-5} \\ {\text {Negative}} \end{array} \nonumber$

The denominator:

$\begin{array}{l} {x+3} \\ {-4+3} \\ {-1} \\ {\text {Negative}} \end{array} \nonumber$

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

Since a negative divided by a negative is positive, mark the quotient positive in the interval $$(-\infty,-3)$$ . $\text {Interval } (-3,1) \nonumber$

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

$\begin{array}{l} {x-1} \\ {0-1} \\ {-1} \\ {\text {Negative}} \end{array} \nonumber$

$\begin{array}{l} {x+3} \\ {0+3} \\ {3} \\ {\text {Positive}} \end{array} \nonumber$

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

Since a negative divided by a positive is negative, the quotient is marked negative in the interval $$(-3,1)$$. $\text {Interval }(1, \infty) \nonumber$

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

$\begin{array}{l} {x-1} \\ {2-1} \\ {1} \\ {\text {Positive}} \end{array} \nonumber$

$\begin{array}{l} {x+3} \\ {2+3} \\ {5} \\ {\text {Positive}} \end{array} \nonumber$

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

Since a positive divided by a positive is positive, mark the quotient positive in the interval $$(1, \infty)$$. 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 $$(-\infty,-3)$$ and $$(1, \infty)$$are solutions.

But what about the critical points?

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

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

## Exercise $$\PageIndex{1}$$

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

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

## Exercise $$\PageIndex{2}$$

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

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

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 $$\PageIndex{2}$$

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

$\dfrac{4 x}{x-6}<1 \nonumber$

Subtract 1 to get zero on the right.

$\dfrac{4 x}{x-6}-1<0 \nonumber$

Rewrite 1 as a fraction using the LCD.

$\dfrac{4 x}{x-6}-\frac{x-6}{x-6}<0 \nonumber$

Subtract the numerators and place the difference over the common denominator.

$\dfrac{4 x-(x-6)}{x-6}<0 \nonumber$

$\dfrac{3 x+6}{x-6}<0 \nonumber$

Factor the numerator to show all factors.

$\dfrac{3(x+2)}{x-6}<0 \nonumber$

Find the critical points.

The quotient will be zero when the numerator is zero. The quotient is undefined when the denominator is zero.

$\begin{array}{rlrl} {x+2} & {=0} & {x-6} & {=0} \\ {x} & {=-2} & {x} & {=6} \end{array} \nonumber$

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).

## Exercise $$\PageIndex{3}$$

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

$$\left(-\dfrac{3}{2}, 3\right)$$

## Exercise $$\PageIndex{4}$$

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

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

## Example $$\PageIndex{3}$$

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

The inequality is in the correct form.

$\dfrac{5}{x^{2}-2 x-15}>0 \nonumber$

Factor the denominator.

$\dfrac{5}{(x+3)(x-5)}>0 \nonumber$

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{aligned} &(x+3)(x-5)=0\\ &x=-3, x=5 \end{aligned} \nonumber 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.

$(-\infty,-3) \cup(5, \infty) \nonumber$

## Exercise $$\PageIndex{5}$$

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

$$(-\infty,-4) \cup(2, \infty)$$

## Exercise $$\PageIndex{6}$$

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

$$(-\infty,-4) \cup(3, \infty)$$

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

## Example $$\PageIndex{4}$$

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

$\dfrac{1}{3}-\dfrac{2}{x^{2}}<\dfrac{5}{3 x} \nonumber$

Subtract $$\dfrac{5}{3 x}$$ to get zero on the right.

$\dfrac{1}{3}-\dfrac{2}{x^{2}}-\dfrac{5}{3 x}<0 \nonumber$

Rewrite to get each fraction with the LCD $$3x^2$$.

$\dfrac{1 \cdot x^{2}}{3 \cdot x^{2}}-\dfrac{2 \cdot 3}{x^{2} \cdot 3}-\dfrac{5 \cdot x}{3 x-x}<0 \nonumber$

$\dfrac{x^{2}}{3 x^{2}}-\dfrac{6}{3 x^{2}}-\dfrac{5 x}{3 x^{2}}<0 \nonumber$

$\dfrac{x^{2}-5 x-6}{3 x^{2}}<0 \nonumber$

Factor the numerator.

$\dfrac{(x-6)(x+1)}{3 x^{2}}<0 \nonumber$

$\begin{array}{rlrl} {3 x^{2}=0} && {x-6=0} && {x+1=0} \\ {x=0} && {x=6} && {x=-1} \end{array} \nonumber$ 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 $$(-1,0) \cup(0,6)$$ intervals, $$(-1,0)$$ and $$(0,6)$$.

## Exercise $$\PageIndex{7}$$

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

## Exercise $$\PageIndex{8}$$

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

## 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 $$\PageIndex{5}$$

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

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

$R(x) \leq 0 \nonumber$

Substitute the rational expression for $$R(x)$$.

$\dfrac{x+3}{x-5} \leq 0 \quad x \neq 5 \nonumber$

$\begin{array}{rlrl} {x+3=0} && {x-5=0} \\ {x=-3} && {x=5} \end{array} \nonumber$ 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.

$[-3,5) \nonumber$

## Exercise $$\PageIndex{9}$$

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

## Exercise $$\PageIndex{10}$$

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

In economics, the function $$C(x)$$ is used to represent the cost of producing $$x$$ units of a commodity. The average cost per unit can be found by dividing $$C(x)$$ by the number of items $$x$$. Then, the average cost per unit is $$c(x)=\dfrac{C(x)}{x}). ## Example \(\PageIndex{6}$$

The function$$C(x)=10 x+3000$$ represents the cost to produce $$x$$, number of items. Find:

• The average cost function, $$c(x)$$
• How many items should be produced so that the average cost is less than 40. • $C(x)=10 x+3000 \nonumber$ The average cost function is $$c(x)=\dfrac{C(x)}{x})$$. To find the average cost function, divide the cost function by $$x$$. \begin{aligned} &c(x)=\dfrac{C(x)}{x}\\ &c(x)=\dfrac{10 x+3000}{x} \end{aligned} \nonumber The average cost function is $$c(x)=\dfrac{10 x+3000}{x}$$ • We want the function $$c(x)$$ to be less than 40. $c(x)<40 \nonumber$ Substitute the rational expression forc(x). $\dfrac{10 x+3000}{x}<40, \quad x \neq 0 \nonumber$ Subtract 40 to get 0 on the right. $\dfrac{10 x+3000}{x}-40<0 \nonumber$ Rewrite the left side as one quotient by finding the LCD and performing the subtraction. \begin{aligned} \dfrac{10 x+3000}{x}-40\left(\dfrac{x}{x}\right) &<0\\ \dfrac{10 x+3000}{x}-\dfrac{40 x}{x} &<0\\ \dfrac{10 x+3000-40 x}{x} &<0 \\ \dfrac{-30 x+3000}{x} &<0 \end{aligned} \nonumber $\begin{array}{ll} {\dfrac{-30(x-100)}{x}<0} \\ {-30(x-100)=0} && {x=0} \end{array} \nonumber$ $\begin{array}{rl} {-30 \neq 0} & {x-100=0} \\ &{x=100} \end{array} \nonumber$ More than 100 items must be produced to keep the average cost below40 per item.

## Exercise $$\PageIndex{11}$$

The function$$C(x)=20 x+6000$$ represents the cost to produce $$x$$, number of items. Find:

• How many items should be produced so that the average cost is less than $60. • $$c(x)=\dfrac{20 x+6000}{x}$$ • More than 150 items must be produced to keep the average cost below$60 per item.

## Exercise $$\PageIndex{12}$$

The function$$C(x)=5 x+900$$ represents the cost to produce $$x$$, number of items. Find:

• How many items should be produced so that the average cost is less than $20. • $$c(x)=\dfrac{5 x+900}{x}$$ • More than 60 items must be produced to keep the average cost below$20 per item.
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## Section 2.13 : Rational Inequalities

Solve each of the following inequalities.

• $$\displaystyle \frac{{4 - x}}{{x + 3}} > 0$$ Solution
• $$\displaystyle \frac{{2z - 5}}{{z - 7}} \le 0$$ Solution
• $$\displaystyle \frac{{{w^2} + 5w - 6}}{{w - 3}} \ge 0$$ Solution
• $$\displaystyle \frac{{3x + 8}}{{x - 1}} Solution • \( \displaystyle u \le \frac{4}{{u - 3}}$$ Solution
• $$\displaystyle \frac{{{t^3} - 6{t^2}}}{{t - 2}} > 0$$ Solution Rational Expressions and Functions

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

Before you get started, take this readiness quiz. 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.

A rational inequality is an inequality that contains 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. 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. 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. The number line is divided into three intervals: 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. Step 5. Determine the intervals where the inequality is correct. Write the solution in interval notation. But what about the critical points? We summarize the steps for easy reference.

• Write the inequality as one quotient on the left and zero on the right.
• Determine the critical points–the points where the rational expression will be zero or undefined.
• 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 numerator and denominator in each interval. Below the number line show the sign of the quotient.
• 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. In the next example, the numerator is always positive, so the sign of the rational expression depends on the sign of the denominator. The next example requires some work to get it into the needed form. 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. We want the function to be less than or equal to 0. More than 100 items must be produced to keep the average cost below ?40 per item. ## Key Concepts

• 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.

## Section Exercises

Practice makes perfect.

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

Write the steps you would use to explain solving rational inequalities to your little brother. ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section. ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

## Chapter Review Exercises

Simplify, multiply, and divide rational expressions.

Determine the Values for Which a Rational Expression is Undefined

In the following exercises, determine the values for which the rational expression is undefined. Simplify Rational Expressions

In the following exercises, simplify. Multiply Rational Expressions

In the following exercises, multiply. Divide Rational Expressions

In the following exercises, divide. Multiply and Divide Rational Functions ## Add and Subtract Rational Expressions

Add and Subtract Rational Expressions with a Common Denominator

In the following exercises, perform the indicated operations. Add and Subtract Rational Expressions Whose Denominators Are Opposites

In the following exercises, add and subtract. Find the Least Common Denominator of Rational Expressions

In the following exercises, find the LCD. Add and Subtract Rational Expressions with Unlike Denominators  ## Simplify Complex Rational Expressions

Simplify a Complex Rational Expression by Writing It as Division Simplify a Complex Rational Expression by Using the LCD ## 7.4 Solve Rational Equations

Solve Rational Equations

In the following exercises, solve. no solution Solve Rational Equations that Involve Functions Solve a Rational Equation for a Specific Variable

In the following exercises, solve for the indicated variable. ## Solve Applications with Rational Equations

Solve Proportions Solve Using Proportions

Rachael had a 21-ounce strawberry shake that has 739 calories. How many calories are there in a 32-ounce shake? Leo went to Mexico over Christmas break and changed ?525 dollars into Mexican pesos. At that time, the exchange rate had ?1 US is equal to 16.25 Mexican pesos. How many Mexican pesos did he get for his trip?

Solve Similar Figure Applications On a map of Europe, Paris, Rome, and Vienna form a triangle whose sides are shown in the figure below. If the actual distance from Rome to Vienna is 700 miles, find the distance from

ⓐ Paris to Rome

ⓑ Paris to Vienna Francesca is 5.75 feet tall. Late one afternoon, her shadow was 8 feet long. At the same time, the shadow of a nearby tree was 32 feet long. Find the height of the tree.

The height of a lighthouse in Pensacola, Florida is 150 feet. Standing next to the statue, 5.5-foot-tall Natasha cast a 1.1-foot shadow. How long would the shadow of the lighthouse be?

Solve Uniform Motion Applications

When making the 5-hour drive home from visiting her parents, Lolo ran into bad weather. She was able to drive 176 miles while the weather was good, but then driving 10 mph slower, went 81 miles when it turned bad. How fast did she drive when the weather was bad? Mark is riding on a plane that can fly 490 miles with a tailwind of 20 mph in the same time that it can fly 350 miles against a tailwind of 20 mph. What is the speed of the plane?

Josue can ride his bicycle 8 mph faster than Arjun can ride his bike. It takes Luke 3 hours longer than Josue to ride 48 miles. How fast can John ride his bike? Curtis was training for a triathlon. He ran 8 kilometers and biked 32 kilometers in a total of 3 hours. His running speed was 8 kilometers per hour less than his biking speed. What was his running speed?

Solve Work Applications

Brandy can frame a room in 1 hour, while Jake takes 4 hours. How long could they frame a room working together? Prem takes 3 hours to mow the lawn while her cousin, Barb, takes 2 hours. How long will it take them working together?

Jeffrey can paint a house in 6 days, but if he gets a helper he can do it in 4 days. How long would it take the helper to paint the house alone? Marta and Deb work together writing a book that takes them 90 days. If Sue worked alone it would take her 120 days. How long would it take Deb to write the book alone?

Solve Direct Variation Problems If the cost of a pizza varies directly with its diameter, and if an 8” diameter pizza costs ?12, how much would a 6” diameter pizza cost?

The distance to stop a car varies directly with the square of its speed. It takes 200 feet to stop a car going 50 mph. How many feet would it take to stop a car going 60 mph? Solve Inverse Variation Problems The number of tickets for a music fundraiser varies inversely with the price of the tickets. If Madelyn has just enough money to purchase 12 tickets for ?6, how many tickets can Madelyn afford to buy if the price increased to ?8? On a string instrument, the length of a string varies inversely with the frequency of its vibrations. If an 11-inch string on a violin has a frequency of 360 cycles per second, what frequency does a 12-inch string have? In the following exercises, solve each rational function inequality and write the solution in interval notation The function ## Practice Test In the following exercises, perform the indicated operation and simplify. In the following exercises, solve each equation. Given the function, Matheus can ride his bike for 30 miles with the wind in the same amount of time that he can go 21 miles against the wind. If the wind’s speed is 6 mph, what is Matheus’ speed on his bike?

Oliver can split a truckload of logs in 8 hours, but working with his dad they can get it done in 3 hours. How long would it take Oliver’s dad working alone to split the logs? The volume of a gas in a container varies inversely with the pressure on the gas. If a container of nitrogen has a volume of 29.5 liters with 2000 psi, what is the volume if the tank has a 14.7 psi rating? Round to the nearest whole number.

The cities of Dayton, Columbus, and Cincinnati form a triangle in southern Ohio. The diagram gives the map distances between these cities in inches. The actual distance from Dayton to Cincinnati is 48 miles. What is the actual distance between Dayton and Columbus?

The distance between Dayton and Columbus is 64 miles.

## Solving Rational Inequalities

A Rational Expression looks like:

## Inequalities

Sometimes we need to solve rational inequalities like these:

Solving inequalities is very like solving equations ... you do most of the same things.

These are the steps:

• the "=0" points (roots), and
• "vertical asymptotes" (where the function is undefined)
• in between the "points of interest", the function is either greater than zero (>0) or less than zero (<0)
• then pick a test value to find out which it is (>0 or <0)

Here is an example:

## Example: 3x−10 x−4 > 2

First , let us simplify!

But You Cannot Multiply By (x−4)

Because "x−4" could be positive or negative ... we don't know if we should change the direction of the inequality or not. This is all explained on Solving Inequalities .

Instead, bring "2" to the left:

3x−10 x−4 − 2 > 0

Then multiply 2 by (x−4)/(x−4):

3x−10 x−4 − 2 x−4 x−4 > 0

Now we have a common denominator, let's bring it all together:

3x−10 − 2(x−4) x−4 > 0

x−2 x−4 > 0

Second , let us find "points of interest".

At x=2 we have: (0)/(x−4) > 0 , which is a "=0" point, or root

At x=4 we have: (x−2)/(0) > 0 , which is undefined

Third , do test points to see what it does in between:

• x−2 = −2, which is negative
• x−4 = −4, which is also negative
• So (x−2)/(x−4) must be positive

We can do the same for x=3 and x=5 , and end up with these results:

That gives us a complete picture!

And where is it > 0 ?

• Less than 2
• More than 4

So our result is:

(−∞, 2) U (4, +∞)

We did all that without drawing a plot!

But here is the plot of (x−2)/(x−4) so you can see:

## Solving Rational Inequalities Practice Test

To solve a rational inequality, we first write the inequality in the format of a rational expression on one side and zero on the other. We will then find the boundaries or endpoints by setting the numerator and denominator each equal to zero and finding solutions for the equations. We can use our boundaries or endpoints to set up intervals on a number line and use test values to determine which intervals satisfy the inequality.

• Demonstrate the ability to solve a rational inequality
• Demonstrate the ability to write an inequality solution in interval notation
• Demonstrate the ability to graph an interval on the number line

Instructions: solve each inequality, write in interval notation, graph.

$$a)\hspace{.2em}\frac{x - 7}{x - 1}+ 5 > 0$$

$$b)\hspace{.2em}\frac{x + 5}{x - 4}- 1 < 0$$

Watch the Step by Step Video Solution   |   View the Written Solution

$$a)\hspace{.2em}\frac{x + 6}{x - 5}> 1$$

$$b)\hspace{.2em}\frac{5}{x - 1}> \frac{8}{2x + 7}$$

$$a)\hspace{.2em}\frac{3}{x + 2}≤ \frac{4}{x - 9}$$

$$b)\hspace{.2em}\frac{x^2 + 4x - 21}{x^2 - 6x + 9}> 0$$

$$a)\hspace{.2em}\frac{x^2 + 5x + 4}{x^2 + 5x + 6}< 0$$

$$b)\hspace{.2em}\frac{6}{x}> 6x - 5$$

$$a)\hspace{.2em}\frac{x^2 - 4x + 4}{x - 2}≥ 1$$

$$b)\hspace{.2em}\frac{x + 6}{x^2 - 5x - 24}≥ 0$$

$$a)\hspace{.2em}x < 1 \hspace{.2em}\text{or}\hspace{.2em}x > 2$$ $$(-\infty, 1) ∪ (2, \infty)$$

$$b)\hspace{.2em}x < 4$$ $$(-\infty, 4)$$

Watch the Step by Step Video Solution

$$a)\hspace{.2em}x > 5$$ $$(5, \infty)$$

$$b)\hspace{.2em}{-}\frac{43}{2}< x < -\frac{7}{2}\hspace{.2em}\text{or}\hspace{.2em}x > 1$$ $$\left(-\frac{43}{2}, -\frac{7}{2}\right) ∪ (1, \infty)$$

$$a)\hspace{.2em}{-}35 ≤ x < -2 \hspace{.2em}\text{or}\hspace{.2em}x > 9$$ $$[-35, -2) ∪ (9, \infty)$$

$$b)\hspace{.2em}x < -7 \hspace{.2em}\text{or}\hspace{.2em}x > 3$$ $$(-\infty, -7) ∪ (3, \infty)$$

$$a)\hspace{.2em}{-}4 < x < -3 \hspace{.2em}\text{or}\hspace{.2em}{-}2 < x < -1$$ $$(-4, -3) ∪ (-2, -1)$$

$$b)\hspace{.2em}x < -\frac{2}{3}\hspace{.2em}\text{or}\hspace{.2em}0 < x < \frac{3}{2}$$ $$\left(-\infty, -\frac{2}{3}\right) ∪ \left(0, \frac{3}{2}\right)$$

$$a)\hspace{.2em}x ≥ 3$$ $$[3, \infty)$$

$$b)\hspace{.2em}{-}6 ≤ x < -3 \hspace{.2em}\text{or}\hspace{.2em}x > 8$$ $$[-6, -3) ∪ (8, \infty)$$ ## We apologize for the inconvenience...

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