Chabot Mathematics §4.3a Absolute Value Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot College Mathematics 1 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt Review § 4.2 MTH 55 Any QUESTIONS About • §4.2 → InEqualities & Problem-Solving Any QUESTIONS About HomeWork • §4.2 → HW-12 Chabot College Mathematics 2 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt Absolute Value The absolute value of x denoted |x|, is defined as x 0 x x; x 0 x x. 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 Chabot College Mathematics 3 0 5 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt Graph y = |x| 6 Make T-table x -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 Chabot College Mathematics 4 5 y = |x | 6 5 4 3 2 1 0 1 2 3 4 5 6 y 4 3 2 x 1 0 -6 -5 -4 -3 -2 -1 0 1 2 3 -1 -2 -3 -4 -5 file =XY_Plot_0211.xls -6 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 4 5 6 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 Chabot College Mathematics 5 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt Example Absolute Value Calcs Simplify, leaving as little as possible inside the absolute-value signs 8x 2 a. |7x| b. |−8y| c. |6x | d. 2 4 x SOLUTION a. |7x| = 7 x 7 x b. |−8y| = 8 y 8 y 2 2 2 2 6x c. |6x | = 6 x 6 x 2 2 8x 2 d. . 2 x x 4 x x Chabot College Mathematics 6 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 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 Chabot College Mathematics 7 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 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} Chabot College Mathematics 8 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 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} Chabot College Mathematics 9 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 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 Chabot College Mathematics 10 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 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. Chabot College Mathematics 11 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 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 2x = −6 or 2x = 4 x = −3 or x=2 Chabot College Mathematics 12 Absolute-value principle Thus The solution set is {−3, 2}. Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 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 Chabot College Mathematics 13 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 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. 2x + 3 = 5 or 2x + 3 = –5 • Solve 2x = 2 or 2x = –8 Equation x=1 or x = –4 Set • Graphing the Solutions –5 –4 Chabot College Mathematics 14 –3 –2 –1 0 1 2 3 4 5 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 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 Chabot College Mathematics 15 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 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. Chabot College Mathematics 16 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 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 3x – 5 = 8 + 4x OR 3x – 5 = –(8 + 4x) Need to solve Both Eqns for x Chabot College Mathematics 17 This assumes these numbers are opposites. Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt Example 2 AbsVal Expressions 1. 3x – 5 = 8 + 4x –13 + 3x = 4x –13 = x 2. 3x – 5 = –(8 + 4x) 3x 5 8 4 x 7 x 5 8 7 x 3 3 x 7 Chabot College Mathematics 18 Thus For Eqn |3x – 5| = |8 + 4x| The solutions are • −13 • −3/7 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 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. Chabot College Mathematics 19 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 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 Chabot College Mathematics 20 ) 3 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 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 Chabot College Mathematics 21 ] −3 [ 3 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt Basic Absolute Value Eqns Absolute Value Equivalent Solution Equation x k k 0 x 0 x k k 0 State x k or x k x0 Set -k,k 0 • Examples x 7 x 7 or x 7 x 0 x0 x 13 Chabot College Mathematics 22 -7 ,7 0 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 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) Chabot College Mathematics 23 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt Example Catering Costs Translate Cathrine's Cost 50 x Carry Out Chabot College Mathematics 24 Is Less Than Johnson's Cost 30 x 100 50 x 100 30 x 20 x 100 x5 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt Example Catering Costs Check 50 5 30 5 100 ? ? 250 150 100 250 250 STATE For values of x < 5 hr, Catherine’s Catering will cost less than Johnson’s Chabot College Mathematics 25 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt WhiteBoard Work Problems From §4.3 Exercise Set • 8, 24, 34, 38, 40, 48 y Graph of Absolute Value Function Chabot College Mathematics 26 x 0 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt All Done for Today Cool Catering Chabot College Mathematics 27 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt Chabot Mathematics Appendix r s r s r s 2 2 Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu – Chabot College Mathematics 28 Bruce Mayer, PE BMayer@ChabotCollege.edu • MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt