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

4.2 Review § Any QUESTIONS About Any QUESTIONS About HomeWork MTH 55 Review § Any QUESTIONS About §4.2 → InEqualities & Problem-Solving Any QUESTIONS About HomeWork §4.2 → HW-9

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 5 –5

Graph y = |x| Make T-table

Absolute Value Properties |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

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

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 Equation Set 2x + 3 = 5 or 2x + 3 = –5 2x = 2 or 2x = –8 x = 1 or x = –4 Graphing the Solutions –5 –4 –3 –2 –1 1 2 3 4 5

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.

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

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 3x – 5 = 8 + 4x –13 + 3x = 4x –13 = x 3x – 5 = –(8 + 4x) Thus For Eqn |3x – 5| = |8 + 4x| The solutions are −13 −3/7

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.

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 −3 3 ] [

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 Carry Out

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 Absolute Value Function

All Done for Today Cool Catering

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