[email protected] MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical

[email protected] • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

§4.3b AbsVal§4.3b AbsValInEqualitiesInEqualities



Review §Review §

Any QUESTIONS About• §4.3a → Absolute Value

Any QUESTIONS About HomeWork• §4.3a → HW-13

4.3 MTH 55



Solving AbsVal InEqual with <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)



Example Example AbsVal & < 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 −3 < x < 9 (add +3 to all sides)



Example Example AbsVal & < AbsVal & <

SOLUTION: |x − 3| < 6• Thus the Solution → −3 < x < 9

• Solution in Graphical Form

10-9 -7 -5 -3 -1 1 3 5 7 9-10 -8 -4 0 4 8-10 -2 6-6 102( )

• Set-builder notation: {x| −3 < x < 9}

• Interval notation: (−3, 9)



Example Example |2 |2xx −− 3| + 8 3| + 8 << 5 5

Given InEquality: |2x − 3| + 8 < 5• solve, graph the solution set, and write the


SOLUTION Isolate the absolute value

• |2x − 3| + 8 < 5

• |2x − 3| < −3 ???



Example Example |2 |2xx −− 3| + 8 3| + 8 << 5 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.



Solving AbsVal InEqual with >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



Example Example AbsVal & > AbsVal & >

Given InEquality: |x + 7| > 5• solve, graph the solution set, and write the


SOLUTION: convert to a compound inequality and

solve each• |x + 7| > 5 →

• x + 7 < −5 or x + 7 > 5



Example Example AbsVal & > AbsVal & >

SOLUTION : |x + 7| > 5• x + 7 < −5 or x + 7 > 5

– The Addition Principle Produces Solutions

• x < −12 or x > −2

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

) (

• Set-builder notation: {x| x < −2 or x > −2}

• Interval notation: (−, −12) U (−2, ).



Example Example |4 |4xx + 7| – 9 + 7| – 9 >> –12 –12

Given InEquality: |4x + 7| − 9 > −12• solve, graph the solution set, and write the


SOLUTION Isolate the absolute value

• |4x + 7| – 9 > –12

• |4x + 7| > –3 ???



Example Example |4 |4xx + 7| + 7| −− 9 9 >> – 12 – 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: (−, )



Summary: Solve |Summary: Solve |axax + + bb| | >> kk

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.

p q



Example Example |2 |2xx + 3| + 3| >> 5 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



Example Example |2 |2xx + 3| > 5 + 3| > 5

Solve the Compound InEquality2x + 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 –3 –2 –1 0 1 2 3 4 5



Summary: Solve |Summary: Solve |axax + + bb| | << kk

Let k be a positive real number, and p and q be real numbers.

To solve |ax + b| < k, solve the three-part “and” inequality

–k < ax + b < k The solution set is of the form (p, q),

a single interval.

p q



Example Example |2 |2xx + 3| + 3| << 5 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



Example Example |2 |2xx + 3| + 3| << 5 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 –3 –2 –1 0 1 2 3 4 5



Caution for AbsVal vs <>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.



Caution for AbsVal vs <>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



Absolute Value Special CasesAbsolute 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.



Example Example Special Cases 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.



Example Example Special Cases 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.




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: (−, ).




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




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}.



WhiteBoard WorkWhiteBoard Work

Problems From §4.3 Exercise Set• 88 (ppt), 94 (ppt), 98 (ppt), 56, 62, 78

Albany, NYYearly TemperatureData



P4.3-88 P4.3-88 |4 |4 − − xx| < 5| < 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



P4.3-88 P4.3-88 |4 |4 − − xx| < 5| < 5

( )

Ans in SET & INTERVAL Form• {x| −1 < x < 9}

• (−1, 9)



P4.3-88 P4.3-88 |4 |4 − − xx| < 5| < 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)



P4.3-94 P4.3-94 Albany, NY Temps 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)



P4.3-94 P4.3-94 Albany, NY Temps 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



P4.3-98 P4.3-98 Machining Machining ToleranceTolerance 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)



P4.3-92 P4.3-92 Machining Machining ToleranceTolerance 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)



All Done for TodayAll Done for Today

JR EwingFrom

“Dallas”



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

AppendiAppendixx

–

srsrsr 22



Graph Graph yy = | = |xx||

Make T-tablex y = |x |

-6 6-5 5-4 4-3 3-2 2-1 10 01 12 23 34 45 56 6

x

y

-6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

file =XY_Plot_0211.xls



Weather UnderGnd – Albany, NYWeather UnderGnd – Albany, NY

Documents

[email protected] MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical