Download LESSON 1 REVIEW OF SOLVING NONLINEAR INEQUALITIES

LESSON 1 SOLVING NONLINEAR INEQUALITIES
In this lesson, we will make use of the Axiom of Trichotomy given below.
Axiom of Trichotomy A real number can only be one of the following: positive,
negative, or zero.
NOTE: When you substitute a real number in for the variable in a nonlinear
expression, you will either get another real number (which is either positive,
negative, or zero) or something that is undefined as a real number.
For example, when we replace the x in the nonlinear expression
7  3x
by
x  5 x  24
2
7  12
5
5

obtained by
16  20  24 12 . The resulting
12
real number is negative. If we replace the x by  9 in the expression, we get the
17
7  27
34 17


real number
obtained by
. This number is positive. If
81  45  24 12
6
6
7
you replace the x by
in the expression, you will get the real number zero
3
77
0
0


 0 . Finally, if we replace
obtained by 49 35
49 105 216
62

 24



9
3
9
9
9
9
the x by 3, we get an undefined real number since we get division by zero obtained
79
2

by
.
9  15  24
0
4, we get the real number 
Thus, to solve a nonlinear inequality, we will find all the real numbers that make a
nonlinear expression equal to zero. We will also have to find all the numbers that
make the nonlinear expression undefined. Thus, all the remaining real numbers,
when substituted for the variable in the nonlinear expression, would make the
resulting real number either be positive or negative. Thus, a nonlinear expression
has the ability to change signs at the real numbers where the expression is either
zero or undefined.
We will determine when a nonlinear expression is positive and negative using the
following three steps:
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1320
Step 1 Find all the real numbers that make the nonlinear expression equal zero
and all the real numbers that make the expression undefined.
Step 2 Plot all the numbers found in Step 1 on the real number line.
Step 3 Using the real number line in Step 2, identify the open intervals determined
by the plotted numbers. For each open interval, pick a real number that is in the
interval. We will call this number the “test value” for the interval. Substitute the
test value for the variable in the nonlinear expression. Whatever sign the
expression has for this test value, the expression will have the same sign for any
number in the open interval.
Examples Solve the following nonlinear inequalities.
1.
x 2  3x  40  0
Step 1:
Find when the nonlinear expression x 2  3x  40 is equal to
zero. That is, solve the equation x 2  3x  40  0 .
x 2  3x  40  0  ( x  5 ) ( x  8 )  0  x   5 , x  8
Find when the nonlinear expression x 2  3x  40 is undefined.
The expression x 2  3x  40 is defined for all real numbers x.
Step 2:
Plot all the numbers found in Step 1 on the real number line.

+

5
Step 3:
+
Sign of x 2  3x  40

8
Use the real number line to identify the open intervals determined
by the plotted numbers. Pick a test value for each open interval.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1320
Sign of x 2  3x  40 = ( x  5 ) ( x  8 )
Interval
Test Value
(  ,  5)
6
(  5, 8)
0
( 0  5) ( 0  8)  (  ) (  )  
(8,  )
9
(9  5) (9  8)  (  ) (  )  
(  6  5) (  6  8)  (  ) (  )  
Answer: (  5 , 8 )
2.
6x
 0
3x  14
NOTE: This is a two part problem. One part of the problem is to solve the
6x
 0 . The other part of the problem is to solve
nonlinear inequality
3x  14
6x
 0.
the equation
3x  14
We will use the three step method to solve the nonlinear inequality
6x
 0:
3x  14
Step 1:
Find when the nonlinear expression
6x
is equal to zero.
3 x  14
6x
 0 . The fraction is equal to
3x  14
zero if and only if the numerator of the fraction is equal to zero.
That is, solve the equation
That is,
6x
 0  6x0  x6
3 x  14
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1320
6x
Find when the nonlinear expression
is undefined. The
3 x  14
fraction is undefined if and only if the denominator of the
fraction is equal to zero.
That is,
Step 2:
6x
14
undefined  3x  14  0  x 
3 x  14
3
Plot all the numbers found in Step 1 on the real number line.

Step 3:

+


14
3
6
Sign of
6x
3 x  14
Use the real number line to identify the open intervals determined
by the plotted numbers. Pick a test value for each open interval.
Sign of
6x
3 x  14
Interval
Test Value
14 

  , 
3

0
60
()

 
0  14
()
 14

, 6

 3

5
65
()

 
15  14
()
( 6,  )
7
67
()

 
21  14
()
Thus, the solution for the nonlinear inequality
6x
 0 is the set of real
3x  14
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1320
6x
14 

 0
  ( 6 ,  ) . The solution for
numbers given by    ,
3x  14
3 

6x
 0 is the set
was found in Step 1 above. Thus, the solution for
3x  14
6 . Putting these two solutions together, we have that the solution for
6x
14 

 0 is the set of real numbers    ,
  [6,  ) .
3x  14
3 

14 

  [6,  )
Answer:    ,
3 

3.
4 t 2  15t  14  0
4 t 2  15t  14  0  ( t  2 ) ( 4 t  7 )  0  t   2 , t  
Step 1:
7
4
The expression 4 t 2  15t  14 is defined for all real numbers t.
Step 2:

+

2
+
Sign of 4 t 2  15t  14


7
4
Step 3:
Interval
Test Value
(  ,  2)
3
7

  2,  
4

 1.8
 7

  , 
 4

0
Sign of ( t  2 ) ( 4 t  7 )
(  3  2 ) (  12  7 )  (  ) (  )  
(  1.8  2 ) (  7.2  7 )  (  ) (  )  
(0  2)(0  7)  ( )( )  
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1320
 7

Answer: (   ,  2 )    ,  
 4

4.
7  3x
 0
x 2  5 x  24
7  3x
7
 0  7  3x  0  x 
x  5 x  24
3
Step 1:
2
7  3x
undefined  x 2  5x  24  0 
2
x  5 x  24
( x  8) ( x  3)  0  x   8, x  3
Step 2:

+

+



8
7
3
3
Step 3:
Sign of
Sign of
7  3x
( x  8) ( x  3)
Interval
Test Value
(  ,  8)
9
()
()

 
()()
()
7

  8, 
3

0
()
()

 
()()
()
7

 , 3
3

2.4
()
()

 
()()
()
( 3,  )
4
()
()

 
()()
()
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1320
7  3x
x  5 x  24
2
7

Answer:   8 ,   ( 3 ,  )
3

5.
16  5 y 2  0
NOTE: This is a two part problem. One part of the problem is to solve the
nonlinear inequality 16  5 y 2  0 . The other part of the problem is to solve
the equation 16  5 y 2  0 .
We will use the three step method to solve the nonlinear inequality
16  5 y 2  0 :
4
16
2
 y
16  5 y 2  0  16  5 y 2  y 
Step 1:
5
5
The expression 16  5 y 2 is defined for all real numbers y.

Step 2:

+



4
4
5
5
Sign of 16  5 y 2
Step 3:
Interval
Test Value


  ,  4 

5 

2

 4 ,

5

4 
5 
 4


, 
 5



Sign of 16  5 y 2
16  20  
0
16  0  
2
16  20  
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1320
Thus, the solution for the nonlinear inequality 16  5 y 2  0 is the set of real

4
,
numbers given by  
5

in Step 1 above.

4

,

5


4 

.
5 

4 
. The solution for 16  5 y 2  0 was found

5
6x
 0 is the set
Thus, the solution for
3x  14
Putting these two solutions together, we have that the

4

5
solution for 16  5 y 2  0 is the set of real numbers  

4

5
Answer: 
6.
,
,
4 
.
5 
4 

5 
5w3  20w 2
 0
2w  5
5w3  20w 2 5w 2 ( w  4 )

NOTE: Since 5w  20w  5w ( w  4 ) , then
2w  5
2w  5
3
Step 1:
2
2
5w 2 ( w  4 )
 0  w  0, w   4
2w  5
5w 2 ( w  4 )
5
undefined  w 
2w  5
2
Step 2:

+

+



4
0
5
2
Step 3:
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1320
5w 2 ( w  4 )
Sign of
2w  5
5w 2 ( w  4 )
Sign of
2w  5
Interval
Test Value
( ,  4)
5
()()()
()

 
()
()
(  4, 0)
1
()()()
()

 
()
()
5

 0, 
2

1
()()()
()

 
()
()
5

 , 
2

3
()()()
()

 
()
()
5

Answer: (   ,  4 )   ,  
2

7.
x3 ( x  3) 2 ( 6x  5 )
 0
( 7  x ) 4 ( 4 x  13 ) 5
NOTE: This is a two part problem. One part of the problem is to solve the
x3 ( x  3) 2 ( 6x  5 )
 0 . The other part of the problem
nonlinear inequality
( 7  x ) 4 ( 4 x  13 ) 5
x3 ( x  3) 2 ( 6x  5 )
 0.
is to solve the equation
( 7  x ) 4 ( 4 x  13 ) 5
We will use the three step method to solve the nonlinear inequality
x3 ( x  3) 2 ( 6x  5 )
 0:
( 7  x ) 4 ( 4 x  13 ) 5
Step 1:
x3 ( x  3) 2 ( 6x  5 )
5
 0  x  0 , x   3, x 
4
5
( 7  x ) ( 4 x  13 )
6
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1320
x3 ( x  3 ) 2 ( 6x  5 )
13
x

7
,
x



undefined
( 7  x ) 4 ( 4 x  13 ) 5
4
Step 2:
x3 ( x  3 ) 2 ( 6x  5 )
Sign of
:
( 7  x ) 4 ( 4 x  13 ) 5

+


13
4

+
+
+




3
0
5
6
7
Step 3:
Interval
Test Value
x3 ( x  3 ) 2 ( 6x  5 )
Sign of
( 7  x ) 4 ( 4 x  13 ) 5
13 

  ,  
4

4
()()()
()

 
()()
()
 13

,  3

 4

 3.1
()()()
()

 
()()
()
(  3, 0 )
1
()()()
()

 
()()
()
5

 0, 
6

0.1
()()()
()

 
()()
()
5

 , 7
6

1
()()()
()

 
()()
()
(7, )
8
()()()
()

 
()()
()
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1320
x3 ( x  3) 2 ( 6x  5 )
 0 is
Thus, the solution for the nonlinear inequality
( 7  x ) 4 ( 4 x  13 ) 5
 13

 5

,  3   (  3, 0 )   , 7   ( 7 ,  ) .
the set of real numbers  
 4

 6

x3 ( x  3) 2 ( 6x  5 )
 0 was found in Step 1 above. Thus,
The solution for
( 7  x ) 4 ( 4 x  13 ) 5
x3 ( x  3) 2 ( 6x  5 )
5


0

3
,
0
,

 . Putting
the solution for
is
the
set
6
( 7  x ) 4 ( 4 x  13 ) 5

these two solutions together, we have that the solution for
x3 ( x  3) 2 ( 6x  5 )
 0 is the set of real numbers
( 7  x ) 4 ( 4 x  13 ) 5
 13

5

, 0    , 7   (7, ) .

 4

6

 13

5

, 0    , 7   (7, )
Answer:  
 4

6

Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~janders/1320