LESSON 1 SOLVING NONLINEAR INEQUALITIES. In this lesson, we will make use of the Axiom of Trichotomy given below.

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1 LESSON 1 SOLVING NONLINEAR INEQUALITIES In this lesson, we will make use of the Aiom of Trichotomy given below. Aiom 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 epression, you will either get another real number (which is either positive, negative, or zero) or something that is undefined as a real number. For eample, when we replace the in the nonlinear epression by 1, we get the real number obtained by. The resulting real number is negative. If we replace the by 9 in the epression, we get the 1 1 real number obtained by. This number is positive. If 81 1 you replace the by in the epression, you will get the real number zero obtained by Finally, if we replace the by, we get an undefined real number since we get division by zero obtained 9 by. 9 1 Thus, to solve a nonlinear inequality, we will find all the real numbers that make a nonlinear epression equal to zero. We will also have to find all the numbers that make the nonlinear epression undefined. Thus, all the remaining real numbers, when substituted for the variable in the nonlinear epression, would make the resulting real number either be positive or negative. Thus, a nonlinear epression has the ability to change signs at the real numbers where the epression is either zero or undefined. We will determine when a nonlinear epression is positive and negative using the following three steps:

2 Step 1 Find all the real numbers that make the nonlinear epression equal zero and all the real numbers that make the epression undefined. Step Plot all the numbers found in Step 1 on the real number line. Step Using the real number line in Step, 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 epression. Whatever sign the epression has for this test value, the epression will have the same sign for any number in the open interval. Eamples Solve the following nonlinear inequalities. 1. Step 1: Find when the nonlinear epression is equal to zero. That is, solve the equation. ( )( 8), 8 Find when the nonlinear epression is undefined. The epression is defined for all real numbers. Step : Plot all the numbers found in Step 1 on the real number line. + + Sign of 8 Step : Use the real number line to identify the open intervals determined by the plotted numbers. Pick a test value for each open interval.

3 Interval Test Value Sign of = ( )( 8) (, ) ( )( 8) ( )( ) (, 8) ( )( 8) ( )( ) ( 8, ) 9 ( 9 )(9 8) ( )( ) Answer: (, 8). 1 NOTE: This is a two part problem. One part of the problem is to solve the nonlinear inequality. The other part of the problem is to solve 1 the equation. 1 We will use the three step method to solve the nonlinear inequality : 1 Step 1: Find when the nonlinear epression is equal to zero. 1 That is, solve the equation. The fraction is equal to 1 zero if and only if the numerator of the fraction is equal to zero. That is, 1

4 Find when the nonlinear epression is undefined. The 1 fraction is undefined if and only if the denominator of the fraction is equal to zero. That is, 1 undefined 1 1 Step : Plot all the numbers found in Step 1 on the real number line. + Sign of 1 1 Step : Use the real number line to identify the open intervals determined by the plotted numbers. Pick a test value for each open interval. Interval Test Value Sign of 1 1, 1 ( ) ( ) 1, 1 1 ( ) ( ) (, ) 1 1 ( ) ( ) Thus, the solution for the nonlinear inequality 1 is the set of real

5 1 numbers given by, (, ). The solution for 1 was found in Step 1 above. Thus, the solution for is the set 1. Putting these two solutions together, we have that the solution for 1 is the set of real numbers, [, ). 1 1 Answer:, [, ). t 1t 1 Step 1: t 1t 1 ( t )(t ) t, t The epression t 1t 1 is defined for all real numbers t. Step : + + Sign of t 1t 1 Step : Interval Test Value Sign of ( t )(t ) (, ) ( )( 1 ) ( )( ), 1. 8 ( 1.8 )(. ) ( )( ), ( )( ) ( )( )

6 Answer: (, ),. Step 1: undefined ( 8)( ) 8, Step : + + Sign of 8 Step : Interval Test Value Sign of ( 8)( ) ( ) (, 8) 9 ( )( ) ( ) ( ) ( ) 8, ( )( ) ( ) ( ), ( ). ( )( ) ( ) ( ) ( ) (, ) ( )( ) ( ) ( )

7 Answer: 8, (, ). 1 y NOTE: This is a two part problem. One part of the problem is to solve the nonlinear inequality 1 y. The other part of the problem is to solve the equation 1 y. We will use the three step method to solve the nonlinear inequality 1 y : 1 Step 1: 1 y 1 y y y The epression 1 y is defined for all real numbers y. Step : + Sign of 1 y Step : Interval Test Value Sign of 1 y, 1, 1, 1

8 Thus, the solution for the nonlinear inequality 1 y is the set of real numbers given by,. The solution for 1 y was found in Step 1 above. Thus, the solution for is the set 1,. Putting these two solutions together, we have that the solution for 1 y is the set of real numbers,. Answer:, w w. w NOTE: Since w w w ( w ), then w w w w ( w ) w Step 1: w ( ) ww w, w w ( w ) w undefined w Step : + + Sign of w ( w ) w Step :

9 Interval Test Value Sign of w ( w ) w ( )( )( ) (, ) ( ) ( ) ( ) ( )( )( ) (, ) 1 ( ) ( ) ( ) ( )( )( ), 1 ( ) ( ) ( ), ( )( )( ) ( ) ( ) ( ) Answer: (, ), ( ) ( ). ( ) ( 1) NOTE: This is a two part problem. One part of the problem is to solve the ( ) ( ) nonlinear inequality. The other part of the problem ( ) ( 1) ( ) ( ) is to solve the equation. ( ) ( 1) We will use the three step method to solve the nonlinear inequality ( ) ( ) : ( ) ( 1) Step 1: ( ) ( ) ( ( ) 1)

10 ( ) ( ) ( ( ) 1) undefined, 1 ( ) ( ) Step : Sign of : ( ) ( 1) Step : Interval Test Value ( ) ( ) Sign of ( ) ( 1) 1 ( )( )( ), ( )( ) ( ) ( ) 1, ( )( )( ).1 ( )( ) ( ) ( ) ( )( )( ) (, ) 1 ( )( ) ( ) ( ) ( )( )( ),.1 ( )( ) ( ) ( ), ( )( )( ) 1 ( )( ) ( ) ( ) ( )( )( ) (, ) 8 ( )( ) ( ) ( )

11 ( ) ( ) Thus, the solution for the nonlinear inequality ( ) ( 1) 1 the set of real numbers, (, ), (, ). ( ) ( ) The solution for was found in Step 1 above. Thus, ( ) ( 1) ( ) ( ) the solution for is the set,, ( ) ( 1). Putting these two solutions together, we have that the solution for ( ) ( ) is the set of real numbers ( ) ( 1) 1,, (, ). is 1 Answer:,, (, )