MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer,...

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Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

§4.3a Absolute §4.3a Absolute ValueValue



Review §Review §

Any QUESTIONS About• §4.2 → InEqualities & Problem-Solving

Any QUESTIONS About HomeWork• §4.2 → HW-09

4.2 MTH 55



Absolute ValueAbsolute Value The absolute value of x denoted |x|,

is defined as

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.

0 ; 0 .x x x x x x

0 5–5

5 units from zero 5 units from zerox = –5 or x = 5



Graph Graph yy = | = |xx|| Make T-table

x 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



Absolute Value PropertiesAbsolute Value Properties1. |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



Example Example Absolute Value Calcs Absolute Value Calcs Simplify, leaving as little as possible

inside the absolute-value signsa. |7x| b. |−8y| c. |6x2| d. 2

84xx

SOLUTIONa. |7x| = 7 x 7 x

b. |−8y| = 8 y 8 y

c. |6x2| = 26 x 26 x 26x

d. .2

84xx

2x

2x

2x



Distance & Absolute-ValueDistance & 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



Example Example AbsVal Expressions 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}




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}




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



Absolute Value PrincipleAbsolute 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 = pb. The equation |X| = 0 is equivalent to

the equation X = 0c. The equation |X| = −p has no solution.



Example Example AbsVal Principle 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 = −3 or x = 2

2x = −6 or 2x = 4

Absolute-value principle|2x +1| = 5

|X| = p

2x +1 = −5 or 2x +1 = 5

Thus The solution set is {−3, 2}.



Example Example AbsVal Principle 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



Example Example AbsVal Principle 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.

• Solve EquationSet

2x + 3 = 5 or 2x + 3 = –52x = 2

x = 1

2x = –8

x = –4

or

or

• Graphing the Solutions

–5 –4 –3 –2 –1 0 1 2 3 4 5



Solving 1-AbsVal EquationsSolving 1-AbsVal Equations To solve an equation containing a

single absolute value1. 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



Two AbsVal Expression EqnsTwo 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.



Example Example 2 AbsVal Expressions 2 AbsVal Expressions Solve: |3x – 5| = |8 + 4x|. SOLUTION

• Recall that if |a| = |b| then either they are the same or they are opposites

3x – 5 = 8 + 4x

This assumes these numbers are the same

This assumes these numbers are opposites.

3x – 5 = –(8 + 4x)OR

Need to solve Both Eqns for x



Example Example 2 AbsVal Expressions 2 AbsVal Expressions1. 3x – 5 = 8 + 4x –13 + 3x = 4x –13 = x2. 3x – 5 = –(8 + 4x)

3 5 8 4x x 7 5 8x

7 3x37

x

Thus For Eqn |3x – 5| = |8 + 4x|

The solutions are • −13• −3/7



Solve Eqns of Form |Solve Eqns of Form |axax++bb| = || = |cxcx++dd|| 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.



Inequalities &AbsVal ExpressionsInequalities &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 3( )



Inequalities & AbsVal ExpressionsInequalities & 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 −3 3

] [



Basic Absolute Value EqnsBasic Absolute Value Eqns

0000

or0Set

SolutionState

EquivalentEquation

Value Absolute

kkxxx

-k,kkxkxkkx

13000

777or77

xxx

,-xxx• Examples



Example Example Catering Costs 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)



Example Example Catering Costs Catering Costs Translate

Cathrine'sCost

Is LessThan

Johnson'sCost

1003050 xx

50 100 3020 100

5

x xxx

CarryOut



Example Example Catering Costs Catering Costs Check

250250100150250

100530550

?

?

STATE For values of x < 5 hr, Catherine’s

Catering will cost less than Johnson’s



WhiteBoard WorkWhiteBoard Work

Problems From §4.3 Exercise Set• 8, 24, 34, 38, 40, 48

Graph of AbsoluteValueFunction x

y

0



All Done for TodayAll Done for Today

CoolCatering



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

AppendiAppendixx

–

srsrsr 22

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MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer,...

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