0 1 2 3 4 5 6
0 1 2 3 4 5 6
)
Figure 1.63
Figure 1.64
0 1 2 3 4 5 6
0 1 2 3 4 5 6
]
S U M M A R Y
Interval Notation for Unbounded Intervals
Interval
Set notation Type Graph
5x兩x 7 a6 1a, ⬁2 Open 5x兩x 6 a6 1-⬁, a2 Open
5x兩x Ú a6 [a, ⬁2 Closed 5x兩x … a6 1-⬁, a] Closed Real numbers 1-⬁, ⬁2 Open
a
)
a
)
a
]
a
]
We use a parenthesis when an endpoint of an interval is not included in the solution set and a bracket when an endpoint is included. A bracket is never used next to ⬁ because infinity is not a number. On the graphs above, the number lines are shaded, showing that the solutions include all real numbers in the given interval.
EXAMPLE 1 Interval notation
Write an inequality whose solution set is the given interval.
a. 1-⬁, -92 b. 30, ⬁2 Solution
a. The interval 1-⬁, -92 represents all real numbers less than -9. It is the solu- tion set to x 6 -9.
b. The interval 30, ⬁2 represents all real numbers greater than or equal to 0. It is the solution set to x Ú0.
TRY THIS. Write an inequality whose solution set is 1-⬁, 54. ■
Linear Inequalities
Replacing the equal sign in the general linear equation ax + b= 0 by any of the symbols 6, …, 7, or Ú gives a linear inequality. Two inequalities are equivalent if they have the same solution set. We solve linear inequalities like we solve linear equations by performing operations on each side to get equivalent inequalities.
However, the rules for inequalities are slightly different from the rules for equations.
Adding any real number to both sides of an inequality results in an equivalent inequality. For example, adding 3 to both sides of -46 5 yields -1 68, which is true. Adding or subtracting the same number simply moves the original numbers to the right or left along the number line and does not change their order.
The order of two numbers will also be unchanged when they are multiplied or divided by the same positive real number. For example, 10 6 20, and after divid- ing both numbers by 10 we have 16 2. However, multiplying or dividing any two numbers by a negative number will change their order. For example, -2 6 -1, but after multiplying both numbers by -1 we have 2 7 1. See Fig. 1.65. Likewise, -10 6 20, but after dividing both numbers by -10 we have 1 7 -2. When an in- equality is multiplied or divided by a negative number, the direction of the inequal- ity symbol is reversed. These ideas are stated symbolically in the following box for
6, but they also hold for 7, …, and Ú.
Figure 1.65
0 1 2
–1 –2
Multiplying by –1 changes the order
2 > 1 –2 < –1
Properties of Inequality If A and B are algebraic expressions and C is a nonzero real number, then the inequality A 6 B is equivalent to
1. A { C6 B { C,
2. CA 6 CB (for C positive), CA 7 CB (for C negative), 3. A
C 6 B
C (for C positive), A C 7 B
C (for C negative).
EXAMPLE 2 Solving a linear inequality
Solve -3x - 96 0. Write the solution set in interval notation and graph it.
Solution
Isolate the variable as is done in solving equations.
-3x- 9 60
-3x - 9+ 9 60 + 9 Add 9 to each side.
-3x 69
x 7 -3 Divide each side by -3, reversing the inequality.
The solution set is the interval 1-3, ⬁2 and its graph is shown in Fig. 1.66. Check- ing the solution to an inequality is generally not as simple as checking an equation, because usually there are infinitely many solutions. We can do a “partial check” by checking one number in 1-3, ⬁2 and one number not in 1-3, ⬁2. For example, 0 7 -3 and -3102 - 96 0 is correct, while -6 6 -3 and -31-62 -9 6 0 is incorrect.
TRY THIS. Solve 2 - 5x… 7. Write the solution set in interval notation and
graph it. ■
We can read the solution to an inequality in one variable from a graph in two variables in the same manner that we read the solution to an equation in one variable.
Figure 1.67 shows the graph of y = -3x - 9. From this figure we see that the y- coordinates on this line are negative when the x-coordinates are greater than -3. In other words, -3x -9 6 0 when x 7 -3.
We can also perform operations on each side of an inequality using a variable expression. Addition or subtraction with variable expressions will give equivalent inequalities. However, we must always watch for undefined expressions. Multipli- cation and division with a variable expression are usually avoided because we do not know whether the expression is positive or negative.
EXAMPLE 3 Solving a linear inequality
Solve 12 x - 3Ú 14 x+ 2 and graph the solution set.
Solution
Multiply each side by the LCD to eliminate the fractions.
1
2 x - 3Ú 1 4 x+ 2 4a1
2 x - 3b Ú 4a1
4 x + 2b Multiply each side by 4.
2x- 12 Ú x + 8 x- 12 Ú 8
x Ú 20
The solution set is the interval [20, ⬁2. See Fig. 1.68 for its graph.
The algebraic solution to 12 x - 3Ú 14 x+ 2 proves that the graph of y = 12 x - 3 is at or above the graph of y = 14 x + 2 when x Ú20, as it appears to be in Fig. 1.69. Conversely, the graphs in Fig. 1.69 support the conclusion that x Ú 20 causes the inequality to be true.
TRY THIS. Solve 12 x + 13 … 13 x + 1 and graph the solution set. ■
Compound Inequalities
A compound inequality is a sentence containing two simple inequalities connected with “and” or “or.” The solution to a compound inequality can be an interval of real numbers that does not involve infinity, a bounded interval of real numbers.
For example, the solution set to the compound inequality x Ú 2 and x … 5 is the set of real numbers between 2 and 5, inclusive. This inequality is also written as 2 …x … 5. Its solution set is 5x兩2… x … 56, which is written in interval nota- tion as 32, 54. Because 32, 54 contains both of its endpoints, the interval is closed.
The following summary lists the different types of bounded intervals used in inter- val notation and the graphs of those intervals on a number line.
Figure 1.66
– 2 –1 0
– 3 – 4 – 5 – 6
)
Figure 1.67 y
x –2
–4 –6
2 –2
2 –4
–6
4
–10 –12 y < 0
when x > –3
(–3, 0) y = –3x – 9
Figure 1.68 24 23 22 21 20 19 18
]
Figure 1.69
⫺10 40
⫺10 10
The notation a 6x 6 b is used only when x is between a and b, and a is less than b. We do not write inequalities such as 5 6x 6 3, 4 7x 6 9, or 2 6x 7 8.
The intersection of sets A and B is the set A傽B (read “A intersect B”), where x僆A傽B if and only if x僆A and x僆B. (The symbol僆means “belongs to.”) The union of sets A and B is the set A傼B (read “A union B”), where x僆A傼B if and only if x僆A or x僆B. In solving compound inequalities it is often necessary to find intersections and unions of intervals.
EXAMPLE 4 Intersections and unions of intervals
Let A= 11, 52, B = 33, 72, and C =16, ⬁2. Write each of the following sets in interval notation.
a. A傼B b. A傽B c. A傼C d. A傽C Solution
a. Graph both intervals on the number line, as shown in Fig. 1.70(a). The union of two intervals is the set of points that are in one, the other, or both intervals. For a union, nothing is omitted. The union consists of all points shaded in the figure.
So A傼B = 11, 72.
b. The intersection of A and B is the set of points that belong to both intervals.
The intersection consists of the points that are shaded twice in Fig. 1.70(a). So A傽B = 33, 52.
S U M M A R Y
Interval Notation for Bounded Intervals
Interval
Set notation Type Graph
5x兩a6 x 6 b6 1a, b2 Open 5x兩a… x … b6 [a, b] Closed 5x兩a… x 6 b6 [a, b2 Half open or
half closed
5x兩a6 x … b6 1a, b] Half open or
half closed
a b
] ]
a b
) )
a b
) ]
a b
) ]
c. Graph both intervals on the number line, as shown in Fig. 1.70(b). For a union, nothing is omitted. So A傼C = 11, 52傼16, ⬁2. Note that A傼C cannot be written as single interval.
d. Since there are no points shaded twice in Fig. 1.70(b), A傽C = ⭋.
TRY THIS. Find A傼B and A傽B if A = 11, 62 and B = 34, 92. ■ The solution set to a compound inequality using the connector “or” is the union of the two solution sets, and the solution set to a compound inequality using “and” is the intersection of the two solution sets.
(a) (b) Figure 1.70
)
2 3 4 7 8
1
0 5 6
A )
)
) ] B ) )
2 3 4 7 8
1
0 5 6
C
) )
A
)
) ) )
EXAMPLE 5 Solving compound inequalities
Solve each compound inequality. Write the solution set using interval notation and graph it.
a. 2x -3 7 5 and 4 - x… 3 b. 4- 3x 6 -2 or 31x - 22… -6 c. -4… 3x- 1 65
Solution
a. 2x -3 7 5 and 4 - x …3
2x 7 8 and -x … -1
x 7 4 and x Ú1
Graph 14, ⬁2 and 31, ⬁2 on the number line, as shown in Fig. 1.71(a). The inter- section of the intervals is the set of points that are shaded twice in Fig. 1.71(a).
So the intersection is the interval 14, ⬁2, and 14, ⬁2 is the solution set to the compound inequality. Its graph is shown in Fig. 1.71(b).
(a) (b) Figure 1.71
2 3 4 5 6
1 –1 0 – 2
] )
2 3 4 5 6
1 0 –1 – 2
)
b. 4 - 3x 6 -2 or 31x - 22 … -6 -3x6 -6 or x - 2… -2 x7 2 or x … 0
The union of the intervals 12, ⬁2 and 1- ⬁, 04 consists of all points that are shaded in Fig. 1.72. This set cannot be written as a single interval. So the solu- tion set is 1-⬁, 04傼12, ⬁2. Its graph is shown in Fig. 1.72.
c. We could write -4 …3x - 16 5 as the compound inequality -4 …3x - 1 and 3x -1 6 5, and then solve each simple inequality. Since each is solved using the same sequence of steps, we can solve the original inequality without separating it:
-4… 3x- 1 65
-4 + 1… 3x- 1 +1 6 5+ 1 Add 1 to each part of the inequality.
-3… 3x6 6 -3
3 … 3x 3 6 6
3 Divide each part by 3.
-1… x 6 2
The solution set is the half-open interval 3-1, 22, graphed in Fig. 1.73.
TRY THIS. Solve 2x 7 -4 and 4 - x Ú 0 and graph the solution set. ■
It is possible that all real numbers satisfy a compound inequality or no real numbers satisfy a compound inequality.
EXAMPLE 6 Solving compound inequalities
Solve each compound inequality.
a. 3x -9 … 9 or 4 -x … 3 b. - 2
3 x 64 and 3 4 x 6 -6 Figure 1.72
2 3 4 5 6
1 0 –1 – 2
) ]
Figure 1.73
2 3 4 5 6
1 –1 0 – 2
) ]
Solution
a. Solve each simple inequality and find the union of their solution sets:
3x- 9 …9 or 4 -x … 3 3x …18 or -x … -1
x …6 or x Ú 1
The union of 1-⬁, 64 and 31, ⬁2 is the set of all real numbers, 1-⬁, ⬁2.
b. Solve each simple inequality and find the intersection of their solution sets:
- 2
3 x 6 4 and 3
4 x 6 -6 a- 3
2b a- 2
3 xb 7 a- 3
2b4 and a4 3b a3
4 xb 6 a4 3b1-62
x 7 -6 and x 6 -8
Since 1-6, ⬁2傽1-⬁, -82 = ⭋, there is no solution to the compound inequality.
TRY THIS. Solve 3x +2 7 -1 and 5 6 -3- 4x. ■
Absolute Value Inequalities
We studied absolute value equations in Section 1.1. Absolute value inequalities are closely related to absolute value equations. Remember that the absolute value of a number is its distance from 0 on the number line. The equation 0x0 = 3 means that x is exactly three units from 0 on the number line. Since 3 and -3 are three units from 0, both 3 and -3 satisfy the equation. The inequality 0x0 63 means that x is less than three units from 0. See Fig. 1.74. Any number between -3 and 3 is less than three units from 0. So 0x0 6 3 is equivalent to -36 x 6 3 and the solution set is the open interval 1-3, 32.
The inequality 0x0 75 means that x is more than five units from 0 on the num- ber line, which is equivalent to the compound inequality x 7 5 or x 6 -5. See Fig. 1.75. So the solution to 0x0 75 is the union of two intervals, 1-⬁, -52傼15, ⬁2.
These ideas about absolute value inequalities are summarized as follows.
Figure 1.75
0 5 10 15
– 5 –10 –15
) )
x is more than
5 units from 0 x is more than 5 units from 0
S U M M A R Y
Basic Absolute Value Inequalities (for k + 0)
Absolute Solution set
value Equivalent in interval Graph of inequality statement notation solution set
0x0 7 k x6 -k or x 7 k 1-⬁, -k2傼1k, ⬁2 0x0 Ú k x… -k or x Ú k 1-⬁, -k4傼3k, ⬁2 0x0 6 k -k6 x 6 k 1-k, k2
0x0 … k -k… x … k 3-k, k4
–k k
) )
–k k
] ]
–k k
) )
–k k
] ]
In the next example we use the rules for basic absolute value inequalities to solve more complicated absolute value inequalities.
Figure 1.74
–1 1 2 3 4 5
– 2 – 3 – 4
– 5 0
x is less than 3 units from 0
) )
EXAMPLE 7 Absolute value inequalities
Solve each absolute value inequality and graph the solution set.
a. 03x+ 20 6 7 b. -204 - x0 … -4 c. 07x- 90 Ú -3 Solution
a. 03x + 20 6 7
-7 6 3x + 26 7 Write the equivalent compound inequality.
-7- 26 3x +2 - 26 7 -2 Subtract 2 from each part.
-96 3x 6 5
-3 6x 6 5
3 Divide each part by 3.
The solution set is the open interval 1-3, 532. The graph is shown in Fig. 1.76.
b. -204 -x0 … -4 Divide each side by -2, reversing
04 -x0 Ú 2 the inequality.
4- x … -2 or 4 - x Ú 2 Write the equivalent compound inequality.
-x … -6 or -x Ú -2
x Ú 6 or x … 2 Multiply each side by -1.
The solution set is 1-⬁, 24傼36, ⬁2. Its graph is shown in Fig. 1.77. Check 8, 4, and 0 in the original inequality. If x = 8, we get -204 - 80 … -4, which is true. If x =4, we get -204- 40 … -4, which is false. If x= 0, we get
-204 -00 … -4, which is true.
c. The expression 07x -90 has a nonnegative value for every real number x. So the inequality 07x - 90 Ú -3 is satisfied by every real number. The solution set is 1-⬁, ⬁2, and its graph is shown in Fig. 1.78.
TRY THIS. Solve 0x - 60 - 3 … -2 and graph the solution set. ■
Modeling with Inequalities
Inequalities occur in applications just as equations do. In fact, in real life, equality in anything is usually the exception. The solution to a problem involving an inequality is generally an interval of real numbers. In this case we often ask for the range of values that solve the problem.
EXAMPLE 8 An application involving inequality
Remington scored 74 on his midterm exam in history. If he is to get a B, the average of his midterm and final exam must be between 80 and 89 inclusive. In what range must his final exam score lie for him to get a B in the course? Both exams have a maximum of 100 points possible.
Solution
Let x represent Remington’s final exam score. The average of his midterm and final must satisfy the following inequality.
80 … 74 + x
2 …89
160 … 74 +x … 178 86 … x… 104 Figure 1.76
–1 0
– 2 – 3 – 4
– 5 1 2 3
5– 3
) )
Figure 1.77
1 2 6 7
0 3 4 5 8
] ]
Figure 1.78
–1 0 1 2 3 4
– 2 – 3 – 4
His final exam score must lie in the interval [86, 104]. Since 100 is the maximum possible score, we can shorten the interval to [86, 100].
TRY THIS. Kelly’s commissions for her first two sales of the Whirlwind Vacuum Cleaner were $80 and $90. In what range must her third commission lie for her aver-
age to be between $100 and $110? ■
When discussing the error made in a measurement, we may refer to the absolute error or the relative error. For example, if L is the actual length of an object and x is the length determined by a measurement, then the absolute error is 0x -L0 and the relative error is 0x - L0 >L.
EXAMPLE 9 Application of absolute value inequality
A technician is testing a scale with a 50-lb block of steel. The scale passes this test if the relative error when weighing this block is less than 0.1%. If x is the reading on the scale, then for what values of x does the scale pass this test?
Solution
If the relative error must be less than 0.1%, then x must satisfy the following in- equality:
0x- 500
50 60.001 Solve the inequality for x:
0x - 500 6 0.05 -0.05 6 x - 50 6 0.05 49.956 x 6 50.05
So the scale passes the test if it shows a weight in the interval (49.95, 50.05).
TRY THIS. A gas pump is certified as accurate if the relative error when dispens- ing 10 gallons of gas is less than 1%. If x is the actual amount of gas dispensed, then for what values of x is the pump certified as accurate? ■
1. The inequality -3 6x + 6 is equivalent to x +6 7 -3.
2. The inequality -2x6 -6 is equivalent to -2x-2 6 -6-2. 3. The smallest real number that satisfies x 7 12 is 13.
4. The number -6 satisfies 0x - 60 7 -1.
5. 1-⬁, -32傽1-⬁, -22 = 1-⬁, -22 6. 15, ⬁2傽1-⬁, -32= 1-3, 52
7. All negative numbers satisfy 0x- 20 6 0.
8. The compound inequality x 6 -3 or x 7 3 is equiva- lent to 0x0 6 -3.
9. The inequality 0x0 + 26 5 is equivalent to -5 6 x+ 2 65.
10. The fact that the difference between your age y and my age m is at most 5 years is written 0y - m0 … 5.
FOR thought... True or False? Explain.
Fill in the blank.
1. The set 5x兩x 7a6 is written in notation as 1a, ⬁2.
2. An interval does not include its endpoints.
3. A interval includes its endpoints.
4. An interval that involves the infinity symbol is an _______
interval.
5. Two simple inequalities connected with “and” or “or” is a _______ inequality.
6. The of sets A and B is the set of all elements that are in both A and B.
For each interval write an inequality whose solution set is the interval, and for each inequality, write the solution set in interval notation. See the summary of interval notation for unbounded intervals on page 82.
7. 1-⬁, 122 8. 1-⬁, -34 9. 3-7, ⬁2 10. 11.2, ⬁2 11. x Ú -8 12. x6 54 13. x 6p>2 14. x Ú 13
Solve each inequality. Write the solution set using interval notation and graph it.
15. 3x -67 9 16. 2x+ 166 17. 7- 5x… -3 18. -1- 4xÚ 7 19. 1
2 x- 46 1
3 x +5 20. 1 2-x7 x
3+1 4 21. 7-3x
2 Ú -3 22. 5 -x 3 … -2 23. 2x- 3
-5 Ú0 24. 5 -3x -7 … 0 25. -213x- 22Ú4-x 26. -5x …31x -92 Solve each inequality by reading the accompanying graph.
27. 1.8x +6.36 0 28. 1.2x- 3Ú0
29. 6.3 -4.5x Ú0 30. -5.1 -1.7x 60