MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Chabot Mathematics §4.3a Absolute Value

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 2 Bruce Mayer, PE Chabot College Mathematics Review §  Any QUESTIONS About §4.2 → InEqualities & Problem-Solving  Any QUESTIONS About HomeWork §4.2 → HW MTH 55

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 3 Bruce Mayer, PE Chabot College Mathematics 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. 05–5 5 units from zero x = –5 or x = 5

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 4 Bruce Mayer, PE Chabot College Mathematics Graph y = |x|  Make T-table

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 5 Bruce Mayer, PE Chabot College Mathematics Absolute Value Properties 1.|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

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 6 Bruce Mayer, PE Chabot College Mathematics Example  Absolute Value Calcs  Simplify, leaving as little as possible inside the absolute-value signs a. |7x|b. |−8y|c. |6x 2 |d.  SOLUTION a.|7x| = b.|−8y| = c.|6x 2 | = d..

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 7 Bruce Mayer, PE Chabot College Mathematics 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

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 8 Bruce Mayer, PE Chabot College Mathematics 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}

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 9 Bruce Mayer, PE Chabot College Mathematics 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}

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 10 Bruce Mayer, PE Chabot College Mathematics 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

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 11 Bruce Mayer, PE Chabot College Mathematics Absolute 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 = p b.The equation |X| = 0 is equivalent to the equation X = 0 c.The equation |X| = −p has no solution.

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 12 Bruce Mayer, PE Chabot College Mathematics 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 = −3 or x = 2 2x = −6 or 2x = 4 Absolute-value principle |2x +1| = 5 |X| = p 2x +1 = −5 or 2x +1 = 5 TThus The solution set is {−3, 2}.

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 13 Bruce Mayer, PE Chabot College Mathematics 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

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 14 Bruce Mayer, PE Chabot College Mathematics 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 or2x + 3 = –5 2x = 2 x = 1 2x = –8 x = –4 or Graphing the Solutions –5–4–3–2–

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 15 Bruce Mayer, PE Chabot College Mathematics Solving 1-AbsVal Equations  To solve an equation containing a single absolute value 1.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

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 16 Bruce Mayer, PE Chabot College Mathematics 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.

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

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 18 Bruce Mayer, PE Chabot College Mathematics Example  2 AbsVal Expressions 1.3x – 5 = 8 + 4x –13 + 3x = 4x –13 = x 2.3x – 5 = –(8 + 4x)  Thus For Eqn |3x – 5| = |8 + 4x|  The solutions are −13 −3/7

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 19 Bruce Mayer, PE Chabot College Mathematics Solve Eqns of Form |ax+b| = |cx+d|  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.

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 20 Bruce Mayer, PE Chabot College Mathematics 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 ( )

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 21 Bruce Mayer, PE Chabot College Mathematics 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 ] [

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 22 Bruce Mayer, PE Chabot College Mathematics Basic Absolute Value Eqns Examples 

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 23 Bruce Mayer, PE Chabot College Mathematics 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)

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 24 Bruce Mayer, PE Chabot College Mathematics Example  Catering Costs  Translate  Carry Out

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 25 Bruce Mayer, PE Chabot College Mathematics Example  Catering Costs  Check   STATE  For values of x < 5 hr, Catherine’s Catering will cost less than Johnson’s

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 26 Bruce Mayer, PE Chabot College Mathematics WhiteBoard Work  Problems From §4.3 Exercise Set 8, 24, 34, 38, 40, 48  Graph of Absolute Value Function

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 27 Bruce Mayer, PE Chabot College Mathematics All Done for Today Cool Catering

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 28 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Chabot Mathematics Appendix –