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November 28, 2013
 to solve inequalities that involve radical 
functions
 A radical function looks like 
n y 
radicant 
index 
 When , we have a square root fuction. 
n  2 
n  3 
 When , we have a cube root function. 
 When , we have a fourth root function. 
n  4 
 When n  5 
, we have a fifth root function, 
and so on…
Given a radical function , 
Case 1: 
If n is even, then and 
Case 2: 
If n is odd, then and 
Furthermore, 
n y 
0 n y  0y  
n y R yR 
 n 
n y  y
 Which of the following equations does not 
have a solution? Why? 
Because 
x  2  0
 What can you say about the solutions of the 
following inequalities? 
4 
8 
x 
3  5   
2 
   
3 
x 
x 
2  4   
2
 Do the following inequalities have solutions? 
What is the lower bound for each of the 
inequalities? 
4 
3 5 
3 
1 3 
x 
x 
x 
  
  
  
x 
0   3  
5 
4 
x 
0    
3 
x 
0   1  
3 
Solve the inequalities.
 Suppose the following inequalities are valid, 
then what must be true about each of the 
inequalities? 
4 2 
6 
x x 
6 5 
3 6 
1 
x 
x 
   
   
   
4 2 
6 
x x 
6 0 
3 0 
0 
x 
x 
  
  
  
Solve the inequalities.
 Let’s solve the following inequalities 
4 
2 3 3 
3 2 5 
2 4 
2 
x 
x 
x 
x 
  
  
  
 
 Up until now, we have discussed the fact 
that for 
Case 1: 
If 
k 
f x g x 
a if g x then 
b if g x and f x then f x g x 
Case 2: 
If 
n  2k 
2 
2 
( ) ( ) 
( ) ( ) 0, 
( ) ( ) 0 ( ) 0, ( ) k 
( ) 
 
  
   
2 
k 
f ( x ) g ( x 
) 
a if g x then f x 
b if g x then f x g 2 
x 
( ) ( ) 0, ( ) 0 
( ) ( ) 0, ( ) k 
( ) 
 
  
 
 Do you remember the fact that if , 
then and ? 
n yR y R  
 Because of this, we have less rules for solving 
inequalities whose index is odd. 
 In fact, 
n  2k 1 
k k 
2  1 2  
1 
f x g x f x g x 
( )  ( )  ( )  
( ) 
k k 
2  1 2  
1 
f x g x f x g x 
( )  ( )  ( )  
( )
 Solve each of the following inequalities whose 
index is odd. 
3 
  
2 2 
3 
9 2 
1  2   
3 
1 
1 
x 
x 
x 
x 
 
  
Answers: 
[1, ) 
( 14, 14) 
 
 
  
     
  
  
1 
0, 
2 
x

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Solving inequalities that involve radical functions (Nov 28, 2013)

  • 2.  to solve inequalities that involve radical functions
  • 3.  A radical function looks like n y radicant index  When , we have a square root fuction. n  2 n  3  When , we have a cube root function.  When , we have a fourth root function. n  4  When n  5 , we have a fifth root function, and so on…
  • 4. Given a radical function , Case 1: If n is even, then and Case 2: If n is odd, then and Furthermore, n y 0 n y  0y  n y R yR  n n y  y
  • 5.  Which of the following equations does not have a solution? Why? Because x  2  0
  • 6.  What can you say about the solutions of the following inequalities? 4 8 x 3  5   2    3 x x 2  4   2
  • 7.  Do the following inequalities have solutions? What is the lower bound for each of the inequalities? 4 3 5 3 1 3 x x x       x 0   3  5 4 x 0    3 x 0   1  3 Solve the inequalities.
  • 8.  Suppose the following inequalities are valid, then what must be true about each of the inequalities? 4 2 6 x x 6 5 3 6 1 x x          4 2 6 x x 6 0 3 0 0 x x       Solve the inequalities.
  • 9.  Let’s solve the following inequalities 4 2 3 3 3 2 5 2 4 2 x x x x        
  • 10.  Up until now, we have discussed the fact that for Case 1: If k f x g x a if g x then b if g x and f x then f x g x Case 2: If n  2k 2 2 ( ) ( ) ( ) ( ) 0, ( ) ( ) 0 ( ) 0, ( ) k ( )       2 k f ( x ) g ( x ) a if g x then f x b if g x then f x g 2 x ( ) ( ) 0, ( ) 0 ( ) ( ) 0, ( ) k ( )     
  • 11.  Do you remember the fact that if , then and ? n yR y R   Because of this, we have less rules for solving inequalities whose index is odd.  In fact, n  2k 1 k k 2  1 2  1 f x g x f x g x ( )  ( )  ( )  ( ) k k 2  1 2  1 f x g x f x g x ( )  ( )  ( )  ( )
  • 12.  Solve each of the following inequalities whose index is odd. 3   2 2 3 9 2 1  2   3 1 1 x x x x    Answers: [1, ) ( 14, 14)              1 0, 2 x