11.4 nonlinear inequalities in one variable math, statistics & physics 1 first we find the...

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



( ))

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)

(



Interval

Test Point

3x(x-2) Sign Is 3x(x-2) < 0 satisfied?

(-∞,0 -1 3(-1)(-1-2)=3(-1)(-2) + No

(0,2) 1 3(1)(1-2)=3(1)(-1) - Yes

(2,∞) 4 3(4)(4-2)=3(4)(2) + No

(( ))



( )

The solution set

= (0,2)



First we find the zeroes of the polynomial; that’s the solutions of the equation:(x+2)(x-1)(x-5) = 0



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}

)) ) (((



Interval Test Point

(x+2)(x-1)(x-5) Sign Is (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

)( ( ())



] [ ]

The solution set

= (-∞,-2) U (1, 5) U {-2,1,5}= (-∞,-2] U [1, 5]



First we find the zeroes of the polynomial; that’s the solutions of the equation:2x2 + 5x – 12 = 0



42

3

04032

0)4)(32(

01252 2

xOrx

xOrx

xx

xx



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 2x2+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 2x2+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 }

( )) (



Interval

Test Point

2x2+5x-12=(x-1.5)(x+4)

Sign Is 2x2+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

))) )((



[]

The solution set

= (-∞,-4) U (1.5,∞) U { -4 , 1.5}

= (-∞,-4] U [1.5,∞)



Solution

First We find the zeroes of the polynomial 9x3 + 72x2 -25x -200

Let:

9x3 + 72x2 -25x -200 = 0



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)

)) ) (( (



Interval Test Point

(x+8)(x+5/3)(x-5/3) Sign Is (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

() (() )



)( (

The solution set:

= (-8, - 5/3) U (5/3 , ∞ )



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.



01

72

01

2250

1

)1(25

01

)1(2

1

502

1

5

21

5

x

xx

x

x

xx

x

xx

x



1

01:

:2

13

2

7

072:

:1

72

x

xLet

definednotispressionexthewhichanumbersThe

x

xLetx

xxpressionetheofzeroesThe



( ))

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).

(



Interval Test Point

(2x+7)/(x+1) Sign Is (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

(( ))



( )

The solution set

= (-7/2,-1)



01

72

01

2250

1

)1(25

01

)1(2

1

502

1

5

21

5

x

xx

x

x

xx

x

xx

x

21

5

x



1

01:

:2

13

2

7

072:

:1

72

x

xLet

definednotispressionexthewhichanumbersThe

x

xLetx




( ))

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

(



Interval Test Point

(2x+7)/(x+1) Sign Is (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

(( ))



( )

The solution set

= [-7/2,-1)



10

010:

:

10

010:

:10

10

x

xLet

definednotispressionexthewhichatnumbersThe

x

xLetx




() )

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 }

(



Interval Test Point

(x+10)/(x-10) Sign Is (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

(( ))



(

The solution set= (-∞,-10) U (10, ∞ ) U { -10 }

= (-∞,-10] U (10, ∞ )

]

11.4 nonlinear inequalities in one variable math, statistics & physics 1 first we find the...

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