Lesson Solving inequalities for rational functions with non-zero right side

Solving inequalities for rational functions with non-zero right side

Problem 1

 
  

  


 < 2.      (1)


1. Assume that x - 6 > 0, i.e. x > 6.
   
   Multiply both sides of (1) by (x-6), which is positive in this case. You will get an inequality                   

   x < 2*(x-6)  --->  x < 2x - 12  --->  12 > x.

   Thus he solution in this case is the set of real numbers {x | 6 < x < 12}, i.e the interval (6,12).


2. Assume that x - 6 < 0, i.e. x < 6.
   
   Multiply both sides of (1) by (x-6), which is negative in this case. You will get an inequality

   x > 2*(x-6)  --->  x > 2x - 12  --->  12 > x.   

      (Notice that I changed the inequality sign when multiplied by negative number!)

   Thus the solution in this case is the set of real {x | x < 6}, i.e the semi-infinite interval (,).

Answer. The solution is the union of two intervals: (,) U (,).


The plot of the function  f(x) =   is shown in Figure 1.

 
  





    Figure 1. Plot y = 

  
  

Problem 2

 
  

  


 < 3.      (2)


1. Assume that x > 0.
   
   Multiply both sides of (2) by 3x, which is positive in this case. You will get an inequality                       

   x - 5 < 9x  --->  -5 < 8x  --->  x > .

   Thus he solution in this case is the set of real numbers {x | x > 0}, i.e the interval (,).


2. Assume that x < 0.
   
   Multiply both sides of (2) by 3x, which is negative in this case. You will get an inequality

   x -5 > 9x  --->  -5 > 8x  --->  x < .   

      (Notice that I changed the inequality sign when multiplied by negative number!)

   Thus the solution in this case is the set of real {x | x <  }, i.e the semi-infinite interval (,).

Answer. The solution is the union of two intervals: (,) U (,).


The plot of the function f(x) =  is shown in Figure 2.

 
  





    Figure 2. Plot y = 

  
  

Problem 3

 
  

  


 > 4.      (3)


1. First, let us assume that x+1 > 0.
   In other words, we will consider now real numbers { x | x > -1 }. 

   Multiply both side of (1) by (x+1), which is positive in this case. Then you will get inequality                   

   3x+2 > 4*(x+1)  --->  3x+2 > 4x+4  --->  2-4 > 4x-3x  --->  -2 > x.

   Thus we obtain this: if x > -1, then x < -2. 
   
   It is, surely, absurd. 
   So, in the domain x > -1 there is no solution to (1).


2. Next, let us consider the interval x < -1. In this interval, the denominator (x+1) is negative.

   Multiply both side of (3) by (x+1), which is negative now. Then you will get

   3x+2 < 4*(x+1).     (4) 

   Notice that I changed the sign ">" of the inequality to the opposite sign "<", when 
   multiplied both sides of (1) by negative number (x+1).

   Further, (2) implies  3x+2 < 4x+4  --->  2-4 < 4x-3x  --->  -2 < x,   or   x > -2.
   
   Thus we obtain this: if x < -1, then x > -2.

   It means that the set of real numbers -2 < x < -1 satisfies the inequality (3).

   It is the solution of the inequality (3).

Answer. The solution to (3) is the interval (-2,-1).


The plot of the function    is shown in Figure 3.

 
  





    Figure 3. Plot y = 

  
  

Problem 4

 
  

  


 > 2.      (5)


1. First, let us assume that x-2 > 0.
   In other words, we will look now for solutions in the domain  { x | x > 2 }. 

   Multiply both side of (5) by (x-2), which is positive in this case. Then you will get inequality                   

   3x+1 > 2*(x-2)  --->  3x+1 > 2x-4  --->  3x - 2x > -4 -1  --->  x > -5.

   Thus we obtain this: the solution is the intersection of two sets: {x| x > 2} and {x| x > -5}. 
   
   This intersection is the set {x| x > 2}.
   
   So, the interval (,) is the solution to (5), under the assumption that x > 2.


2. Next, let us consider the domain x < 2. In this domain, the denominator (x-2) is negative.

   Multiply both side of (1) by (x-2), which is negative now. Then you will get

   3x+1 < 2*(x-2).     (6) 

   Notice that I changed the sign ">" of the original inequality to the opposite sign "<", when 
   multiplied both sides of (5) by negative number (x-2).

   Now, (6) implies  3x+1 < 2x-4  --->  3x - 2x < -4 -1  --->  x < -5.
   
   Thus by analyzing the domain x < 2 we obtain the solution  x < -5.

   By collecting the results of n.1 and n.2 you get the full solution set.
   It is the union (,) U (,).

   The problem is solved.

Answer. The solution to  (5)  is the union  (,) U (,).


The plot of the function    is shown in Figure 4.

 
  





    Figure 4. Plot y =