MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Chabot Mathematics §8.5 PolyNom InEqualities

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 2 Bruce Mayer, PE Chabot College Mathematics Review §  Any QUESTIONS About §8.4 → Equations in Quadratic Form  Any QUESTIONS About HomeWork §8.3 → HW MTH 55

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 3 Bruce Mayer, PE Chabot College Mathematics PolyNomial InEquality  PolyNomial InEqualities:  Second-Degree polynomial inequalities in one variable are called Quadratic inequalities.  To solve polynomial inequalities, focus attention on where the outputs of a polynomial function are positive and where they are negative

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 4 Bruce Mayer, PE Chabot College Mathematics Example  Solve  SOLUTION: Consider the related function y = f(x) = x 2 + 2x − 8 and its graph. x y y = x 2 + 2x – 8 positive y-values  Since the graph opens upward, the y-values are positive outside the interval formed by the x-Intercepts

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 5 Bruce Mayer, PE Chabot College Mathematics Example  Solve  SOLUTION: Thus y > 0 for x-values OUTSIDE of the region described the x-intercepts x y y = x 2 + 2x – 8 positive y-values  It follows that the solution set of the inequality is

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 6 Bruce Mayer, PE Chabot College Mathematics StreamLine Solution  In the next example, simplify the process by concentrating on the sign of a polynomial function over each interval formed by the x-intercepts.  We will do this by tracking the sign of each factor. By looking at how many positive or negative factors are being multiplied, we will be able to determine the sign of the polynomial function.

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 7 Bruce Mayer, PE Chabot College Mathematics Example  Solve  SOLUTION - first solve the related equation f (x) = 0 x 3 − 9x = 0 x(x − 3)(x + 3) = 0 x = 0 or x − 3 = 0 or x + 3 = 0 x = 0 or x = 3 or x = −3. Factoring Using the principle of zero products

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 8 Bruce Mayer, PE Chabot College Mathematics Example  Solve  The function f(x) has zeros at −3, 0 & 3.  We will use the factorization f(x) = x(x − 3)(x + 3).  The product x(x − 3)(x + 3) is positive or negative, depending on the signs of x, x − 3, and x + 3.  This is easily determined using a chart

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 9 Bruce Mayer, PE Chabot College Mathematics Example  Solve  The Sign Chart for f(x) = x(x – 3)(x + 3) Sign of x: Sign of x – 3: Sign of x + 3: Sign of product x(x – 3)(x + 3) –3 0 3 Interval: – – – – – – – –  Note that ANY Negative No. < 0

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 10 Bruce Mayer, PE Chabot College Mathematics Example  Solve  A product is negative when it has an odd number of negative factors.  Since the sign allows for EQuality, the ENDpoints −3, 0, and 3 ARE solutions.  The chart then displays solution set:

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 11 Bruce Mayer, PE Chabot College Mathematics Solve PolyNom InEq by Factoring 1.Add or subtract to get 0 on one side of a “f(x) Greater/Less Than” InEquality  e.g.; 3x 2 < 7 − 5x  3x 2 + 5x − 7 < 0 2.Next solve the related f(x) = 0 polynomial equation by factoring. 3.Use the numbers found in step (1) to divide the number line into intervals. 4.Using a test value (or Test-Pt) from each interval, determine the sign of each factor over that interval.

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 12 Bruce Mayer, PE Chabot College Mathematics Solve PolyNom InEq by Factoring 5.Determine the sign of the product of the factors over each interval. 6.Select the interval(s) for which the inequality is satisfied and write set-builder notation or interval notation for the solution set. 7.Include the endpoints of the intervals whenever ≤ or ≥ describes the InEquality

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 13 Bruce Mayer, PE Chabot College Mathematics Example  Solve  SOLUTION – Obtain Zero on One Side the InEquality  Next Solve the Associated Equation by Quadratic Formula ── ─

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 14 Bruce Mayer, PE Chabot College Mathematics Example  Solve  Isolating x from Quadratic Formula  This divides the No. line into 3 intervals so –1–2–3– 4– 4 00  Pick “Round Numbers as Test Points −3, 0, and 4

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 15 Bruce Mayer, PE Chabot College Mathematics Example  Solve  Constructing the Sign Table: x 2 −2x−7 > 0 IntervalPointValueResult −38+ 0−7−7– 41+

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 16 Bruce Mayer, PE Chabot College Mathematics Example  Solve  The Sign Intervals on the Number Line –1–2–3– 4– 4 00– – – – – – – – –  Thus for x 2 > 2x +7  x 2 − 2x − 7 > 0 is POSITIVE (exceeds Zero) in intervals ( ) –1–2–3– 4– 4

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 17 Bruce Mayer, PE Chabot College Mathematics Example  Solve x 2 + x ≤ 12  Solution: write the related equation x 2 + x − 12 = 0 (x − 3)(x + 4) = 0 x = 3 or x = −4 Set the quadratic expression equal to 0 Factor Use the zero-products theorem  Plot the Break-Points” on a number line to create boundaries between the three intervals IIIIII

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 18 Bruce Mayer, PE Chabot College Mathematics Example  Solve x 2 + x ≤ 12  Choose a test number (or Test-Pt) from each interval and substitute that value into x 2 + x − 12 ≤ 0 Test  For ( − , − 4], choose − 5 For [ − 4, 3] choose 0 For [3,  ) we choose 4 (–5) 2 + (–5) – 12 = 25 – 5 – 12 = 8 False (0) 2 + (0) – 12 = 0 – 12 = –12 True (4) 2 + (4) – 12 = – 12 = 8 False IIIIII

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 19 Bruce Mayer, PE Chabot College Mathematics Example  Solve x 2 + x ≤ 12  The Table Reveals the at the CENTRAL interval is the Solution Last time the solution was the EXTERIOR Intervals  The solution is the interval [−4, 3] [ ]

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 20 Bruce Mayer, PE Chabot College Mathematics Example  Solve x 2 + 5x > 0  Solution: write & Solve the related equation  Plot the Break-Pts on the No. line to form boundaries between the 3 intervals ) ( IIIIII

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 21 Bruce Mayer, PE Chabot College Mathematics Example  Solve x 2 + 5x > 0  Choose a Test-Point from each interval and substitue that value into The InEquality x 2 + 5x > 0 For ( − , − 5), choose − 6 For ( − 5, 0) choose − 1 For (0,  ) choose 1 (–6) 2 + 5(–6) > 0 36 – 30 > 0 6 > 0 True (–1) 2 + 5(–1) > 0 1 – 5 > 0 –4 > 0 False (1) 2 + 5(1) > > 0 6 > 0 True

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 22 Bruce Mayer, PE Chabot College Mathematics Example  Solve x 2 + 5x > 0  The Sign Table Reveals that the EXTERIOR intervals are the Solution  The solution in interval Notation (−∞, −5) U (0, ∞)  The Solution on the Number Line ) (

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 23 Bruce Mayer, PE Chabot College Mathematics Example  (x + 3)(x + 1)(x – 2) < 0  Solution: write the related equation (x + 3)(x + 1)(x – 2) = 0 x + 3 = 0 or x + 1 = 0 or x − 2 = 0 x = −3 or x = −1 or x = 2 Write Related Eqn  Plot the Break-Points” on a number line to create FOUR intervals Using Zero-Products Theorem Solving for x

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 24 Bruce Mayer, PE Chabot College Mathematics Example  (x + 3)(x + 1)(x – 2) < 0  Solution: Create Sign/Truth Chart  Thus the Soln Set: (−∞, −3) U (−1, 2) On Number Line: Interval ( ,  3)(  3,  1)(  1, 2)(2,  ) Test Number 44 22 03 Test Results  18 < 0 4 < 0  6 < 0 24 < 0 True or FalseTrueFalseTrueFalse ))(

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 25 Bruce Mayer, PE Chabot College Mathematics Example  Skid Marks  A car involved in an accident left skid marks over 75 feet long. Under the road conditions at the accident, the distance d (in feet) it takes a car traveling v miles per hour to stop is given by the equation  The accident occurred in a 25 mph zone.  Was the driver Speeding?

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 26 Bruce Mayer, PE Chabot College Mathematics Example  Skid Marks  SOLUTION: Solve the InEquality: (stopping distance) > 75 feet, or  The Solns divide the number line into 3 intervals

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 27 Bruce Mayer, PE Chabot College Mathematics Example  Skid Marks  Make Sign-Chart, and Divide No. Line 04030–50– – – – – – – – – – – – – – – – IntervalPointValueResult (–∞, –50)– (–50, 30)0–75– (30, ∞)4045+

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 28 Bruce Mayer, PE Chabot College Mathematics Example  Skid Marks  ++++ on No. Line indicates Speeding 04030–50– – – – – – – – – – – – – – – –  For this situation, we look at only the positive values of v. Note that the numbers corresponding to speeds between 0 and 30 mph (i.e., 0 ≤ v ≤ 30 ) are not solutions of 0.05v 2 +v  Thus, the car was traveling more than 30 miles per hour  Excessive Speed

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 29 Bruce Mayer, PE Chabot College Mathematics Example  Solve x 4 ≤ 1  SOLUTION: Solve Related Eqn for x  These Break- Points divide the number line into 3 intervals

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 30 Bruce Mayer, PE Chabot College Mathematics Example  Solve x 4 ≤ 1  Make Sign Chart & Associated No. Line IntervalPoint Value of Result (−∞, –1)−2−215+ (–1, 1)0−1−1– (1, ∞) –1–2–3 00– – –

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 31 Bruce Mayer, PE Chabot College Mathematics Example  Solve x 4 ≤ 1  The Sign-Regions on No. Line 12043–1–2–3 00– – –  The solution set consists of all x between −1 & 1, including both −1 & –1–2 ][ {−1 ≤ x ≤ 1} or [−1, 1]

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 32 Bruce Mayer, PE Chabot College Mathematics ONE-Sign Theorem  If a polynomial equation has NO REAL solution, then the polynomial is either ALWAYS positive or always negative  Example  Use One-Sign Theorem to Solve InEquality:

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 33 Bruce Mayer, PE Chabot College Mathematics Example  1-Sign Theorem  SOLUTION: Solve Related Eqn  The Expression does NOT factor so use Quadratic Formula First Test Discriminant for Real Solutions  Discriminant is NEGATIVE → NO Real-Number Solution

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 34 Bruce Mayer, PE Chabot College Mathematics Example  1-Sign Theorem  Since the discriminant is negative, there are no real roots.  Use 0 as a test point, which yields 2.  The inequality is always positive, the solution set is ALL REAL Numbers Using Interval Notion: (−∞, ∞).

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 35 Bruce Mayer, PE Chabot College Mathematics WhiteBoard Work  Problems From §8.5 Exercise Set 16, 28, 34, 68  x 2 − x − 6 < 0 Graphically

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 36 Bruce Mayer, PE Chabot College Mathematics All Done for Today Acapulco Cliff Diving

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 37 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Chabot Mathematics Appendix –

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 38 Bruce Mayer, PE Chabot College Mathematics Graph y = |x|  Make T-table

MTH55_Lec-54_sec_8-5a_PolyNom_InEqual.ppt 39 Bruce Mayer, PE Chabot College Mathematics