copyright © 2011 pearson education, inc. more on inequalities, absolute value, and functions...

Copyright © 2011 Pearson Education, Inc.

More on Inequalities, Absolute Value, and Functions

CHAPTER

8.1 Compound Inequalities8.2 Equations Involving Absolute Value8.3 Inequalities Involving Absolute Value8.4 Functions and Graphing8.5 Function Operations

88


Compound Inequalities8.18.1

1. Solve compound inequalities involving and.2. Solve compound inequalities involving or.

3Copyright © 2011 Pearson Education, Inc.

Compound Inequality: Two inequalities joined by either and or or.

Examples: x > 3 and x 82 x or x > 4

Intersection: For two sets A and B, the intersection of A and B, symbolized by A B, is a set containing only elements that are in both A and B.


Example 1For the compound inequality x > 5 and x < 2, graph the solution set and write the compound inequality without “and” if possible. Then write in set-builder notation and in interval notation.Solution The set is the region of intersection.

x > 5

x < 2

x > 5 and x < 2

(

)

( )


continued

x > 5 and x < 2

Without “and”: 5 < x < 2

Set-builder notation: {x| 5 < x < 2}

Interval notation: (5, 2) Warning: Be careful not to confuse the interval notation with an ordered pair.


Example 2For the inequality graph the solution set. Then write the solution set in set-builder notation and in interval notation.

Solution Solve each inequality in the compound inequality.

2 1 3 3 12,x and x

2 1 3x 2 4 x

2 x

3 12x 4x and


continued

[

)

[ )

2 x

4x

2 and 4x x Without “and”: 2 x < 4

Set-builder notation: {x| 2 x < 4}

Interval notation: [2, 4)


Solving Compound Inequalities Involving andTo solve a compound inequality involving and,1. Solve each inequality in the compound

inequality.2. The solution set will be the intersection of the

individual solution sets.

9


Example 3a

Solution

4 2 8x

For the compound inequality, graph the solution set. Then write the solution set in set-builder notation and in interval notation.

2 4 and 2 8x x 2 and 6x x

](

Set-builder notation: {x| 2 < x 6}Interval notation: (2, 6]

10


Example 3b

Solution

5 2 3 and 2 4 8x x

For the compound inequality, graph the solution set. Then write the solution set in set-builder notation and in interval notation.

5 2 3 and 2 4 8x x

5 5 and 2 4x x

1 and 2x x

11


continued1 and 2x x

(

[

Solution set:[

Set-builder notation: {x|x 2}Interval notation: [2, ∞)


Solution

For the compound inequality, graph the solution set. Then write the solution set in set-builder notation and in interval notation. 5 10 4 3.x and x

5 1.x and x 5 10 4 3x and x

Since no number is greater than 5 and less than 1, the solution set is the empty set .

Example 3c

Set builder notation: { } or

Interval notation: We do not write interval notation because there are no values in the solution set.

.-10 -5 0 5 10


Union: For two sets A and B, the union of A and B, symbolized by A B, is a set containing every element in A or in B.

Solving Compound Inequalities Involving orTo solve a compound inequality involving or,1. Solve each inequality in the compound

inequality.2. The solution set will be the union of the

individual solution sets.


Example 4aFor the compound inequality, graph the solution set. Then write the solution set in set-builder notation and in interval notation. 2 1 3 3 3.x or x Solution2 1 3x

2 2 x 1 x

3 3x 1x or

[

)

[)


continued

Solution set:

Set-builder notation: {x|x < 1 or x 1}

Interval notation: (, 1) [1, )

1 orx 1x


Example 4bFor the compound inequality, graph the solution set. Then write the solution set in set-builder notation and in interval notation. 3 4 8 or 4 1 19x x Solution3 4 8x

3 12 x 4 x

or

]

]

4 1 19x 4 20x

5x

]Solution set:


continued

Solution set:

Set-builder notation: {x|x ≤ 5}

Interval notation: (, 5]

4 orx 5x


Example 4cFor the compound inequality, graph the solution set. Then write the solution set in set-builder notation and in interval notation. 5 13 18 or 6 12 36x x Solution

5 13 18x 5 5 x

1 x or

]

(

6 12 36x 6 48x

8x

Solution set:


continued

Solution set:

Set-builder notation: {x|x is a real number}, or

Interval notation: (, )

1 orx 8x

.


Solve:

a)

b)

c)

d)

4 1 0.x

3 1x x

3 1x x

1 3x x or x

4 1x x or x

8.1


Solve:

a)

b)

c)

d)

5 2 4 0.x or x

7 4x x

4 7x x

7 4x x or x

4 7x x or x

8.1


Equations Involving Absolute Value8.28.2

1. Solve equations involving absolute value.


Absolute Value PropertyIf |x| = a, where x is a variable or an expression and a 0, then x = a or x = a.


Example 1Solve. a. |2x +1| = 5 b. |3 – 4x| = –10 Solutiona.

b. |3 – 4x| = –10 The solution are –3 and 2.

x = –3 or x = 2

2x = –6 or 2x = 4

2x +1 = –5 or 2x +1 = 5

This equation has the absolute value equal to a negative number. Because the absolute value of every real number is a positive number or zero, this equation has no solution.


Example 2Solve. |3x + 4| + 3 = 11 Solution|3x + 4| + 3 = 11 Subtract 3 from both sides to isolate the absolute

value. |3x + 4| = 83x + 4 = 8 or 3x + 4 = –8

3x = 4 or 3x = –12 x = 4/3 or x = –4

The solutions are 4/3 and 4.


Solving Equations Containing a Single Absolute ValueTo solve an equation containing a single absolute value,1. Isolate the absolute value so that the equation is

in the form |ax + b| = c. If c > 0, proceed to Steps 2 and 3. If c < 0, the equation has no solution.2. Separate the absolute value into two equations, ax + b = c and ax + b = c.3. Solve both equations.


Solving Equations in the Form |ax + b| = |cx + d|To solve an equation in the form |ax + b| = |cx + d|,1. Separate the absolute value equation into two equations: ax + b = cx + d and ax + b = (cx + d).2. Solve both equations.


Example 3a Solve: |3x – 5| = |8 + 4x|.

Solution

3x – 5 = 8 + 4x or 3x – 5 = (8 + 4x)

–13 + 3x = 4x

–13 = x

The solutions are 13 and 3/7.

3 5 8 4x x 7 5 8x

7 3x 3

7x


Example 3b Solve: |2x – 9| = |3 − 2x|.

Solution

2x – 9 = 3 − 2x or 2x – 9 = (3 − 2x)

4x = 12 2x – 9 = 3 + 2x

x = 3

The absolute value equation has only one solution, 3.

– 9 = 3


Solve:

a)

b)

c)

d)

3 5.x

2, 8

8,2

2,8

2,8

8.2


Solve:

a)

b)

c) all real numbers

d) no solution

2 5 4x

912 2x x

912 2x x

8.2


Inequalities Involving Absolute Value8.38.3

1. Solve absolute value inequalities involving less than.2. Solve absolute value inequalities involving greater than.


Solving Inequalities in the Form |x| < a, where a > 0To solve an inequality in the form |x| < a, where a > 0,1. Rewrite the inequality as a compound inequality

involving and: x >a and x < a (or use a < x < a).2. Solve the compound inequality. Similarly, to solve |x| a, we write x a and

x a (or a x a).


Example 1aFor the inequality, solve, graph the solution set, and write the solution set in both set-builder and interval notation. |x| < 9.Solution

x > −9 and x < 9So our graph is as follows:

Set-builder notation: {x|9 < x < 9} Interval notation (9, 9)

( )


Example 1bFor the inequality, solve, graph the solution set, and write the solution set in both set-builder and interval notation. |x − 4| ≤ 5Solution

A number line solution:

Set-builder notation: {x|1 ≤ x ≤ 9} Interval notation [1, 9]

][

x – 4 −5 and x – 4 ≤ 5

x −1 and x ≤ 9


Example 1cFor the inequality, solve, graph the solution set, and write the solution set in both set-builder and interval notation. |x – 3| < 6Solution

|x – 3| < 6 6 < x – 3 < 6 Rewrite as a compound inequality.

3 < x < 9 Add 3 to each part of the inequality.


Set-builder notation: {x|3 < x < 9} Interval notation (3, 9).

)(


Example 1eFor the inequality, solve, graph the solution set, and write the solution set in both set-builder and interval notation. |2x – 3| + 8 < 5.SolutionIsolate the absolute value.

|2x – 3| + 8 < 5 |2x – 3| < –3

Since the absolute value cannot be less than a negative number, this inequality has no solution: .Set builder notation: { } or

Interval notation: We do not write interval notation because there are no values in the solution set.

.


Solving Inequalities in the Form |x| > a, where a > 0To solve the inequality in the form x| a, where a > 0,

1. Rewrite the inequality as a compound inequality involving or: x < a or x > a.

2. Solve the compound inequality. Similarly, to solve |x| a, we would write x a

or x a.


Example 2aFor the inequality, solve, graph the solution set, and write the solution set in both set-builder and interval notation. |x + 7| > 5.SolutionWe convert to a compound inequality and solve each.

|x + 7| > 5x + 7 < 5 or x + 7 > 5 x < 12


Set-builder notation: {x| x < 12 or x > 2}Interval notation: (, 12) (2, ).

2x

5-14 -12 -10 -8 -6 -4 -2 0 2 4-15 -13 -9 -5 -1 3-15 -7 1-11 5-3) (


Example 2dFor the inequality, solve, graph the solution set, and write the solution set in both set-builder and interval notation. |4x + 7| 9 > 12.SolutionIsolate the absolute value |4x + 7| 9 > 12

|4x + 7| > 3This inequality indicates that the absolute value is greater than a negative number. Since the absolute value of every real number is either positive or 0, the solution set is . Set-builder notation: {x|x is a real number} or Interval notation: (, ).

5-14 -12 -10 -8 -6 -4 -2 0 2 4-15 -13 -9 -5 -1 3-15 -7 1-11 5-3

.


Solve:

a)

b)

c)

d)

2 1 7.x

4 4x x

4 3x x or x

4 3x x

3 3x x

8.3


Solve:

a)

b)

c)

d)

2 8 12.x

| 2 or 2x x x

2x x

2 2x x

2x x

8.3


Functions and Graphing8.48.4

1. Identify the domain and range of a relation and determine if the relation is a function.

2. Find the value of a function.3. Graph functions.


Relation: A set of ordered pairs.

Domain: The set of all input values (x-values) for a relation.

Range: The set of all output values (y-values) for a relation.

Function: A relation in which every value in the domain is paired with exactly one value in the range.


Example 1aIdentify the domain and range of the relation, then determine whether it is a function. Birthdate Family memberMarch 1 DonnaApril 17 DennisSept. 3 CatherineOctober 9 Denise

Nancy

The relation is not a function because an element in the domain, Sept. 3, is assigned to two names in the range.

Domain: {March 1, April 17, Sept 3, Oct 9}

Range: {Donna, Dennis, Catherine, Denise, Nancy}


Example 1bIdentify the domain and range of the relation, then determine whether it is a function. {(−4, −1), (−2, 1), (0, 0), (2, −1), (4, 2)}

The relation is a function because every value in the domain is paired with only one value in the range.

Domain: {−4, −2, 0, 2, 4}

Range: {−1, 1, 0, −1, 2}

Solution


continued

For each graph, identify the domain and range. Then state whether each relation is a function.c. d.

Domain: {x|x 1}Range: all real numbers

Not a functionDomain: all real numbersRange: {y 1}

Function


Example 2

For the function f(x) = 3x – 5, find the following.a. f(2) b. f(4)

Solutiona. f(2) = 3x – 5

= 3(2) – 5 = 6 – 5 = 1

b. f(4) = 3x – 5

= 3(4) – 5

= 12 – 5

= 17


Example 3

Graph:

SolutionWe use (0, 2) as one ordered pairand then use the slope, , to findthe second ordered pair. Recall that slope indicates the “rise” and “run” from any point on the line to another point on the line.

32

4f x x

3

4

32

4 y x


Example 4Graph. f(x) = 2x2

Solution We create a table of ordered pairs, plotthe points, and connect with a smooth curve.

x f(x)

2 8

1 2

0 0

1 2

2 8


Example 5bGraph. f(x) = |x| + 2Solution We create a table of ordered pairs, plotthe points, and connect the points.

x f(x)

4 6

2 4

0 2

2 4

4 6

x

y

-10 -8 -6 -4 -2 2 4 6 8 10

-10

-8

-6

-4

-2

2

4

6

8

10

0


Example 5cGraph. f(x) = |x − 3| + 2Solution We create a table of ordered pairs, plotthe points, and connect the points.

x f(x)

0 5

2 3

3 2

4 3

6 5

x

y

-10 -8 -6 -4 -2 2 4 6 8 10

-10

-8

-6

-4

-2

2

4

6

8

10

0


How would you graph the function y = 3x – 2?

a) Plot (0, –2), down 2, right 3

b) Plot (0, –2), down 3, right 1

c) Plot (0, 2), up 3, left 2

d) Plot (0, –2), up 3, right 1

8.4


Given f(x) = −3x + 5, find f(−2).

a) −1

b) 2

c) 11

d) 13

8.4


Function Operations8.58.5

1. Add or subtract functions.2. Multiply functions.3. Divide functions.


Adding or Subtracting FunctionsThe sum of two functions, f + g, is founded by (f + g)(x) = f(x) + g(x).

The difference of two functions, f – g, is founded by (f – g)(x) = f(x) – g(x).


Example 1

Given f(x) = 3x + 1 and g(x) = 5x + 2, find the following.

a. f + g b. f g c. (f – g)(−2)Solutiona. f + g = f(x) + g(x)

= (3x + 1) + (5x + 2) = 8x + 3 b. f − g = f(x) − g(x)

= (3x + 1) − (5x + 2)

= − 2x − 1

c. Replace x with −2.

(f − g)(−2) = −2(−2) + 1

= 4 + 1

= 5


Multiplying Functions

The product of two functions, f • g, is founded by (f • g)(x) = f(x)g(x).


Example 2

Given f(x) = 2x + 7 and g(x) = x − 4, find f • g.

Solution

f • g = f(x)g(x) = (2x + 7)(x − 4)

= 2x2 − 8x + 7x – 28

= 2x2 − x – 28


Dividing Functions

The quotient of two functions, f / g, is founded by

/ , where 0.

f xf g x g x

g x


Example 3

Given f(x) = 16x3 −12x2 + 8x and g(x) = 4x, find f / g.

Solution

3 216 12 8

/4

f x x x xf g x

g x x

3 216 12 8

4 4 4

x x x

x x x

24 3 2x x


Given f(x) = 4x – 1 and g(x) = 5x + 2, what is f + g?

a) x + 4

b) x − 4

c) 9x + 1

d) 9x – 1

8.5


Given f(x) = 3x – 2 and g(x) = 5x – 1, what is f • g?

a) 15x2 − 13x + 2

b) 15x2 − 13x − 2

c) 15x2 − 7x + 2

d) 15x2 − 7x − 2

8.5

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