1. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 1 First we find the zeroes of the polynomial; that’s the solutions of the equation: 3x(x-2) = 0

2. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 2 ( ) ) The points x=0 and x=2 divide the real line into 3 intervals: (-∞,0), (0,2), (2,∞) as shown on the figure below: We investigate the sign of the of the polynomial x(x-2) on each of these intervals, by using any point inside the each interval ( test point). The solution set will be the union of all intervals on which the given inequality 3x(x-2) < 0 is satisfied ( that’s the intervals on which the polynomial is negative) (

3. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 3 Interv al Test Point 3x(x-2)SignIs 3x(x-2) < 0 satisfied? (-∞,03(-1)(-1-2)=3(-1)(-2) + No (0,2)13(1)(1-2)=3(1)(-1) -Yes (2,∞)43(4)(4-2)=3(4)(2) +No ( ( ) )

4. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 4 ( ) The solution set = (0,2)

5. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 5 First we find the zeroes of the polynomial; that’s the solutions of the equation: (x+2)(x-1)(x-5) = 0

6. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 6 The points x=-2, x=1 and x=5 divide the real line into 4 intervals: (-∞,-2), (-2,1), (1,5) and (5,∞) as shown on the figure below. We investigate the sign of the of the polynomial (x+2)(x-1)(x-5) on each of these intervals, by using any point inside the each interval ( test point). The solution set will be the union of all intervals on which the given inequality (x+2)(x-1)(x-5) < 0 is satisfied ( that’s the intervals on which the polynomial is negative) and the set of the zeroes of that { -2, 1, 5} ) ) ) ( ( (

7. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 7 IntervalTest Point (x+2)(x-1)(x-5)SignIs (x+2)(x-1)(x-5) < 0 satisfied? (-∞,-2)-3(-3+2)(-3-1)(-3-5) =(-1)(-4)(-8) - Yes (-2,1)0(0+2)(0-1)(0-5) =(2)(-1)(-5) +No (1,5)2(2+2)(2-1)(2-5) =(4)(1)(-3) -Yes (5,∞)6(6+2)(6-1)(6-5) =(8)(5)(1) +No ) ( ( ( ) )

8. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 8 ] [ ] The solution set = (-∞,-2) U (1, 5) U {-2,1,5} = (-∞,-2] U [1, 5]

9. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 9 First we find the zeroes of the polynomial; that’s the solutions of the equation: 2x 2 + 5x – 12 = 0

10. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 10

11. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 11 The points x=-4 and x = 3/2 = 1.5 divide the real line into 3 intervals: (-∞,-4), (-4,1.5), (1.5,∞) as shown on the figure below. We investigate the sign of the of the polynomial 2x 2 +5x-12 on each of these intervals, by using any point inside the interval ( test point). The solution set will be the union of all intervals on which the given inequality 2x 2 +5x-12 > 0 is satisfied (that’s the intervals on which the polynomial is positive and the set of the zeroes of that polynomial { 1.5, -4 } ( ) ) (

12. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 12 Interva l Test Point 2x 2 +5x-12 =(x-1.5)(x+4) SignIs 2x 2 +5x-12 > 0 satisfied? (-∞,-4)-5(-5-1.5) (-5+4) =(-6.5)(-4) + Yes (-4,1.5)0(0-1.5) (0+4) =(-1.5)(4) -No (1.5,∞)2(2-1.5) (2+4) =(0.5)(6) +Yes ) ) ) )( (

13. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 13 [] The solution set = (-∞,-4) U (1.5,∞) U { -4, 1.5} = (-∞,-4] U [1.5,∞)

14. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 14 Solution First We find the zeroes of the polynomial 9x 3 + 72x 2 -25x -200 Let: 9x 3 + 72x 2 -25x -200 = 0

15. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 15

16. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 16

17. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 17 The points x=-8, x=-5/3= - 1 2/3 and x= 1 2/3 divide the real line into 4 intervals: (-∞,-8), (-8,-5/3), (-5/3,5/3) and (5/3,∞) as shown on the figure below. We investigate the sign of the of the polynomial (x+8)(x+5/3)(x- 5/3) on each of these intervals, by using any point inside the interval ( test point). The solution set will be the union of all intervals on which the given inequality (x+8)(x+5/3)(x-5/3) > 0 is satisfied ( that’s the intervals on which the polynomial is positive) ) ) ) ( ( (

18. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 18 IntervalTest Point (x+8)(x+5/3)(x-5/3)SignIs (x+8)(x+5/3)(x-5/3>0 satisfied? (-∞,-8)-9 (-9+8)(-9+5/3)(-9-5/3) =(-)(-)(-) - No (-8,-5/3)-2 (-2+8)(-2+5/3)(-2-5/3) =(+)(-)(-) +Yes (-5/3,5/3 )0 (0+8)(0+5/3)(0-5/3) =(+)(+)(-) -No (5/3,∞)2 (2+8)(2+5/3)(2-5/3) (+)(+)(+) +Yes () ( () )

19. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 19 )( ( The solution set: = (-8, - 5/3) U (5/3, ∞ )

20. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 20

21. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 21 First we find: 1.. The zeroes of the rational expression ( the zeroes of the polynomial of the numerator that are not zeroes of the polynomial of the denominator) 2.. The numbers at which the expression is not defined ( the zeroes of the polynomial of the denominator) The numbers obtained in (1) and (2) divide the real line into open intervals. We investigate the sign of the of the rational expression on each of these intervals, by using any point inside the interval ( test point). The solution set will be: 1.. For the cases “>” or “<” : The union of all intervals on which the given inequality is satisfied. 2.. For the cases “≥” or “≤ : The union of these intervals and the set of the zeroes of the expression.

22. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 22

23. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 23

24. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 24 ()) The points x=-1 and x=-7/2 divide the real line into 3 intervals: (-∞,-7/2), (-7/2,-1), (-1,∞) as shown on the figure below. We investigate the sign of the of the rational expression (2x+7)/(x+1)on each of these intervals, by using any point inside the each interval ( test point). The solution set will be the union of all intervals on which the (2x+7)/(x+1) < 0 is satisfied ( that’s the intervals on which the polynomial is negative). (

25. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 25 IntervalTest Point (2x+7)/(x+1)SignIs ( 2x+7)/(x+1) < 0 satisfied? (-∞,-7/2)-4(-8+7)/(-4+1) =(-1)(-3) + No (-7/2,-1)-2(-4+7)/(-2+1) =(3)(-1) -Yes (-1,∞)0(0+7)/(0+1) =(7)(1) +No ( ( ))

26. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 26 ( ) The solution set = (-7/2,-1)

27. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 27

28. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 28

29. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 29 ()) The points x=-1 and x=-7/2 divide the real line into 3 intervals: (-∞,-7/2), (-7/2,-1), (-1,∞) as shown on the figure below. We investigate the sign of the of the rational expression (2x+7)/(x+1)on each of these intervals, by using any point inside the each interval ( test point). The solution set will be the union of all intervals on which the (2x+7)/(x+1) < 0 is satisfied ( that’s the intervals on which the polynomial is negative) and the set { -7/2 } of the zeroes of he rational expression (

30. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 30 IntervalTest Point (2x+7)/(x+1)SignIs ( 2x+7)/(x+1) < 0 satisfied? (-∞,-7/2)-4(-8+7)/(-4+1) =(-1)(-3) + No (-7/2,-1)-2(-4+7)/(-2+1) =(3)(-1) -Yes (-1,∞)0(0+7)/(0+1) =(7)(1) +No ( ( ))

31. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 31 ( ) The solution set = [-7/2,-1)

32. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 32

33. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 33 () ) The points x=-10 and x=10divide the real line into 3 intervals: (-∞,-10), (-10,10), (10,∞) as shown on the figure below. We investigate the sign of the of the rational expression (x+10)/(x-10) on each of these intervals, by using any point inside the each interval ( test point). The solution set will be the union of all intervals on which the inequality (2x+7)/(x+1) > 0 is satisfied ( that’s the intervals on which the polynomial is positive) and the set of the zeros of the expression, namely {-10 } (

34. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 34 IntervalTest Point (x+10)/(x-10)SignIs ( 2x+7)/(x+1) < 0 satisfied? (-∞,-10)-11(-11+10)/(-11-10) =(-1)/(-21) + Yes (-10,10)0(0+10)/(0-10) =10/(-10) -No (10,∞)11(11+10)/(11-10) 21/1 +Yes ( ( ) )

35. 11.4 Nonlinear Inequalities in One Variable Math, Statistics & Physics 35 ( The solution set = (-∞,-10) U (10, ∞ ) U { -10 } = (-∞,-10] U (10, ∞ ) ]