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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
Copyright © 2011 Pearson Education, Inc.
Compound Inequalities8.18.1
1. Solve compound inequalities involving and.2. Solve compound inequalities involving or.
Slide 8 - 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.
Slide 8 - 4Copyright © 2011 Pearson Education, Inc.
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
(
)
( )
Slide 8 - 5Copyright © 2011 Pearson Education, Inc.
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.
Slide 8 - 6Copyright © 2011 Pearson Education, Inc.
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
Slide 8 - 7Copyright © 2011 Pearson Education, Inc.
continued
[
)
[ )
2 x
4x
2 and 4x x Without “and”: 2 x < 4
Set-builder notation: {x| 2 x < 4}
Interval notation: [2, 4)
Slide 8 - 8Copyright © 2011 Pearson Education, Inc.
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.
Slide 8 - 9
Copyright © 2011 Pearson Education, Inc.
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]
Slide 8 - 10
Copyright © 2011 Pearson Education, Inc.
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
Slide 8 - 11
Copyright © 2011 Pearson Education, Inc.
continued1 and 2x x
(
[
Solution set:[
Set-builder notation: {x|x 2}Interval notation: [2, ∞)
Slide 8 - 12Copyright © 2011 Pearson Education, Inc.
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
Slide 8 - 13Copyright © 2011 Pearson Education, Inc.
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.
Slide 8 - 14Copyright © 2011 Pearson Education, Inc.
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
[
)
[)
Slide 8 - 15Copyright © 2011 Pearson Education, Inc.
continued
Solution set:
Set-builder notation: {x|x < 1 or x 1}
Interval notation: (, 1) [1, )
1 orx 1x
Slide 8 - 16Copyright © 2011 Pearson Education, Inc.
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:
Slide 8 - 17Copyright © 2011 Pearson Education, Inc.
continued
Solution set:
Set-builder notation: {x|x ≤ 5}
Interval notation: (, 5]
4 orx 5x
Slide 8 - 18Copyright © 2011 Pearson Education, Inc.
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:
Slide 8 - 19Copyright © 2011 Pearson Education, Inc.
continued
Solution set:
Set-builder notation: {x|x is a real number}, or
Interval notation: (, )
1 orx 8x
.
Slide 2 - 20Copyright © 2011 Pearson Education, Inc.
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
Slide 2 - 21Copyright © 2011 Pearson Education, Inc.
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
Slide 2 - 22Copyright © 2011 Pearson Education, Inc.
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
Slide 2 - 23Copyright © 2011 Pearson Education, Inc.
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
Copyright © 2011 Pearson Education, Inc.
Equations Involving Absolute Value8.28.2
1. Solve equations involving absolute value.
Slide 8 - 25Copyright © 2011 Pearson Education, Inc.
Absolute Value PropertyIf |x| = a, where x is a variable or an expression and a 0, then x = a or x = a.
Slide 8 - 26Copyright © 2011 Pearson Education, Inc.
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.
Slide 8 - 27Copyright © 2011 Pearson Education, Inc.
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.
Slide 8 - 28Copyright © 2011 Pearson Education, Inc.
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.
Slide 8 - 29Copyright © 2011 Pearson Education, Inc.
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.
Slide 8 - 30Copyright © 2011 Pearson Education, Inc.
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
Slide 8 - 31Copyright © 2011 Pearson Education, Inc.
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
Slide 2 - 32Copyright © 2011 Pearson Education, Inc.
Solve:
a)
b)
c)
d)
3 5.x
2, 8
8,2
2,8
2,8
8.2
Slide 2 - 33Copyright © 2011 Pearson Education, Inc.
Solve:
a)
b)
c)
d)
3 5.x
2, 8
8,2
2,8
2,8
8.2
Slide 2 - 34Copyright © 2011 Pearson Education, Inc.
Solve:
a)
b)
c) all real numbers
d) no solution
2 5 4x
912 2x x
912 2x x
8.2
Slide 2 - 35Copyright © 2011 Pearson Education, Inc.
Solve:
a)
b)
c) all real numbers
d) no solution
2 5 4x
912 2x x
912 2x x
8.2
Copyright © 2011 Pearson Education, Inc.
Inequalities Involving Absolute Value8.38.3
1. Solve absolute value inequalities involving less than.2. Solve absolute value inequalities involving greater than.
Slide 8 - 37Copyright © 2011 Pearson Education, Inc.
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).
Slide 8 - 38Copyright © 2011 Pearson Education, Inc.
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)
( )
Slide 8 - 39Copyright © 2011 Pearson Education, Inc.
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
Slide 8 - 40Copyright © 2011 Pearson Education, Inc.
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.
A number line solution:
Set-builder notation: {x|3 < x < 9} Interval notation (3, 9).
)(
Slide 8 - 41Copyright © 2011 Pearson Education, Inc.
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.
.
Slide 8 - 42Copyright © 2011 Pearson Education, Inc.
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.
Slide 8 - 43Copyright © 2011 Pearson Education, Inc.
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
A number line solution:
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) (
Slide 8 - 44Copyright © 2011 Pearson Education, Inc.
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
.
Slide 2 - 45Copyright © 2011 Pearson Education, Inc.
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
Slide 2 - 46Copyright © 2011 Pearson Education, Inc.
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
Slide 2 - 47Copyright © 2011 Pearson Education, Inc.
Solve:
a)
b)
c)
d)
2 8 12.x
| 2 or 2x x x
2x x
2 2x x
2x x
8.3
Slide 2 - 48Copyright © 2011 Pearson Education, Inc.
Solve:
a)
b)
c)
d)
2 8 12.x
| 2 or 2x x x
2x x
2 2x x
2x x
8.3
Copyright © 2011 Pearson Education, Inc.
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.
Slide 8 - 50Copyright © 2011 Pearson Education, Inc.
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.
Slide 8- 51Copyright © 2011 Pearson Education, Inc.
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}
Slide 8- 52Copyright © 2011 Pearson Education, Inc.
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
Slide 8- 53Copyright © 2011 Pearson Education, Inc.
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
Slide 8- 54Copyright © 2011 Pearson Education, Inc.
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
Slide 8- 55Copyright © 2011 Pearson Education, Inc.
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
Slide 8- 56Copyright © 2011 Pearson Education, Inc.
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
Slide 8- 57Copyright © 2011 Pearson Education, Inc.
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
Slide 8- 58Copyright © 2011 Pearson Education, Inc.
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
Slide 3- 59Copyright © 2011 Pearson Education, Inc.
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
Slide 3- 60Copyright © 2011 Pearson Education, Inc.
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
Slide 3- 61Copyright © 2011 Pearson Education, Inc.
Given f(x) = −3x + 5, find f(−2).
a) −1
b) 2
c) 11
d) 13
8.4
Slide 3- 62Copyright © 2011 Pearson Education, Inc.
Given f(x) = −3x + 5, find f(−2).
a) −1
b) 2
c) 11
d) 13
8.4
Copyright © 2011 Pearson Education, Inc.
Function Operations8.58.5
1. Add or subtract functions.2. Multiply functions.3. Divide functions.
Slide 8 - 64Copyright © 2011 Pearson Education, Inc.
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).
Slide 8- 65Copyright © 2011 Pearson Education, Inc.
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
Slide 8 - 66Copyright © 2011 Pearson Education, Inc.
Multiplying Functions
The product of two functions, f • g, is founded by (f • g)(x) = f(x)g(x).
Slide 8 - 67Copyright © 2011 Pearson Education, Inc.
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
Slide 8 - 68Copyright © 2011 Pearson Education, Inc.
Dividing Functions
The quotient of two functions, f / g, is founded by
/ , where 0.
f xf g x g x
g x
Slide 8 - 69Copyright © 2011 Pearson Education, Inc.
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
Slide 3- 70Copyright © 2011 Pearson Education, Inc.
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
Slide 3- 71Copyright © 2011 Pearson Education, Inc.
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
Slide 3- 72Copyright © 2011 Pearson Education, Inc.
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
Slide 3- 73Copyright © 2011 Pearson Education, Inc.
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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