Expert: Paul Klarreich - 4/7/2008

Question
Hi Paul!

This is from the homework of my Mathematical Analysis I course. I don't know how to do this kind of exercise where I have an inequation between two absolute values.

Problem:

Find all the values of X that verify /X-1/ <= /X+3/

Note: "/" represents the absolute value).

I will really appreciate any explanation.

Thanks for your time!

Regards,
Villafana Liñan, Vanesa.

Answer
Questioner:   Vanesa Villafana Liñan
Category:  Calculus
Private:  No
 
Subject:  Absolute Value (Inequation)
Question:  Hi Paul!

This is from the homework of my Mathematical Analysis I course. I don't know how to do this kind of exercise where I have an inequation between two absolute values.

Problem:

Find all the values of X that verify /X-1/ <= /X+3/

Note: "/" represents the absolute value).

I will really appreciate any explanation.

Thanks for your time!

Regards,
Villafana Liñan, Vanesa.
......................................
Hi, Vanesa,

The definition of an absolute value expression always has a break.  The form is: (I use the vertical bar, '|' for it.)


| something | = { the something, when  the something >= 0
               { the opposite of the something, when  the something < 0

So, for example, if the something is (x - 9),

| (x - 9) | = {   (x - 9), when  (x - 9) >= 0
             { - (x - 9), when  (x - 9) < 0


Now you have two of these somethings.  So you will have two 'breaks'.  One is at x = 1, and one at x = -3.

 Region 1        -3   Region 2          +1  Region 3
-------------------+----------------------+----------------
Both things (-)     (x-3) is +,(x-1)is(-)   Both (+)

So you must rewrite your inequality differently in each region.

Region 1:  x < -3

|x-1| <= |x+3|  rewrites to:

-(x-1) <= -(x+3) and solves:
-x + 1 <= -x - 3
 + 1 <=  - 3
which is never true
.......................
Region 2:  x > -3 and x < 1

|x-1| <= |x+3|  rewrites to:

-(x-1) <= +(x+3) and solves:
- x + 1 <= x + 3
-2x <= 2
x <= -1.

Which means x must be between -1 and 1.  
.......................

Region 3:  x > 1

|x-1| <= |x+3|  rewrites to:

+(x-1) <= +(x+3)

x - 1 <= x + 3

-1 <= 3
which is always true.  So your inequality will be true in all parts of region 3.
...............................
What is your final answer?  Combine the intervals [-1, 1] and [1, inf.] and you have   -1 <= x <= inf, or just  x >= -1