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Chabot Mathematics
§4.3a Absolute
Value
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Chabot College Mathematics
1
Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Review § 4.2
MTH 55
 Any QUESTIONS About
• §4.2 → InEqualities & Problem-Solving
 Any QUESTIONS About HomeWork
• §4.2 → HW-12
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Absolute Value
 The absolute value of x denoted |x|,
is defined as
x  0  x  x;
x  0  x   x.
 The absolute value of x represents the
distance from x to 0 on the number line
• e.g.; the solutions of |x| = 5 are 5 and −5.
5 units from zero
5 units from zero
x = –5 or x = 5
–5
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0
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Graph y = |x|
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 Make T-table
x
-6
-5
-4
-3
-2
-1
0
1
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3
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5
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y = |x |
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5
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3
2
1
0
1
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y
4
3
2
x
1
0
-6
-5
-4
-3
-2
-1
0
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-1
-2
-3
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-5
file =XY_Plot_0211.xls
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Absolute Value Properties
1. |ab| = |a |· |b| for any real numbers a & b
•
The absolute value of a product is the
product of the absolute values
2. |a/b| = |a|/|b| for any real numbers a & b  0
•
The absolute value of a quotient is the
quotient of the absolute values
3. |−a| = |a| for any real number a
•
The absolute value of the opposite of a number is
the same as the absolute value of the number
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Example  Absolute Value Calcs
 Simplify, leaving as little as possible
inside the absolute-value signs
8x
2
a. |7x|
b. |−8y| c. |6x | d.
2
4 x

SOLUTION
a. |7x| = 7  x
7 x
b. |−8y| = 8  y  8 y
2
2
2
2

6x
c. |6x | = 6  x  6 x
2
2
8x
2
d. .


 
2
x
x
4 x
x
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Bruce Mayer, PE
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Distance & Absolute-Value
 For any real numbers a and b, the
distance between them is |a – b|
 Example  Find the distance between
−12 and −56 on the number line
 SOLUTION
• |−12 − (−56)| = |+44| = 44
• Or
• |−56 − (−12)| = |−44| = 44
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Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  AbsVal Expressions
 Find the Solution-Sets for
a) |x| = 6
b) |x| = 0
c) |x| = −2
 SOLUTION a) |x| = 6
 We interpret |x| = 6 to mean that the
number x is 6 units from zero on a
number line.
 Thus the solution set is {−6, 6}
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Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  AbsVal Expressions
 Find the Solution-Sets for
a) |x| = 6
b) |x| = 0
c) |x| = –2
 SOLUTION b) |x| = 0
 We interpret |x| = 0 to mean that x is 0
units from zero on a number line. The
only number that satisfies this criteria is
zero itself.
 Thus the solution set is {0}
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Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  AbsVal Expressions
 Find the Solution-Sets for
a) |x| = 6
b) |x| = 0
c) |x| = −2
 SOLUTION c) |x| = −2
 Since distance is always NonNegative,
|x| = −2 has NO solution.
 Thus the solution set is Ø
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Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Absolute Value Principle
 For any positive number p and
any algebraic expression X:
a. The solutions of |X| = p are those
numbers that satisfy X = −p or X = p
b. The equation |X| = 0 is equivalent to
the equation X = 0
c. The equation |X| = −p has no solution.
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Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  AbsVal Principle
 Solve: a) |2x+1| = 5; b) |3 − 4x| = −10
 SOLUTION a) |2x + 1| = 5
• use the absolute-value principle, replacing
X with 2x + 1 and p with 5. Then we solve
each equation separately
|X| = p
|2x +1| = 5
2x +1 = −5 or 2x +1 = 5
2x = −6 or
2x = 4
x = −3 or
x=2
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Absolute-value principle
 Thus The solution
set is {−3, 2}.
Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  AbsVal Principle
 Solve: a) |2x+1| = 5; b) |3 − 4x| = −10
 SOLUTION b) |3 − 4x| = −10
• The absolute-value principle reminds us
that absolute value is always nonnegative.
• So the equation |3 − 4x| = −10 has
NO solution.
• Thus The solution set is Ø
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Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  AbsVal Principle
 Solve |2x + 3| = 5
 SOLUTION
• For |2x + 3| to equal 5, 2x + 3 must be 5
units from 0 on the no. line. This can happen
only when 2x + 3 = 5 or 2x + 3 = −5.
2x + 3 = 5
or
2x + 3 = –5
• Solve
2x = 2
or
2x = –8
Equation
x=1
or
x = –4
Set
• Graphing the Solutions
–5
–4
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–3
–2
–1
0
1
2
3
4
5
Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Solving 1-AbsVal Equations

To 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 for x
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Bruce Mayer, PE
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Two AbsVal Expression Eqns
 Sometimes an equation has TWO
absolute-value expressions.
 Consider |a| = |b|. This means that a
and b are the same distance from zero.
 If a and b are the same distance from
zero, then either they are the same
number or they are opposites.
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Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  2 AbsVal Expressions
 Solve: |3x – 5| = |8 + 4x|.
 SOLUTION
• Recall that if |a| = |b| then either they are
the same or they are opposites
This assumes these
numbers are the same
3x – 5 = 8 + 4x

OR
3x – 5 = –(8 + 4x)
Need to solve Both Eqns for x
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This assumes these
numbers are opposites.
Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  2 AbsVal Expressions
1. 3x – 5 = 8 + 4x 
–13 + 3x = 4x
–13 = x

2. 3x – 5 = –(8 + 4x)
3x  5  8  4 x
7 x  5  8
7 x  3
3
x
7
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Thus For Eqn
|3x – 5| = |8 + 4x|
The solutions are
•
−13
•
−3/7
Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Solve Eqns of 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.
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[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Inequalities &AbsVal Expressions
 Example  Solve: |x| < 3 Then graph
 SOLUTION
• The solutions of |x| < 3 are all numbers whose
distance from zero is less than 3. By
substituting we find that numbers such as
−2, −1, −1/2, 0, 1/3, 1, and 2 are all solutions.
• The solution set is {x| −3 < x < 3}. In interval
notation, the solution set is (−3, 3). The graph:
(
-3
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)
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Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Inequalities & AbsVal Expressions
 Example  Solve |x| ≥ 3 Then Graph
 SOLUTION
• The solutions of |x| ≥ 3 are all numbers whose
distance from zero is at least 3 units. The
solution set is {x| x ≤ −3 or x ≥ 3}
• In interval notation, the solution set is
(−, −3] U [3, )
• The Solution
Graph
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]
−3
[
3
Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Basic Absolute Value Eqns
Absolute Value
Equivalent
Solution
Equation
x  k k  0 
State
x  k or x  k
Set
-k,k
x 0
x  k k  0
x0
•
x 7
x  7 or x  7
x 0
x0
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
Examples 
x  13
0
-7 ,7
0

Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  Catering Costs
 Johnson Catering charges $100 plus
$30 per hour to cater an event.
Catherine’s Catering charges a straight
$50 per hour rate. For what lengths of
time does it cost less to hire
Catherine’s Catering?
 Familiarize → LET
• x ≡ the Catering time in hours
• TotalCost = (OneTime Charge) plus
(Hourly Rate)·(Catering Time)
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[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  Catering Costs
 Translate
Cathrine's
Cost
50 x
 Carry
Out
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Is Less
Than

Johnson's
Cost
30 x  100
50 x  100  30 x
20 x  100
x5
Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  Catering Costs
 Check
50  5  30  5  100
?
?
250 150  100
250  250 
 STATE
 For values of x < 5 hr, Catherine’s
Catering will cost less than Johnson’s
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WhiteBoard Work
 Problems From §4.3 Exercise Set
• 8, 24, 34, 38, 40, 48
y

Graph of
Absolute
Value
Function
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x
0
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[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
All Done for Today
Cool
Catering
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Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
–
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt