Download Inequalities

10TH EDITION
COLLEGE ALGEBRA
LIAL
HORNSBY
SCHNEIDER
1.7 - 1
1.7 Inequalities
Linear Inequalities
Three-Part Inequalities
Quadratic Inequalities
Rational Inequalities
1.7 - 2
Properties of Inequality
For real numbers a, b and c:
1. If a < b, then a + c < b + c,
2. If a < b and if c > 0, then ac < bc,
3. If a < b and if c < 0, then ac > bc.
Replacing < with >, ≤ , or ≥ results in similar properties.
1.7 - 3
Motion Problems
Note Multiplication may be replaced
by division in properties 2 and 3.
Always remember to reverse the
direction of the inequality symbol when
multiplying or dividing by a negative
number.
1.7 - 4
Linear Inequality in One
Variable
A linear inequality in one variable is an
inequality that can be written in the
form
ax  b  0,
where a and b are real numbers with
a ≠ 0. (Any of the symbols ≥, <, or ≤
may also be used.)
1.7 - 5
Example 1
SOLVING A LINEAR INEQUALITY
Solve 3 x  5  7
Solution
3 x  5  7
3 x  5  5  7  5
3 x  12
Don’t forget to
reverse the
symbol here.
3 x 12

3
3
Subtract 5.
Divide by –3; reverse
direction of the
inequality symbol
when multiplying or
dividing by a negative
number.
1.7 - 6
Interval
Notation
Open
interval
{x  x > a}
{x  a < x < b}
{x  x < b}
(a, )
(a, b)
(– , b)
{x  x ≥ a}
{x  a < x ≤ b}
{x  a ≤ x < b}
{x  x≤ b}
[a, )
(a, b]
[a, b)
(– , b]
Other
interval
Graph
(
a
(
a
(
Set
(
Type of
Interval
b
b
[
a
(
a
[
a
]
b
]
)
b
b
1.7 - 7
Set
Interval
Notation
Closed
interval
{xa ≤ x ≤ b}
[a, b]
Disjoint
interval
{xx < a or x > b}
(– , a) (b, )
All real
numbers {xx is a real numbers}
Graph
]
b
[
a
(
Type of
Interval
a
(
b
(– , )
1.7 - 8
Example 2
SOLVING A LINEAR INEQUALITY
Solve 4 – 3x ≤ 7 + 2x. Give the solution
set in interval notation and graph it.
Solution
4  3 x  7  2x
4  3 x  4  7  2x  4
Subtract 4.
3 x  3  2x
3 x  2x  3  2x  2x
Subtract 2x.
5 x  3
1.7 - 9
Example 2
SOLVING A LINEAR INEQUALITY
Solve 4 – 3x ≤ 7 + 2x. Give the solution
set in interval notation and graph it.
Solution
5 x  3
5 x 3

5 x 5
Divide by –5; reverse the
direction of the inequality
symbol.
3
x
5
[

3
5
0
 3 
In interval notation the solution set is   ,   .
 5 
1.7 - 10
Example 3
SOLVING A THREE-PART
INEQUALITY
Solve –2 < 5 + 3x < 20.
Solution
2  5  3 x  20
2  5  5  3 x  5  20  5
7  3 x  15
Divide each
part by 3.
7 3 x 15


3
3
3
7
 x5
3
Subtract 5 from
each part.
The solution set is
 7 
the interval   ,5  .
 3 
1.7 - 11
Example 4
FINDING THE BREAK-EVEN
POINT
If the revenue and cost of a certain product
are given by
R  4 x and C  2x  1000,
where x is the number of units produced and
sold, at what production level does R at least
equal C?
Solution Set R ≥ C and solve for x.
1.7 - 12
Example 4
FINDING THE BREAK-EVEN
POINT
Solution Set R ≥ C and solve for x.
R C
4 x  2 x  1000
Substitute.
2 x  1000
Subtract 2x.
x  500
Divide by 2.
The break-even point is at x = 500. This
product will at least break even only if the
number of units produced and sold is in the
interval [500, ).
1.7 - 13
Quadratic Inequalities
A quadratic inequality is an inequality
that can be written in the form
ax 2  bx  c  0
for real numbers a, b, and c with a ≠ 0.
(The symbol < can be replaced with >,
≤, or ≥.)
1.7 - 14
Solving a Quadratic Inequality
Step 1 Solve the corresponding quadratic
equation.
Step 2 Identify the intervals determined by
the solutions of the equation.
Step 3 Use a test value from each interval
to determine which intervals form
the solution set.
1.7 - 15
Example 5
SOLVING A QUADRATIC
INEQULITY
Solve x 2  x  12  0
Solution
Step 1 Find the values of x that satisfy
x2 – x – 12 = 0
x  x  12  0
( x  3)( x  4)  0
2
x  3  0 or
x40
Corresponding quadratic
equation
Factor.
Zero-factor
property
1.7 - 16
Example 5
SOLVING A QUADRATIC
INEQULITY
Solve x 2  x  12  0
Solution
Step 1
x  3  0 or
x  3
or
x40
Zero-factor
property
x4
1.7 - 17
Example 5
SOLVING A QUADRATIC
INEQULITY
Step 2 The two numbers –3 and 4 divide
the number line into three intervals. If a
value in Interval A makes the polynomial
negative, then all values in Interval A will
make the polynomial negative.
Interval A
Interval B
Interval C
(–, – 3)
(– 3, 4)
(4, )
–3
0
4
1.7 - 18
Example 5
SOLVING A QUADRATIC
INEQULITY
Step3 Choose a test value from each interval.
Is x2 –x – 12< 0
Test
Interval
Value
True or False?
–4
(–4)2 – 4(–4) – 12 < 0 ?
A: (– , – 3)
8 < 0 False
B: (– 3, 4)
02 – 0 – 12 < 0 ?
0
–12 < 0 True
52 – 5 – 12 < 0 ?
C: (4, )
5
8 < 0 False
Since the values in Interval B make the inequality
true, the solution set is (– 3, 4).
1.7 - 19
Example 6
SOLVING QUADRATIC
INEQULAITY
Solve 2 x 2  5 x  12  0.
Solution
Step 1 Find the values of x that satisfy this
equation.
Corresponding
2
2 x  5 x  12  0
quadratic equation
(2 x  3)( x  4)  0
2x  3  0
or
x40
Factor.
Zero-factor
property
1.7 - 20
Example 6
SOLVING QUADRATIC
INEQULAITY
Solve 2 x 2  5 x  12  0.
Solution
Step 1 Find the values of x that satisfy this
equation.
2x  3  0
or
x40
3
x
2
or
x  4
Zero-factor
property
1.7 - 21
Example 6
SOLVING QUADRATIC
INEQULAITY
Solve 2 x 2  5 x  12  0.
Solution
Step 2 The values form the intervals on
the number line.
Interval A
Interval B
(–, – 4)
(– 4, 3/2)
–4
Interval C
(3/2, )
0
3/2
1.7 - 22
Example 6
SOLVING QUADRATIC
INEQULAITY
Step3 Choose a test value from each interval.
Is 2x2 + 5x – 12 ≥ 0
Test
Interval
Value
True or False?
–5
2(–4)2 +5(–5) – 12 ≥ 0 ?
A: (– , – 4)
13 ≥ 0 True
B: (4, 3/2)
2(0)2 +5(0) – 12 ≥ 0 ?
0
–12 ≥ 0 False
2(2)2 + 5(2) – 12 ≥ 0 ?
C: (3/2, )
2
6 ≥ 0 True
The values in Intervals A and B make the inequality true, the
solution set is the union of the intervals.  ,  4    3 ,  
 8


1.7 - 23
Inequalities
Note Inequalities that use the
symbols < and > are called strict
inequalities; ≤ and ≥ are used in nonstrict
inequalities. The solutions of the equation
in Example 5 were not included in the
solution set since the inequality was a strict
inequality. In Example 6, the solutions of
the equation were included in the solution
set because of the nonstrict inequality.
1.7 - 24
Example 7
SOLVING A PROBLEM INVOLVING
THE HEIGHT OF A PROJECTILE
If a projectile is launched from ground level
with an initial velocity of 96 ft per sec, its
height in feet t seconds after launching is s
feet, where
2
s  16t  96t .
When will the projectile be greater than 80 ft
above ground level?
1.7 - 25
SOLVING A PROBLEM INVOLVING
THE HEIGHT OF A PROJECTILE
Example 7
Solution
16t  96t  80
2
16t  96t  80  0
2
t  6t  5  0
2
Reverse direction
of the inequality
symbol.
Set s greater than
80.
Subtract 80.
Divide by – 16.
Now solve the
corresponding equation.
1.7 - 26
Example 7
SOLVING A PROBLEM INVOLVING
THE HEIGHT OF A PROJECTILE
Solution
t  6t  5  0
2
(t  1)(t  5)  0
t 1
or t  5
Factor.
Zero-factor property
1.7 - 27
Example 7
SOLVING A PROBLEM INVOLVING
THE HEIGHT OF A PROJECTILE
Solution
Interval B Interval C
Interval A
(– , 1)
(5, )
(1, 5)
0
1
5
Use the procedure of Examples 5 and 6 to
determine that values in Interval B, (1, 5) ,
satisfy the inequality. The projectile is
greater than 80 ft above ground level
between 1 and 5 sec after it is launched.
1.7 - 28
Solving a Rational Inequality
Step 1 Rewrite the inequality, if necessary, so that
0 is on one side and there is a single fraction on the
other side.
Step 2 Determine the values that will cause either
the numerator or the denominator of the rational
expression to equal 0. These values determine the
intervals of the number line to consider.
1.7 - 29
Solving a Rational Inequality
Step 3 Use a test value from each interval to
determine which intervals form the solution set.
A value causing the denominator to equal zero will
never be included in the solution set. If the
inequality is strict, any value causing the numerator
to equal zero will be excluded; if nonstrict, any such
value will be included.
1.7 - 30
Caution Solving a rational inequality such as
5
1
x4
by multiplying both sides by x + 4 to obtain
5 ≥ x + 4 requires considering two cases, since
the sign of x + 4 depends on the value of x. If
x + 4 were negative, then the inequality
symbol must be reversed. The procedure
described in the next two examples eliminates
the need for considering separate cases.
1.7 - 31
Example 8
SOLVING A RATIONAL INEQUALITY
5
 1.
Solve
x4
Solution
Step 1
5
1 0
x4
5
x4

0
x4 x4
Subtract 1 so that 0 is on
one side.
Use x + 4 as the common
denominator.
1.7 - 32
Example 8
SOLVING A RATIONAL INEQUALITY
5
 1.
Solve
x4
Solution
Step 1
5
x4

0
x4 x4
Note the
careful use
of
parentheses.
5  ( x  4)
0
x4
1 x
0
x4
Use x + 4 as the common
denominator.
Write as a single fraction.
Combine terms in the
numerator; be careful
with signs.
1.7 - 33
Example 8
SOLVING A RATIONAL INEQUALITY
5
 1.
Solve
x4
Solution
Step 2 The quotient possibly changes sign
only where x-values make the numerator or
denominator 0. This occurs at
1 x  0
or x  4  0
x  1 or
x  4
1.7 - 34
Example 8
SOLVING A RATIONAL INEQUALITY
5
 1.
Solve
x4
Solution
Step 2
Interval A
(–, – 4)
Interval B
Interval C
(1, )
(– 4, 1)
–4
0
1
1.7 - 35
Example 8
SOLVING A RATIONAL INEQUALITY
Step3 Choose test values.
Test
Interval
Value
–5
A: (– , – 4)
B: (–4, 1)
0
C: (1, )
2
5
Is
 1True or False?
x4
5
 1 ? 5  1 False
5  4
5
 1 ? 5  1 True
04
4
5
5
1 ?
 1 False
24
6
The values in the interval (–4, 1) satisfy the original inequality. The value 1
makes the nonstrict inequality true, so it must be included in the solution
set. Since –4 makes the denominator 0, it must be excluded. The
solutions set is (–4, 2].
1.7 - 36
Caution As suggested by Example 8,
be careful with the endpoints of the
intervals when solving rational
inequalities.
1.7 - 37
Example 9
SOLVING A RATIONAL
INEQULAITY
2x  1
 5.
Solve
3x  4
Solution
2x  1
5 0
3x  4
2 x  1 5(3 x  4)

0
3x  4
3x  4
2x  1  5(3 x  4)
0
3x  4
Subtract 5.
Common denominator
is 3x + 4.
Write as a single
fraction.
1.7 - 38
Example 9
SOLVING A RATIONAL
INEQULAITY
2x  1
 5.
Solve
3x  4
Solution
2x  1  5(3 x  4)
0
3x  4
Be careful
with signs.
2x  1  15 x  20
0
3x  4
13 x  21
0
3x  4
Write as a single
fraction.
Distributive property
Combine terms in the
numerator.
1.7 - 39
Example 9
SOLVING A RATIONAL
INEQULAITY
2x  1
 5.
Solve
3x  4
Solution
Set the numerator and denominator equal to
0 and solve the resulting equations to get the
values of x where sign changes may occur.
13 x  21  0
21
x
13
or
or
3x  4  0
4
x
3
1.7 - 40
Example 9
SOLVING A RATIONAL
INEQULAITY
2x  1
 5.
Solve
3x  4
Solution
Interval A
Interval B
21



,



13 

 21 4 
  , 
 13 3 

21
13
Interval C
 4 
  , 
 3 

4
3
1.7 - 41
Example 9
SOLVING A RATIONAL
INEQULAITY
Solution
Now choose from the intervals and verify that:
–2 from Interval A makes the inequality true;
–1.5 from Interval B makes the inequality false;
0
from Interval C makes the inequality true.
Because of the < symbol, neither endpoint satisfies
the inequality, so the solution set is
21  4 

 ,      ,   .
13   3 

1.7 - 42