Download MTH55_Lec-18_sec_4-3b_AbsVal_InEqualities

Chabot Mathematics
§4.3b AbsVal
InEqualities
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
Review § 4.3
MTH 55
 Any QUESTIONS About
• §4.3a → Absolute Value
 Any QUESTIONS About HomeWork
• §4.3a → HW-13
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Solving AbsVal InEqual with <

Solving Inequalities in the
Form |x| < a, where a > 0
1. Rewrite as a compound inequality
involving “and”: x > −a AND x < a.
•
Can also write as: −a < x < a
2. Solve the compound inequality.

Similarly, to solve |x|  a, we would
write x  −a and x  a (or −a  x  a)
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Example  AbsVal & <
 Given InEquality: |x − 3| < 6
• solve, graph the solution set, and write the
solution set in both set-builder and interval
notation
 SOLUTION
• |x – 3| < 6 → x − 3 > −6 and x − 3 < 6
• thus
−6 < x − 3 < 6
– ReWritten as Compound InEquality
• So
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−3 < x < 9
(add +3 to all sides)
Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  AbsVal & <
 SOLUTION: |x − 3| < 6
• Thus the Solution → −3 < x < 9
• Solution in Graphical Form
-10 -9
-8
-7
-6
-5
-4
(
-3
)
-2
-1
0
1
2
3
4
5
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7
8
9
10
• Set-builder notation: {x| −3 < x < 9}
• Interval notation: (−3, 9)
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Example  |2x − 3| + 8 < 5
 Given InEquality: |2x − 3| + 8 < 5
• solve, graph the solution set, and write the
solution set in both set-builder and interval
notation
 SOLUTION
 Isolate the absolute value
• |2x − 3| + 8 < 5
• |2x − 3| < −3 ???
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Example  |2x − 3| + 8 < 5
 The InEquality Simplified to:
|2x − 3| < −3
 Since the absolute value cannot be less
than a negative number, this inequality
has NO solution: Ø
• No Graph
• Set-Builder Notation → {Ø}
• Interval notation: We do not write interval
notation because there are no values
in the solution set.
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[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Solving AbsVal InEqual with >

Solving Inequalities in the
Form |x| > a, where a > 0
1. Rewrite 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
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Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  AbsVal & >
 Given InEquality: |x + 7| > 5
• solve, graph the solution set, and write the
solution set in both set-builder and interval
notation
 SOLUTION:
 convert to a compound inequality and
solve each
• |x + 7| > 5 →
• x + 7 < −5
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or
x+7>5
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Example  AbsVal & >
 SOLUTION : |x + 7| > 5
• x + 7 < −5
or
x+7>5
– The Addition Principle Produces Solutions
•
x < −12
• The
Graph
or
)
x > −2
(
-15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1
0
1
2
3
4
5
• Set-builder notation: {x| x < −2 or x > −2}
• Interval notation: (−, −12) U (−2, ).
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Example  |4x + 7| – 9 > –12
 Given InEquality: |4x + 7| − 9 > −12
• solve, graph the solution set, and write the
solution set in both set-builder and interval
notation
 SOLUTION
 Isolate the absolute value
• |4x + 7| – 9 > –12
• |4x + 7| > –3 ???
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Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  |4x + 7| − 9 > – 12
 The InEquality Simplified to:
|4x + 7| > −3
 This 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 All Real Numbers, 
• The Graph is then the entire Number Line
• Set-builder notation: {x|x is a real number}
• Interval notation: (−, )
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Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Summary: Solve |ax + b| > k
 Let k be a positive real number, and p
and q be real numbers.
 To solve |ax + b| > k, solve the following
compound inequality
ax + b > k OR ax + b < −k.
 The solution set is of the form
(−, p)U(q, ), which consists of two
Separate intervals.
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p
q
Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  |2x + 3| > 5
 By the Previous Slide this absolute
value inequality is rewritten as
2x + 3 > 5
or
2x + 3 < −5
 The expression 2x + 3 must represent a
number that is more than 5 units from 0
on either side of the number line.
• Use this analysis to solve the compound
inequality Above
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Example  |2x + 3| > 5
 Solve the Compound InEquality
2x + 3 > 5
or
2x + 3 < −5
2x > 2
or
2x < −8
x>1
or
x < −4
 The solution set is (–, –4)U(1, ). Notice
that the graph consists of two intervals.
–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
Summary: Solve |ax + b| < k
 Let k be a positive real number, and p
and q be real numbers.
 To solve |ax + b| < k, solve the threepart “and” inequality
–k < ax + b < k
 The solution set is of the form (p, q),
a single interval.
p
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q
Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  |2x + 3| < 5



By the Previous Slide this absolute
value inequality is rewritten as
−5 < 2x + 3
and
2x + 3 < 5
In 3-Part form
−5 < 2x + 3 < 5
Solving for x
−5 < 2x + 3 < 5
−8 < 2x < 2
−4 < x < 1
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Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  |2x + 3| < 5
 Thus the Solution: −4 < x < 1
 We can Check that the solution set is
(−4, 1), so the graph consists of the
single Interval:
–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
Caution for AbsVal vs <>

When solving absolute value
inequalities of the types > & <
remember the following:
1. The methods described apply when the
constant is alone on one side of the
equation or inequality and is positive.
2. Absolute value equations and absolute
value inequalities of the form |ax + b| > k
translate into “or” compound statements.
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Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Caution for AbsVal vs <>

When solving absolute value
inequalities of the types > & <
remember the following:
3. Absolute value inequalities of the form
|ax + b| < k translate into “and” compound
statements, which may be written as
three-part inequalities.
4. An “or” statement cannot be written in
three parts. It would be incorrect to use
−5 > 2x + 3 > 5 in the > Example, because
this would imply that − 5 > 5, which is false
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Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Absolute Value Special Cases
1. The absolute value of an
expression can never be
negative: |a| ≥ 0 for ALL
real numbers a.
2. The absolute value of an
expression equals 0 ONLY when
the expression is equal to 0.
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Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  Special Cases
 Solve: |2n + 3| = −7
 Solution:
• See Case 1 in the preceding slide.
• Since the absolute value of an expression
can never be negative, there are NO
solutions for this equation
• The solution set is Ø.
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Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  Special Cases
 Solve: |6w − 1| = 0
 Solution:
• See Case 2 in the preceding slide.
• The absolute value of the expression
6w − 1 will equal 0 only if
6w − 1 = 0
• The solution of this equation is 1/6. Thus,
the solution set of the original equation is
{1/6}, with just one element.
– Check by substitution.
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Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  Special Cases
 Solve: |x| ≥ −2
 Solution:
• The absolute value of a number is
always greater than or equal to 0.
• Thus, |x| ≥ −2 is true for all real numbers.
• The solution set is then entire number line:
(−, ).
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Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  Special Cases
 Solve: |x + 5| − 1 < −8
 Solution:
• Add 1 to each side to get the absolute
value expression alone on one side.
|x + 5| < −7
• There is no number whose absolute value
is less than −7, so this inequality has no
solution.
• The solution set is Ø
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Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Example  Special Cases
 Solve: |x − 9| + 2 ≤ 2
 Solution:
• Subtracting 2 from each side gives
|x − 9| ≤ 0
• The value of |x − 9| will never be less than
0. However, |x − 9| will equal 0 when x = 9.
• Therefore, the solution set is {9}.
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WhiteBoard Work
 Problems From §4.3 Exercise Set
• 88 (ppt), 94 (ppt), 98 (ppt), 56, 62, 78

Albany, NY
Yearly
Temperature
Data
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P4.3-88  |4 − x| < 5
 Solve |4 − x| < 5 by x-y Graph
 GRAPH SOLUTION:
 On graph find the region where the
y = f(x)= |4 − x| function lies
BELOW (i.e., is Less Than) the
y = f(x) = 5 Horizontal Line
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P4.3-88  |4 − x| < 5
 Ans in SET & INTERVAL Form
• {x| −1 < x < 9}
• (−1, 9)
(
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)
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P4.3-88  |4 − x| < 5
 Solve |4 − x| < 5
• 4 −x < 5
and
4−x > −5
• −x < 1
and
−x > −9
– Multiply Both expressions by −1, REMBERING to
REVERSE the direction of the InEqual signs
• x>−1
and
x < 9 OR
• −1 < x
and
x<9
• So the Compound Soln: (−1 < x < 9)
• In Set notation: {x| −1 < x < 9}
• Interval notation: (−1, 9)
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Bruce Mayer, PE
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P4.3-94  Albany, NY Temps
 InEquality for Albany, NY Monthly Avg
Temperature, T, in °F
|T − 50 °F| ≤ 22 °F
 Solve and Interpret
 SOLUTION
• |T−50| ≤ 22 → −22 ≤ T−50 and T−50 ≤ 22
• Or in 3-part form −22 ≤ T−50 ≤ 22
• Add 50 to each Part (addition principle)
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Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
P4.3-94  Albany, NY Temps
 Albany, NY Temps → |T − 50 °F| ≤ 22 °F
 SOLUTION
• (50−22) ≤ (T−50+50) ≤ (22+50)
• Or ANS → 28 ≤ T ≤ 72
 INTERPRETATION:
• The monthly avg temperature in Albany, NY
ranges from 28 °F in Winter to 72 °F in
Summer
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Bruce Mayer, PE
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P4.3-98  Machining Tolerance
 Length, x, of a Machine Part in cm
|x − 9.4cm| ≤ 0.01cm
 Solve and Interpret
 SOLUTION
• |x−9.4| ≤ 0.01 →
−0.01 ≤ x−9.4 and x−9.4 ≤ 0.01
• Or in 3-part form −0.01 ≤ x−9.4 ≤ 0.01
• Add 9.4 to each Part (addition principle)
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Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
P4.3-92  Machining Tolerance
 Machine Part Tolerance →
|x − 9.4cm| ≤ 0.01cm
 SOLUTION
• (−0.01+9.4) ≤ (x−9.4+9.4) ≤ (0.01+9.4)
• Or ANS → 9.39 ≤ x ≤ 9.41
 INTERPRETATION:
• The expected dimensions of the finished
machine part are between 93.9mm and
94.1mm (i.e. the dim is 94mm ±0.1mm)
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All Done for Today
JR Ewing
From
“Dallas”
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Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
–
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Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt
Graph y = |x|
6
 Make T-table
x
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y = |x |
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y
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x
0
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file =XY_Plot_0211.xls
-6
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4
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Weather UnderGnd – Albany, NY
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Bruce Mayer, PE
[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt