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 §4.3b AbsVal InEqualities

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 2 Bruce Mayer, PE Chabot College Mathematics Review §  Any QUESTIONS About §4.3a → Absolute Value  Any QUESTIONS About HomeWork §4.3a → HW MTH 55

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 3 Bruce Mayer, PE Chabot College Mathematics Solving AbsVal InEqual with <  Solving Inequalities in the Form |x| 0 1.Rewrite as a compound inequality involving “and”: x > −a AND x < a. Can also write as: −a < x < a 2.Solve the compound inequality.  Similarly, to solve |x|  a, we would write x  −a and x  a (or −a  x  a)

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 4 Bruce Mayer, PE Chabot College Mathematics Example  AbsVal & <  Given InEquality: |x − 3| < 6 solve, graph the solution set, and write the solution set in both set-builder and interval notation  SOLUTION |x – 3| −6 and x − 3 < 6 thus −6 < x − 3 < 6 –ReWritten as Compound InEquality So −3 < x < 9 (add +3 to all sides)

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 5 Bruce Mayer, PE Chabot College Mathematics Example  AbsVal & <  SOLUTION: |x − 3| < 6 Thus the Solution → −3 < x < 9 Solution in Graphical Form ( ) Set-builder notation: {x| −3 < x < 9} Interval notation: (−3, 9)

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 6 Bruce Mayer, PE Chabot College Mathematics Example  |2x − 3| + 8 < 5  Given InEquality: |2x − 3| + 8 < 5 solve, graph the solution set, and write the solution set in both set-builder and interval notation  SOLUTION  Isolate the absolute value |2x − 3| + 8 < 5 |2x − 3| < −3 ???

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 7 Bruce Mayer, PE Chabot College Mathematics Example  |2x − 3| + 8 < 5  The InEquality Simplified to: |2x − 3| < −3  Since the absolute value cannot be less than a negative number, this inequality has NO solution: Ø No Graph Set-Builder Notation → {Ø} Interval notation: We do not write interval notation because there are no values in the solution set.

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 8 Bruce Mayer, PE Chabot College Mathematics Solving AbsVal InEqual with >  Solving Inequalities in the Form |x| > a, where a > 0 1.Rewrite as a compound inequality involving “or”: x a. 2.Solve the compound inequality  Similarly, to solve |x|  a, we would write x  −a or x  a

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 9 Bruce Mayer, PE Chabot College Mathematics Example  AbsVal & >  Given InEquality: |x + 7| > 5 solve, graph the solution set, and write the solution set in both set-builder and interval notation  SOLUTION:  convert to a compound inequality and solve each |x + 7| > 5 → x + 7 5

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 10 Bruce Mayer, PE Chabot College Mathematics Example  AbsVal & >  SOLUTION : |x + 7| > 5 x –The Addition Principle Produces Solutions x −2 The Graph ) ( Set-builder notation: {x| x −2} Interval notation: (− , −12) U (−2,  ).

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 11 Bruce Mayer, PE Chabot College Mathematics Example  |4x + 7| – 9 > –12  Given InEquality: |4x + 7| − 9 > −12 solve, graph the solution set, and write the solution set in both set-builder and interval notation  SOLUTION  Isolate the absolute value |4x + 7| – 9 > –12 |4x + 7| > –3 ???

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 12 Bruce Mayer, PE Chabot College Mathematics Example  |4x + 7| − 9 > – 12  The InEquality Simplified to: |4x + 7| > −3  This inequality indicates that the absolute value is greater than a negative number. Since the absolute value of every real number is either positive or 0, the solution set is All Real Numbers,  The Graph is then the entire Number Line Set-builder notation: {x|x is a real number} Interval notation: (− ,  )

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 13 Bruce Mayer, PE Chabot College Mathematics Summary: Solve |ax + b| > k  Let k be a positive real number, and p and q be real numbers.  To solve |ax + b| > k, solve the following compound inequality ax + b > k OR ax + b < −k.  The solution set is of the form (− , p)U(q,  ), which consists of two Separate intervals. pq

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 14 Bruce Mayer, PE Chabot College Mathematics Example  |2x + 3| > 5  By the Previous Slide this absolute value inequality is rewritten as 2x + 3 > 5 or 2x + 3 < −5  The expression 2x + 3 must represent a number that is more than 5 units from 0 on either side of the number line. Use this analysis to solve the compound inequality Above

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 15 Bruce Mayer, PE Chabot College Mathematics Example  |2x + 3| > 5  Solve the Compound InEquality 2x + 3 > 5 or2x + 3 < −5 2x > 2 or 2x < −8 x > 1 or x < −4  The solution set is (– , –4)U(1,  ). Notice that the graph consists of two intervals. –5–4–3–2–

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 16 Bruce Mayer, PE Chabot College Mathematics Summary: Solve |ax + b| < k  Let k be a positive real number, and p and q be real numbers.  To solve |ax + b| < k, solve the three- part “and” inequality –k < ax + b < k  The solution set is of the form (p, q), a single interval. pq

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 17 Bruce Mayer, PE Chabot College Mathematics Example  |2x + 3| < 5  By the Previous Slide this absolute value inequality is rewritten as −5 < 2x + 3and 2x + 3 < 5  In 3-Part form −5 < 2x + 3 < 5  Solving for x −5 < 2x + 3 < 5 −8 < 2x < 2 −4 < x < 1

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 18 Bruce Mayer, PE Chabot College Mathematics Example  |2x + 3| < 5  Thus the Solution: −4 < x < 1  We can Check that the solution set is (−4, 1), so the graph consists of the single Interval: –5–4–3–2–

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 19 Bruce Mayer, PE Chabot College Mathematics Caution for AbsVal vs <>  When solving absolute value inequalities of the types > & < remember the following: 1.The methods described apply when the constant is alone on one side of the equation or inequality and is positive. 2.Absolute value equations and absolute value inequalities of the form |ax + b| > k translate into “or” compound statements.

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 20 Bruce Mayer, PE Chabot College Mathematics Caution for AbsVal vs <>  When solving absolute value inequalities of the types > & < remember the following: 3.Absolute value inequalities of the form |ax + b| < k translate into “and” compound statements, which may be written as three-part inequalities. 4.An “or” statement cannot be written in three parts. It would be incorrect to use −5 > 2x + 3 > 5 in the > Example, because this would imply that − 5 > 5, which is false

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 21 Bruce Mayer, PE Chabot College Mathematics Absolute Value Special Cases 1.The absolute value of an expression can never be negative: |a| ≥ 0 for ALL real numbers a. 2.The absolute value of an expression equals 0 ONLY when the expression is equal to 0.

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 22 Bruce Mayer, PE Chabot College Mathematics Example  Special Cases  Solve: |2n + 3| = −7  Solution: See Case 1 in the preceding slide. Since the absolute value of an expression can never be negative, there are NO solutions for this equation The solution set is Ø.

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 23 Bruce Mayer, PE Chabot College Mathematics Example  Special Cases  Solve: |6w − 1| = 0  Solution: See Case 2 in the preceding slide. The absolute value of the expression 6w − 1 will equal 0 only if 6w − 1 = 0 The solution of this equation is 1/6. Thus, the solution set of the original equation is {1/6}, with just one element. –Check by substitution.

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 24 Bruce Mayer, PE Chabot College Mathematics Example  Special Cases  Solve: |x| ≥ −2  Solution: The absolute value of a number is always greater than or equal to 0. Thus, |x| ≥ −2 is true for all real numbers. The solution set is then entire number line: (− ,  ).

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 25 Bruce Mayer, PE Chabot College Mathematics Example  Special Cases  Solve: |x + 5| − 1 < −8  Solution: Add 1 to each side to get the absolute value expression alone on one side. |x + 5| < −7 There is no number whose absolute value is less than −7, so this inequality has no solution. The solution set is Ø

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 26 Bruce Mayer, PE Chabot College Mathematics Example  Special Cases  Solve: |x − 9| + 2 ≤ 2  Solution: Subtracting 2 from each side gives |x − 9| ≤ 0 The value of |x − 9| will never be less than 0. However, |x − 9| will equal 0 when x = 9. Therefore, the solution set is {9}.

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 27 Bruce Mayer, PE Chabot College Mathematics WhiteBoard Work  Problems From §4.3 Exercise Set 88 (ppt), 94 (ppt), 98 (ppt), 56, 62, 78  Albany, NY Yearly Temperature Data

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 28 Bruce Mayer, PE Chabot College Mathematics P  |4 − x| < 5  Solve |4 − x| < 5 by x-y Graph  GRAPH SOLUTION:  On graph find the region where the y = f(x)= |4 − x| function lies BELOW (i.e., is Less Than) the y = f(x) = 5 Horizontal Line

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 29 Bruce Mayer, PE Chabot College Mathematics P  |4 − x| < 5 ()  Ans in SET & INTERVAL Form {x| −1 < x < 9} (−1, 9)

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 30 Bruce Mayer, PE Chabot College Mathematics P  |4 − x| < 5  Solve |4 − x| < 5 4 −x −5 −x −9 – Multiply Both expressions by −1, REMBERING to REVERSE the direction of the InEqual signs x > − 1 and x < 9 OR −1 < x and x < 9 So the Compound Soln: (−1 < x < 9) In Set notation: {x| −1 < x < 9} Interval notation: (−1, 9)

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 31 Bruce Mayer, PE Chabot College Mathematics P  Albany, NY Temps  InEquality for Albany, NY Monthly Avg Temperature, T, in °F |T − 50 °F| ≤ 22 °F  Solve and Interpret  SOLUTION |T−50| ≤ 22 → −22 ≤ T−50 and T−50 ≤ 22 Or in 3-part form −22 ≤ T−50 ≤ 22 Add 50 to each Part (addition principle)

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 32 Bruce Mayer, PE Chabot College Mathematics P  Albany, NY Temps  Albany, NY Temps → |T − 50 °F| ≤ 22 °F  SOLUTION (50−22) ≤ (T−50+50) ≤ (22+50) Or ANS →28 ≤ T ≤ 72  INTERPRETATION: The monthly avg temperature in Albany, NY ranges from 28 °F in Winter to 72 °F in Summer

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 33 Bruce Mayer, PE Chabot College Mathematics P  Machining Tolerance  Length, x, of a Machine Part in cm |x − 9.4cm| ≤ 0.01cm  Solve and Interpret  SOLUTION |x−9.4| ≤ 0.01 → −0.01 ≤ x−9.4 and x−9.4 ≤ 0.01 Or in 3-part form −0.01 ≤ x−9.4 ≤ 0.01 Add 9.4 to each Part (addition principle)

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 34 Bruce Mayer, PE Chabot College Mathematics P  Machining Tolerance  Machine Part Tolerance → |x − 9.4cm| ≤ 0.01cm  SOLUTION (− ) ≤ (x− ) ≤ ( ) Or ANS →9.39 ≤ x ≤ 9.41  INTERPRETATION: The expected dimensions of the finished machine part are between 93.9mm and 94.1mm (i.e. the dim is 94mm ±0.1mm)

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 35 Bruce Mayer, PE Chabot College Mathematics All Done for Today JR Ewing From “Dallas”

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 36 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Chabot Mathematics Appendix –

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 37 Bruce Mayer, PE Chabot College Mathematics Graph y = |x|  Make T-table

MTH55_Lec-17_sec_4-3a_Absolute_Value.ppt 38 Bruce Mayer, PE Chabot College Mathematics Weather UnderGnd – Albany, NY