Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

Chabot Mathematics §4.3a Absolute Value Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

MTH 55 4.2 Review § • Any QUESTIONS About • §4.2 → InEqualities & Problem-Solving • Any QUESTIONS About HomeWork • §4.2 → HW-12

–5 0 5 Absolute 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. 5 units from zero 5 units from zero x = –5 or x = 5

Graph y = |x| • Make T-table

|ab| = |a |· |b|for any real numbers a & b The absolute value of a product is the product of the absolute values |a/b| = |a|/|b| for any real numbers a & b 0 The absolute value of a quotient is the quotient of the absolute values |−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 Absolute Value Properties

Example  Absolute Value Calcs • Simplify, leaving as little as possible inside the absolute-value signs a. |7x| b. |−8y| c. |6x2| d. • SOLUTION • |7x| = • |−8y|= • |6x2|= • .

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

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}

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}

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

Absolute Value Principle • For any positive number p and any algebraic expression X: • The solutions of |X| = p are those numbers that satisfy X = −p or X = p • The equation |X| = 0 is equivalent to the equation X = 0 • The equation |X| = −p has no solution.

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 Absolute-value principle 2x +1 = −5or 2x +1 =5 • Thus The solution set is {−3, 2}. 2x = −6 or 2x = 4 x = −3 or x = 2

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

–5 –4 –3 –2 –1 0 1 2 3 4 5 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. • Solve EquationSet 2x + 3 = 5 or 2x + 3 = –5 2x = 2 or 2x = –8 x = 1 or x = –4 • Graphing the Solutions

Solving 1-AbsVal Equations • To solve an equation containing a single absolute value • 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. • Separate the absolute value into two equations, ax + b = c and ax + b = −c. • Solve both equations for x

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.

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 This assumes these numbers are opposites. OR 3x – 5 = –(8 + 4x) 3x – 5 = 8 + 4x • Need to solve Both Eqns for x

Example  2 AbsVal Expressions • Thus For Eqn |3x – 5| = |8 + 4x| • The solutions are • −13 • −3/7 • 3x – 5 = 8 + 4x –13 + 3x = 4x –13 = x • 3x – 5 = –(8 + 4x)

Solve Eqns of Form |ax+b| = |cx+d| • To solve an equation in the form |ax + b| = |cx + d| • Separate the absolute value equation into two equations ax + b = cx + d, and ax + b = −(cx + d). • Solve both equations.

( ) -3 3 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 3 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

Basic Absolute Value Eqns • Examples 

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)

Example  Catering Costs • Translate • CarryOut

Example  Catering Costs • Check  • STATE • For values of x < 5 hr, Catherine’s Catering will cost less than Johnson’s

WhiteBoard Work • Problems From §4.3 Exercise Set • 8, 24, 34, 38, 40, 48 • Graph of AbsoluteValueFunction

All Done for Today CoolCatering

Chabot Mathematics Appendix Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu –

Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege