7.6 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 7.16

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.

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?

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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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 the 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
  • Greater than 4

So, after rearranging and analysing 3x−10 x−4 > 2 we get the result for x:

(−∞, 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:

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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.

x-5

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.

\frac{3}{2x}>1,\phantom{\rule{0.5em}{0ex}}\frac{2x}{x-3}<4,\phantom{\rule{0.5em}{0ex}}\frac{2x-3}{x-6}\ge x,

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.

x=3,

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.

\frac{x-1}{x+3}\ge 0.

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

\phantom{\rule{5em}{0ex}}\frac{x-1}{x+3}\ge 0

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

x-1=0

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:

\phantom{\rule{10.3em}{0ex}}\left(\text{−}\infty ,-3\right)\phantom{\rule{5em}{0ex}}\left(-3,1\right)\phantom{\rule{5.5em}{0ex}}\left(1,\infty \right)

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.

\mathbf{\text{Interval}}\phantom{\rule{0.2em}{0ex}}\left(\text{−}\infty ,-3\right)

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

\left(\text{−}\infty ,-3\right)

But what about the critical points?

x=1

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.

\frac{4x}{x-6}<1.

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

\frac{5}{{x}^{2}-2x-15}>0.

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

\frac{1}{3}-\frac{2}{{x}^{2}}<\frac{5}{3x}.

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.

R\left(x\right)=\frac{x+3}{x-5},

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

R\left(x\right)=\frac{x-2}{x+4},

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

C\left(x\right)=20x+6000

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.

\frac{x-3}{x+4}\ge 0

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

R\left(x\right)=\frac{x-5}{x-2},

Writing Exercises

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

Answers will vary.

\left(\text{−}\infty ,-2\right]\cup \left[4,\infty \right).

ⓐ 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?

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.

\frac{5a+3}{3a-2}

Simplify Rational Expressions

In the following exercises, simplify.

\frac{18}{24}

Multiply Rational Expressions

In the following exercises, multiply.

\frac{5}{8}·\frac{4}{15}

Divide Rational Expressions

In the following exercises, divide.

\frac{{x}^{2}-4x+12}{{x}^{2}+8x+12}÷\frac{{x}^{2}-36}{3x}

Multiply and Divide Rational Functions

R\left(x\right)=f\left(x\right)·g\left(x\right)

Add and Subtract Rational Expressions

Add and Subtract Rational Expressions with a Common Denominator

In the following exercises, perform the indicated operations.

\frac{7}{15}+\frac{8}{15}

Add and Subtract Rational Expressions Whose Denominators Are Opposites

In the following exercises, add and subtract.

\frac{18w}{6w-1}+\frac{3w-2}{1-6w}

Find the Least Common Denominator of Rational Expressions

In the following exercises, find the LCD.

\frac{7}{{a}^{2}-3a-10},\frac{3a}{{a}^{2}-a-20}

Add and Subtract Rational Expressions with Unlike Denominators

\frac{7}{5a}+\frac{3}{2b}

Add and Subtract Rational Functions

R\left(x\right)=f\left(x\right)+g\left(x\right)

Simplify Complex Rational Expressions

Simplify a Complex Rational Expression by Writing It as Division

\frac{\frac{7x}{x+2}}{\frac{14{x}^{2}}{{x}^{2}-4}}

Simplify a Complex Rational Expression by Using the LCD

\frac{\frac{1}{3}+\frac{1}{8}}{\frac{1}{4}+\frac{1}{12}}

7.4 Solve Rational Equations

Solve Rational Equations

In the following exercises, solve.

\frac{1}{2}+\frac{2}{3}=\frac{1}{x}

no solution

\frac{z}{12}+\frac{z+3}{3z}=\frac{1}{z}

Solve Rational Equations that Involve Functions

f\left(x\right)=\frac{x+2}{{x}^{2}-6x+8},

Solve a Rational Equation for a Specific Variable

In the following exercises, solve for the indicated variable.

\frac{V}{l}=hw

Solve Applications with Rational Equations

Solve Proportions

\frac{x}{4}=\frac{3}{5}

Solve Using Proportions

Rachael had a 21-ounce strawberry shake that has 739 calories. How many calories are there in a 32-ounce shake?

1161

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

\text{Δ}ABC

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

The figure is a triangle formed by Paris, Vienna, and Rome. The distance between Paris and Vienna is 7.7 centimeters. The distance between Vienna and Rome is 7 centimeters. The distance between Rome and Paris is 8.9 centimeters.

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?

45

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?

16

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?

\frac{4}{5}

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?

12

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

y

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?

288

Solve Inverse Variation Problems

m

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?

97

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?

\frac{x-3}{x+4}\le 0

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

\left(\text{−}\infty ,2\right)\cup \left[5,\infty \right)

The function

C\left(x\right)=150x+100,000

Practice Test

\frac{4{a}^{2}b}{12a{b}^{2}}

In the following exercises, perform the indicated operation and simplify.

\frac{4x}{x+2}·\frac{{x}^{2}+5x+6}{12{x}^{2}}

In the following exercises, solve each equation.

\frac{1}{x}+\frac{3}{4}=\frac{5}{8}

Given the function,

R\left(x\right)=\frac{2}{2{x}^{2}+x-15},

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?

4\frac{4}{5}

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 figure is a triangle formed by Cincinnati, Dayton, and Columbus. The distance between Cincinnati and Dayton is 2.4 inches. The distance between Dayton and Columbus is 3.2 inches. The distance between Columbus and Cincinnati is 5.3 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.

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6.6: Solving Polynomial and Rational Inequalities

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Learning Objectives

  • Solve polynomial inequalities.
  • Solve rational inequalities.

Solving Polynomial Inequalities

A polynomial inequality 18 is a mathematical statement that relates a polynomial expression as either less than or greater than another. We can use sign charts to solve polynomial inequalities with one variable.

Example \(\PageIndex{1}\)

Solve \(x(x+3)^{2}(x-4)<0\).

Begin by finding the critical numbers. For a polynomial inequality in standard form, with zero on one side, the critical numbers are the roots. Because \(f (x) = x(x + 3)^{2} (x − 4)\) is given in its factored form the roots are apparent. Here the roots are: \(0, −3\), and \(4\). Because of the strict inequality, plot them using open dots on a number line.

a71bb021558835e8539a91e083f0ed81.png

In this case, the critical numbers partition the number line into four regions. Test values in each region to determine if f is positive or negative. Here we choose test values \(−5, −1, 2\), and \(6\). Remember that we are only concerned with the sign \((+\) or \(−)\) of the result.

\(\begin{aligned} f(\color{OliveGreen}{-5}\color{black}{)}&=(\color{OliveGreen}{-5}\color{black}{)}(\color{OliveGreen}{-5}\color{black}{+}3)^{2}(\color{OliveGreen}{-5}\color{black}{-}4) =(-)(-)^{2}(-)&=+\color{Cerulean} { Positive} \\ f(\color{OliveGreen}{-1}\color{black}{)}&=(\color{OliveGreen}{-1}\color{black}{)}(\color{OliveGreen}{-1}\color{black}{+}3)^{2}(\color{OliveGreen}{-1}\color{black}{-}4) =(-)(+)^{2}(-)&=+\color{Cerulean} { Positive } \\ f(\color{OliveGreen}{2}\color{black}{)}&=(\color{OliveGreen}{2}\color{black}{)}(\color{OliveGreen}{2}\color{black}{+}3)^{2}(\color{OliveGreen}{2}\color{black}{-}4) =(+)(+)^{2}(-)&=-\color{Cerulean} { Negative } \\ f(\color{OliveGreen}{6}\color{black}{)}&=(\color{OliveGreen}{6}\color{black}{)}(\color{OliveGreen}{6}\color{black}{+}3)^{2}(\color{OliveGreen}{6}\color{black}{-}4) =(+)(+)^{2}(+)&=+\color{Cerulean} { Positive } \end{aligned}\)

After testing values we can complete a sign chart.

33002fc1f76e62347cb826a896239684.png

The question asks us to find the values where \(f (x) < 0\), or where the function is negative. From the sign chart we can see that the function is negative for \(x\)-values in between \(0\) and \(4\).

e1a8afef06a88852f99b3e0dd73e8e89.png

We can express this solution set in two ways:

\(\begin{aligned}\{x | 0<&x<4\} &\color{Cerulean} { Set\: notation } \\ (0,&4) &\color{Cerulean} { Interval\: notation }\end{aligned}\)

In this textbook we will continue to present solution sets using interval notation.

Graphing polynomials such as the one in the previous example is beyond the scope of this textbook. However, the graph of this function is provided below. Compare the graph to its corresponding sign chart.

927334bfbbcee080f1fcea1f65bf7fd4.png

Certainly it may not be the case that the polynomial is factored nor that it has zero on one side of the inequality. To model a function using a sign chart, all of the terms should be on one side and zero on the other. The general steps for solving a polynomial inequality are listed in the following example.

Example \(\PageIndex{2}\):

Solve: \(2 x^{4}>3 x^{3}+9 x^{2}\).

Step 1 : Obtain zero on one side of the inequality. In this case, subtract to obtain a polynomial on the left side in standard from.

\(\begin{aligned}2 x^{4}&>3 x^{3}+9 x^{2} \\ 2 x^{4}-3 x^{3}-9 x^{2}&>0\end{aligned}\)

Step 2 : Find the critical numbers. Here we can find the zeros by factoring.

\(2 x^{4}-3 x^{3}-9 x^{2}=0 \) \(x^{2}\left(2 x^{2}-3 x-9\right)=0 \) \(x^{2}(2 x+3)(x-3)=0 \)

There are three solutions, hence, three critical numbers \(−\frac{3}{2}, 0\), and \(3\). The strict inequality indicates that we should use open dots.

ea28c019017cc9de4a3975419d4ad057.png

Step 3 : Create a sign chart. In this case use \(f (x) = x^{2} (2x + 3) (x − 3) \) and test values \(−2, −1, 1\), and \(4\) to determine the sign of the function in each interval.

\(\begin{aligned} f(\color{OliveGreen}{-2}\color{black}{)} &=(\color{OliveGreen}{-2}\color{black}{)}^{2}[2(\color{OliveGreen}{-2}\color{black}{)}+3](\color{OliveGreen}{-2}\color{black}{-}3)&=(-)^{2}(-)(-)=+\\ f(\color{OliveGreen}{-1}\color{black}{)} &=(\color{OliveGreen}{-1}\color{black}{)}^{2}[2(\color{OliveGreen}{-1}\color{black}{)}+3](\color{OliveGreen}{-1}\color{black}{-}3)&=(-)^{2}(+)(-)=-\\ f(\color{OliveGreen}{1}\color{black}{)} &=(\color{OliveGreen}{1}\color{black}{)}^{2}[2(\color{OliveGreen}{1}\color{black}{)}+3](\color{OliveGreen}{1}\color{black}{-}3) &=(+)^{2}(+)(-)=-\\ f(\color{OliveGreen}{4}\color{black}{)} &=(\color{OliveGreen}{4}\color{black}{)}^{2}[2(\color{OliveGreen}{4}\color{black}{)}+3](\color{OliveGreen}{4}\color{black}{-}3) &=(+)^{2}(+)(+)=+\end{aligned}\)

With this information we can complete the sign chart.

68821aefaffdd3824120f5b226732e9c.png

Step 4 : Use the sign chart to answer the question. Here the solution consists of all values for which \(f (x) > 0\). Shade in the values that produce positive results and then express this set in interval notation.

79a5f6364113025a6d9d741608527102.png

\(\left(-\infty,-\frac{3}{2}\right) \cup(3, \infty)\)

Example \(\PageIndex{3}\)

Solve: \(x^{3}+x^{2} \leq 4(x+1)\).

Begin by rewriting the inequality in standard form, with zero on one side.

\(\begin{aligned}x^{3}+x^{2} &\leq 4(x+1) \\ x^{3}+x^{2} &\leq 4 x+4 \\ x^{3}+x^{2}-4 x-4 &\leq 0\end{aligned}\)

Next find the critical numbers of \(f(x)=x^{3}+x^{2}-4 x-4\):

\(\begin{aligned} x^{3}+x^{2}-4 x-4 &=0 \quad\color{Cerulean} { Factor\: by\: grouping.} \\ x^{2}(x+1)-4(x+1) &=0 \\(x+1)\left(x^{2}-4\right) &=0 \\(x+1)(x+2)(x-2) &=0 \end{aligned}\)

The critical numbers are \(−2, −1\), and \(2\). Because of the inclusive inequality \((≤)\) we will plot them using closed dots.

fc3ba0b1069acc9e8d2c5b9b27d74be4.png

Use test values, \(-3\), \(-\frac{3}{2}\), \(0\), and \(3\) to create a sign chart.

\(\begin{aligned} f(\color{OliveGreen}{-3}\color{black}{)}&=(\color{OliveGreen}{-3}\color{black}{+}1)(\color{OliveGreen}{-3}\color{black}{+}2)(\color{OliveGreen}{-3}\color{black}{-}2) &=(-)(-)(-)=- \\ f(\color{OliveGreen}{-\frac{3}{2}}\color{black}{)}&=(\color{OliveGreen}{-\frac{3}{2}}\color{black}{+}1)(\color{OliveGreen}{-\frac{3}{2}}\color{black}{+}2)(\color{OliveGreen}{-\frac{3}{2}}\color{black}{-}2) &=(-)(+)(-)=+\\ f(\color{OliveGreen}{0}\color{black}{)}&=(\color{OliveGreen}{0}\color{black}{+}1)(\color{OliveGreen}{0}\color{black}{+}2)(\color{OliveGreen}{0}\color{black}{-}2)&=(+)(+)(-)=- \\ f(\color{OliveGreen}{3}\color{black}{)}&=(\color{OliveGreen}{3}\color{black}{+}1)(\color{OliveGreen}{3}\color{black}{+}2)(\color{OliveGreen}{3}\color{black}{-}2)&=(+)(+)(+)=+\end{aligned}\)

And we have

5bfba05d6f38b7557f9f59a4737ce8c4.png

Use the sign chart to shade in the values that have negative results \((f (x) ≤ 0)\).

390764da1644572e8578d9b74db8d044.png

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

Exercise \(\PageIndex{1}\)

Solve \(-3 x^{4}+12 x^{3}-9 x^{2}>0\).

www.youtube.com/v/EXpe_0LzbSY

Solving Rational Inequalities

A rational inequality 19 is a mathematical statement that relates a rational expression as either less than or greater than another. Because rational functions have restrictions to the domain we must take care when solving rational inequalities. In addition to the zeros, we will include the restrictions to the domain of the function in the set of critical numbers.

Example \(\PageIndex{4}\)

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

The zeros of a rational function occur when the numerator is zero and the values that produce zero in the denominator are the restrictions. In this case,

\(\begin{array}{c | c}{\text { Roots (Numerator) }} & {\text{Restriction(Denominator)}} \\ {x-4=0 \text { or } x+2=0} & {x-1=0} \\ {\:\:\quad\quad\quad\: x=4 \quad\quad x=-2}& {x=1}\end{array}\)

Therefore the critical numbers are \(−2, 1\), and \(4\). Because of the inclusive inequality \((≥)\) use a closed dot for the roots \({−2, 4}\) and always use an open dot for restrictions \({1}\). Restrictions are never included in the solution set.

4458af7998cfd166906ae58c17564a39.png

Use test values \(x = −4, 0, 2, 6\).

\(\begin{aligned} f(\color{OliveGreen}{-4}\color{black}{)} &=\frac{(\color{OliveGreen}{-4}\color{black}{-}4)(\color{OliveGreen}{-4}\color{black}{+}2)}{(\color{OliveGreen}{-4}\color{black}{-}1)}&=\frac{(-)(-)}{(-)}=-\\ f(\color{OliveGreen}{0}\color{black}{)} &=\frac{(\color{OliveGreen}{0}\color{black}{-}4)(\color{OliveGreen}{0}\color{black}{+}2)}{(\color{OliveGreen}{0}\color{black}{-}1)}&=\frac{(-)(+)}{(-)}=+\\ f(\color{OliveGreen}{2}\color{black}{)} &=\frac{(\color{OliveGreen}{2}\color{black}{-}4)(\color{OliveGreen}{2}\color{black}{+}2)}{(\color{OliveGreen}{2}\color{black}{-}1)}&=\frac{(-)(+)}{(+)}=-\\ f(\color{OliveGreen}{6}\color{black}{)} &=\frac{(\color{OliveGreen}{6}\color{black}{-}4)(\color{OliveGreen}{6}\color{black}{+}2)}{(\color{OliveGreen}{6}\color{black}{-}1)}&=\frac{(+)(+)}{(+)}=+\end{aligned}\)

And then complete the sign chart.

bb71b28e07af7a632b8ba4a9d680dd93.png

The question asks us to find the values for which \(f (x) ≥ 0\), in other words, positive or zero. Shade in the appropriate regions and present the solution set in interval notation.

2d92bf1db240e3aeeefeca075e46c140.png

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

Graphing such rational functions like the one in the previous example is beyond the scope of this textbook. However, the graph of this function is provided below. Compare the graph to its corresponding sign chart.

0b34188938f21cabb2086b4c2f483d7e.png

Notice that the restriction \(x = 1\) corresponds to a vertical asymptote which bounds regions where the function changes from positive to negative. While not included in the solution set, the restriction is a critical number. Before creating a sign chart we must ensure the inequality has a zero on one side. The general steps for solving a rational inequality are outlined in the following example.

Example \(\PageIndex{5}\):

Solve \(\frac{7}{x+3}<2\).

Step 1 : Begin by obtaining zero on the right side.

\(\begin{aligned}\frac{7}{x+3}&<2 \\ \frac{7}{x+3}-2&<0\end{aligned}\)

Step 2 : Determine the critical numbers. The critical numbers are the zeros and restrictions. Begin by simplifying to a single algebraic fraction.

\(\begin{aligned}\frac{7}{x+3}-\frac{2}{1}&<0 \\ \frac{7-2(x+3)}{x+3}&<0 \\ \frac{7-2 x-6}{x+3}&<0 \\ \frac{-2 x+1}{x+3}&<0\end{aligned}\)

Next find the critical numbers. Set the numerator and denominator equal to zero and solve.

\(\begin{array}{c|c} {\text{Root}}&{\text{Restriction}}\\ {-2x+1=0}\\{-2x=-1}&{x+3=0}\\{x=\frac{1}{2}}&\quad\quad\:\:{x=-3} \end{array}\)

In this case, the strict inequality indicates that we should use an open dot for the root.

76bcd5626595946ab7643c0a0a961be3.png

Step 3 : Create a sign chart. Choose test values \(−4, 0\), and \(1\).

\(\begin{aligned} f(\color{OliveGreen}{-4}\color{black}{)} &=\frac{-2(\color{OliveGreen}{-4}\color{black}{)}+1}{\color{OliveGreen}{-4}\color{black}{+}3}&=\frac{+}{-}=-\\ f(\color{OliveGreen}{0}\color{black}{)} &=\frac{-2(\color{OliveGreen}{0}\color{black}{)}+1}{\color{OliveGreen}{0}\color{black}{+}3}&=\frac{+}{+}=+\\ f(\color{OliveGreen}{1}\color{black}{)} &=\frac{-2(\color{OliveGreen}{1}\color{black}{)}+1}{\color{OliveGreen}{1}\color{black}{+}3}&=\frac{-}{+}=-\end{aligned}\)

388805d18d041e0442fbe86fceefaf43.png

Step 4 : Use the sign chart to answer the question. In this example we are looking for the values for which the function is negative, \(f (x) < 0\). Shade the appropriate values and then present your answer using interval notation.

6200ff1c965a7cc7d73ca4adc16da552.png

\((-\infty,-3) \cup\left(\frac{1}{2}, \infty\right)\)

Example \(\PageIndex{6}\):

Solve: \(\frac{1}{x^{2}-4} \leq \frac{1}{2-x}\).

Begin by obtaining zero on the right side.

\(\begin{aligned}\frac{1}{x^{2}-4} &\leq \frac{1}{2-x} \\ \frac{1}{x^{2}-4}-\frac{1}{2-x} &\leq 0\end{aligned}\)

Next simplify the left side to a single algebraic fraction.

\(\begin{array}{r}{\frac{1}{x^{2}-4}-\frac{1}{2-x} \leq 0} \\ {\frac{1}{(x+2)(x-2)}-\frac{1}{-(x-2)} \leq 0} \\ {\frac{1}{(x+2)(x-2)}+\frac{1\color{Cerulean}{(x+2)}}{\color{black}{(x-2)}\color{Cerulean}{(x+2)}}\color{black}{ \leq} 0} \\ {\frac{1+x+2}{(x+2)(x-2)} \leq 0} \\ {\frac{x+3}{(x+2)(x-2)} \leq 0}\end{array}\)

The critical numbers are \(−3, −2\), and \(2\). Note that \(±2\) are restrictions and thus we will use open dots when plotting them on a number line. Because of the inclusive inequality we will use a closed dot at the root \(−3\).

94b6170dfcf8eae5df95a3e23c7dde39.png

Choose test values \(-4, -2\frac{1}{2} = -\frac{5}{2}, 0\), and \(3\).

\(\begin{aligned}f(\color{OliveGreen}{-4}\color{black}{)}&= \frac{\color{OliveGreen}{-4}\color{black}{+}3}{(\color{OliveGreen}{-4}\color{black}{+}2)(\color{OliveGreen}{-4}\color{black}{-}2)}&=\frac{(-)}{(-)(-)}=- \\ f(\color{OliveGreen}{-\frac{5}{2}}\color{black}{)} &= \frac{\color{OliveGreen}{-\frac{5}{2}}\color{black}{+}3}{(\color{OliveGreen}{-\frac{5}{2}}\color{black}{+}2)(\color{OliveGreen}{-\frac{5}{2}}\color{black}{-}2)}&=\frac{(+)}{(-)(-)}=+\\ f(\color{OliveGreen}{0}\color{black}{)}&=\frac{\color{OliveGreen}{0}\color{black}{+}3}{(\color{OliveGreen}{0}\color{black}{+}2)(\color{OliveGreen}{0}\color{black}{-}2)}&=\frac{(+)}{(+)(-)}=- \\ f(\color{OliveGreen}{3}\color{black}{)}&=\frac{\color{OliveGreen}{3}\color{black}{+}3}{(\color{OliveGreen}{3}\color{black}{+}2)(\color{OliveGreen}{3}\color{black}{-}2)} &=\frac{(+)}{(+)(+)}=+\end{aligned}\)

Construct a sign chart.

f00e3b8453a4258174ea2c4ab8089286.png

Answer the question; in this case, find \(x\) where \(f(x) \leq 0\).

d558688914af6ca4360c2ce97e71986d.png

\((-\infty,-3] \cup(-2,2)\)

Exercise \(\PageIndex{2}\)

Solve: \(\frac{2 x^{2}}{2 x^{2}+7 x-4} \geq \frac{x}{x+4}\).

\((-4,0] \cup\left(\frac{1}{2}, \infty\right)\)

www.youtube.com/v/3ljcHJqExBY

Key Takeaways

  • When a polynomial inequality is in standard form, with zero on one side, the roots of the polynomial are the critical numbers. Create a sign chart that models the function and then use it to answer the question.
  • When a rational inequality is written as a single algebraic fraction, with zero on one side, the roots as well as the restrictions are the critical numbers. The values that produce zero in the numerator are the roots, and the values that produce zero in the denominator are the restrictions. Always use open dots for restrictions, regardless of the given inequality, because restrictions are not part of the domain. Create a sign chart that models the function and then use it to answer the question.

Exercise \(\PageIndex{3}\)

Solve. Present answers using interval notation.

  • \(x(x+1)(x-3)>0\)
  • \(x(x-1)(x+4)<0\)
  • \((x+2)(x-5)^{2}<0\)
  • \((x-4)(x+1)^{2} \geq 0\)
  • \((2 x-1)(x+3)(x+2) \leq 0\)
  • \((3 x+2)(x-4)(x-5) \geq 0\)
  • \(x(x+2)(x-5)^{2}<0\)
  • \(x(2 x-5)(x-1)^{2}>0\)
  • \(x(4 x+3)(x-1)^{2} \geq 0\)
  • \((x-1)(x+1)(x-4)^{2}<0\)
  • \((x+5)(x-10)(x-5)^{2} \geq 0\)
  • \((3 x-1)(x-2)(x+2)^{2} \leq 0\)
  • \(-4 x(4 x+9)(x-8)^{2}>0\)
  • \(-x(x-10)(x+7)^{2}>0\)

1. \((-1,0) \cup(3, \infty)\)

3. \((-\infty,-2)\)

5. \((-\infty,-3] \cup\left[-2, \frac{1}{2}\right]\)

7. \((-2,0)\)

9. \(\left(-\infty,-\frac{3}{4}\right] \cup[0, \infty)\)

11. \((-\infty,-5] \cup[5,5] \cup[10, \infty)\)

13. \(\left(-\frac{9}{4}, 0\right)\)

Exercise \(\PageIndex{4}\)

  • \(x^{3}+2 x^{2}-24 x \geq 0\)
  • \(x^{3}-3 x^{2}-18 x \leq 0\)
  • \(4 x^{3}-22 x^{2}-12 x<0\)
  • \(9 x^{3}+30 x^{2}-24 x>0\)
  • \(12 x^{4}+44 x^{3}>80 x^{2}\)
  • \(6 x^{4}+12 x^{3}<48 x^{2}\)
  • \(x\left(x^{2}+25\right)<10 x^{2}\)
  • \(x^{3}>12 x(x-3)\)
  • \(x^{4}-5 x^{2}+4 \leq 0\)
  • \(x^{4}-13 x^{2}+36 \geq 0\)
  • \(x^{4}>3 x^{2}+4\)
  • \(4 x^{4}<3-11 x^{2}\)
  • \(9 x^{3}-3 x^{2}-81 x+27 \leq 0\)
  • \(2 x^{3}+x^{2}-50 x-25 \geq 0\)
  • \(x^{3}-3 x^{2}+9 x-27>0\)
  • \(3 x^{3}+5 x^{2}+12 x+20<0\)

1. \([-6,0] \cup[4, \infty)\)

3. \(\left(-\infty,-\frac{1}{2}\right) \cup(0,6)\)

5. \((-\infty,-5) \cup\left(\frac{4}{3}, \infty\right)\)

7. \((-\infty, 0)\)

9. \([-2,-1] \cup[1,2]\)

11. \((-\infty,-2) \cup(2, \infty)\)

13. \((-\infty,-3] \cup\left[\frac{1}{3}, 3\right]\)

15. \((3, \infty)\)

Exercise \(\PageIndex{5}\)

  • \(\frac{x}{x-3}>0\)
  • \(\frac{x-5}{x}>0\)
  • \(\frac{(x-3)(x+1)}{x}<0\)
  • \(\frac{(x+5)(x+4)}{(x-2)}<0\)
  • \(\frac{(2 x+1)(x+5)}{(x-3)(x-5)} \leq 0\)
  • \(\frac{(3 x-1)(x+6)}{(x-1)(x+9)} \geq 0\)
  • \(\frac{(x-8)(x+8)}{-2 x(x-2)} \geq 0\)
  • \(\frac{(2 x+7)(x+4)}{x(x+5)} \leq 0\)
  • \(\frac{x^{2}}{(2 x+3)(2 x-3)} \leq 0\)
  • \(\frac{(x-4)^{2}}{-x(x+1)}>0\)
  • \(\frac{-5 x(x-2)^{-}}{(x+5)(x-6)} \geq 0\)
  • \(\frac{(3 x-4)(x+5)}{x(x-4)^{2}} \geq 0\)
  • \(\frac{1}{(x-5)^{4}}>0\)
  • \(\frac{1}{(x-5)^{4}}<0\)

1. \((-\infty,-0) \cup(3, \infty)\)

3. \((-\infty,-1) \cup(0,3)\)

5. \(\left[-5,-\frac{1}{2}\right] \cup(3,5)\)

7. \([-8,0) \cup(2,8]\)

9. \(\left(-\frac{3}{2}, \frac{3}{2}\right)\)

11. \((-\infty,-5) \cup[0,6)\)

13. \((-\infty, 5) \cup(5, \infty)\)

Exercise \(\PageIndex{6}\)

  • \(\frac{x^{2}-11 x-12}{x+4}<0\)
  • \(\frac{x^{2}-10 x+24}{x-2}>0\)
  • \(\frac{x^{2}+x-30}{2 x+1} \geq 0\)
  • \(\frac{2 x^{2}+x-3}{x-3} \leq 0\)
  • \(\frac{3 x^{2}-4 x+1}{x^{2}-9} \leq 0\)
  • \(\frac{x^{2}-16}{2 x^{2}-3 x-2} \geq 0\)
  • \(\frac{x^{2}-12 x+20}{x^{2}-10 x+25}>0\)
  • \(\frac{x^{2}+15 x+36}{x^{2}-8 x+16}<0\)
  • \(\frac{8 x^{2}-2 x-1}{2 x^{2}-3 x-14} \leq 0\)
  • \(\frac{4 x^{2}-4 x-15}{x^{2}+4 x-5} \geq 0\)
  • \(\frac{1}{x+5}+\frac{5}{x-1}>0\)
  • \(\frac{5}{x+4}-\frac{1}{x-4}<0\)
  • \(\frac{1}{x+7}>1\)
  • \(\frac{1}{x-1}<-5\)
  • \(x \geq \frac{30}{x-1}\)
  • \(x \leq \frac{1-2 x}{x-2}\)
  • \(\frac{1}{x-1} \leq \frac{2}{x}\)
  • \(\frac{3}{x+1}>-\frac{1}{x}\)
  • \(\frac{4}{x-3} \leq \frac{1}{x+3}\)
  • \(\frac{2 x-9}{x}+\frac{49}{x-8}<0\)
  • \(\frac{x}{2(x+2)}-\frac{1}{x+2} \leq \frac{12}{x(x+2)}\)
  • \(\frac{1}{2 x+1}-\frac{9}{2 x-1}>2\)
  • \(\frac{3 x}{x^{2}-4}-\frac{2}{x-2}<0\)
  • \(\frac{x}{2 x+1}+\frac{4}{2 x^{2}-7 x-4}<0\)
  • \(\frac{x+1}{2 x^{2}+5 x-3} \geq \frac{x}{4 x^{2}-1}\)
  • \(\frac{x^{2}-14}{2 x^{2}-7 x-4} \leq \frac{5}{1+2 x}\)

1. \((-\infty,-4) \cup(-1,12)\)

3. \(\left[-6,-\frac{1}{2}\right) \cup[5, \infty)\)

5. \(\left(-3, \frac{1}{3}\right] \cup[1,3)\)

7. \((-\infty, 2) \cup(10, \infty)\)

9. \(\left(-2,-\frac{1}{4}\right] \cup\left[\frac{1}{2}, \frac{7}{2}\right)\)

11. \((-5,-4) \cup(1, \infty)\)

13. \((-7,-6)\)

15. \([-5,1) \cup[6, \infty)\)

17. \((0,1) \cup[2, \infty)\)

19. \((-\infty, 5] \cup(-3,3)\)

21. \([-4,-2) \cup(0,6]\)

23. \((-\infty,-2) \cup(2,4)\)

25. \(\left(-3,-\frac{1}{2}\right) \cup\left(\frac{1}{2}, \infty\right)\)

Exercise \(\PageIndex{7}\)

  • Does the sign chart for any given polynomial or rational function always alternate? Explain and illustrate your answer with some examples.
  • Write down your own steps for solving a rational inequality and illustrate them with an example. Do your steps also work for a polynomial inequality? Explain.

1. Answer may vary

18 A mathematical statement that relates a polynomial expression as either less than or greater than another.

19 A mathematical statement that relates a rational expression as either less than or greater than another.

IMAGES

  1. 9. Solving Rational Inequalities Example 2

    problem solving rational inequality

  2. How to Solve & Graph Rational Inequalities

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  3. Solving Rational Inequalities

    problem solving rational inequality

  4. Solved: Solve these rational inequalities. (See Examples.)Examp

    problem solving rational inequality

  5. Solving Rational Inequalities Worksheet

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  6. Solving Rational Inequalities Worksheet

    problem solving rational inequality

VIDEO

  1. Rational Inequality Examples

  2. Rational Inequality

  3. Solving Rational Inequality Part 1

  4. Rational Inequality Problem Solving

  5. Rational Inequality

  6. Solving Rational Inequality Part 2

COMMENTS

  1. 9.7: Solve Rational Inequalities

    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.

  2. Algebra

    3x+8 x −1 < −2 3 x + 8 x − 1 < − 2 Solution. u ≤ 4 u −3 u ≤ 4 u − 3 Solution. t3 −6t2 t−2 > 0 t 3 − 6 t 2 t − 2 > 0 Solution. Here is a set of practice problems to accompany the Rational Inequalities section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University.

  3. Solving Rational Inequalities

    The key approach in solving rational inequalities relies on finding the critical values of the rational expression which divide the number line into distinct open intervals. The critical values are simply the zeros of both the numerator and the denominator. You must remember that the zeros of the denominator make the rational expression ...

  4. 7.6 Solve Rational Inequalities

    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.

  5. Solving Rational Inequalities

    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 the 2 by (x−4)/ (x−4): 3x−10 x−4 − 2 x−4 x−4 > 0. Now we have a common denominator, let's ...

  6. Rational inequalities: one side is zero (video)

    Your example looks like a rational equation, Sal has 3 videos on this topic (they are called solving rational equations). I solved your equation and you have to find a common denominator first, which is 21. Then you multiply 7 (x+1) + 3 (x+2), all that over 21, which equals 2. After a few steps, youll have 10x +13=42, then 10x=29, and finally ...

  7. Solving Rational Inequalities

    State your answer using the desired form of notation. Solve: 1. Rewrite the inequality to contain a zero on the right-hand side. Create a single fraction on the left-hand side. Subtract the 2 from both sides. Get common denominator on the left side. 2. Determine any value (s) that makes the numerator equal 0.

  8. Rational inequalities: both sides are not zero

    Either both of these are positive. So they're either both positive, so minus x minus 11 is greater than or equal to 0. This could be equal to 0. Actually, if this is equal to 0, then this whole thing is going to be true. So either that is greater than or equal to 0 and this right here is greater than 0.

  9. 7.7: Solve Rational Inequalities

    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.

  10. Algebra

    In this section we will solve inequalities that involve rational expressions. The process for solving rational inequalities is nearly identical to the process for solving polynomial inequalities with a few minor differences. Let's just jump straight into some examples. Example 1 Solve x +1 x −5 ≤ 0 x + 1 x − 5 ≤ 0 . Show Solution.

  11. Solve Rational Inequalities

    Solve and write the solution in interval notation: Step 1. Write the inequality as one quotient on the left and zero on the right. Our inequality is in this form. Step 2. Determine the critical points—the points where the rational expression will be zero or undefined.

  12. 4.3: Rational Inequalities and Applications

    In this section, we solve equations and inequalities involving rational functions and explore associated application problems. Our first example showcases the critical difference in procedure between solving a rational equation and a rational inequality. Example 4.3.1 4.3. 1. Solve x3−2x+1 x−1 = 12x − 1 x 3 − 2 x + 1 x − 1 = 1 2 x − 1.

  13. Solving Rational Inequalities

    How to Solve Rational Inequalities. We use factoring and sign analysis in these examples in this free math video tutorial by Mario's Math Tutoring.0:06 Intro...

  14. Solving Rational Equations

    http://www.greenemath.com/In this course, we will learn how to solve rational equations, rational inequalities, and word problems that involve rational equat...

  15. Rational Inequality Calculator

    Free rational inequality calculator - solve rational inequalities with all the steps. Type in any inequality to get the solution, steps and graph ... Study Tools AI Math Solver Popular Problems Study Guides Practice Cheat Sheets Calculators Graphing Calculator Geometry Calculator.

  16. 7.7: Solve Rational Inequalities

    In the next example, the numerator is always positive, so the sign of the rational expression depends on the sign of the denominator. Example 7.7.3. Solve and write the solution in interval notation: 5 x2 − 2x − 15 > 0. Solution. The inequality is in the correct form. 5 x2 − 2x − 15 > 0. Factor the denominator.

  17. rational inequalities

    Solve problems from Pre Algebra to Calculus step-by-step . step-by-step. rational inequalities. en. Related Symbolab blog posts. Practice Makes Perfect. Learning math takes practice, lots of practice. Just like running, it takes practice and dedication. If you want...

  18. Algebra

    Section 2.13 : Rational Inequalities. Back to Problem List. 1. Solve the following inequality. 4 −x x +3 >0 4 − x x + 3 > 0. Show All Steps Hide All Steps.

  19. Inequalities Calculator

    The two rules of inequalities are: If the same quantity is added to or subtracted from both sides of an inequality, the inequality remains true. If both sides of an inequality are multiplied or divided by the same positive quantity, the inequality remains true. If we multiply or divide both sides of an inequality by the same negative number, we ...

  20. Solving Inequalities

    The problem above is usually written as a double inequality.-3 < 5 - 2x < 9 stands for -3 < 5 - 2x and 5 - 2x < 9. Note: When we solved the two inequalities separately, the steps in the two problems were the same. Therefore, the double inequality notation may be used to solve the inequalities simultaneously.-3 < 5 - 2x < 9.-8 < -2x < 4. 4 > x > -2.

  21. 9.2: Rational Inequalities

    Step 1. Rewrite the inequality so that only zero is on the right side. Since x − 3 x + 1 > 0 x − 3 x + 1 > 0 already has zero on the right side, this step is done. Step 2. Determine where the rational expression is undefined or equals zero. To obtain where the rational expression equals zero, we set the numerator equal to zero:

  22. PDF The awareness of difficulties in solving rational inequality and a

    "solving a rational inequality is difficult for mathematics students. Many of them make a mistake when solving such problem. Many students had mistaken of type errors in inequality rules. Most of them are solving the rational inequalities by treating it as a rational equation.

  23. 4.3: Rational Inequalities and Applications

    4.3: Rational Inequalities and Applications. In this section, we solve equations and inequalities involving rational functions and explore associated application problems. Our first example showcases the critical difference in procedure between solving a rational equation and a rational inequality.

  24. 6.6: Solving Polynomial and Rational Inequalities

    Solving Rational Inequalities. A rational inequality 19 is a mathematical statement that relates a rational expression as either less than or greater than another. Because rational functions have restrictions to the domain we must take care when solving rational inequalities. In addition to the zeros, we will include the restrictions to the ...