why does |x - 1| - |x - 2| = 1 equate to 2 <(and equal to) x < infinity?

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joshuaG

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The reason why I ask this is that I bought a calculus book and this one question has been boggling me for a while. The book gives very little detail as to how they found it except something to do with distance with absolute values, which is understandable but doesn't explain it very well
 
There are 3 cases for you to examine: x < 1, 1<x<2, and x > 2.

Case A:
If x < 1, then |x-1| = 1-x. That's because the stuff inside the || will be negative, and the absolute value sign will flip it around.
If x < 1, then |x-2| = 2-x. Same reasoning.

So, playing with the left hand side of the orig equation:
|x - 1| - |x - 2|
(1-x) - (2-x)
1 + -x + -2 + x
-1

So, the left side simplifies to -1, which is NEVER equal to 1.

Case B: 1 < x < 2
If 1 < x < 2, then |x-1| = x-1. x=1 is the point where x-1 crosses from negative to positive.
If 1 < x < 2, then |x-2| = 2-x

|x - 1| - |x - 2|
(x-1) - (2-x)
x + -1 + -2 + x
2x - 3

Now 2x-3 = 1, so x = 2. But this contradicts our original statement that 1<x<2, so there are no solutions.

Case C: x > 2
|x-1| = x-1
|x-2| = x-2

|x - 1| - |x - 2|
x - 1 + -x + 2
1

1 = 1 all the time, so as long as x > 2, this works.

There are two other parts numbers to check - the endpoints, x=1 and x=2. x=2 works; x=1 doesn't.
 
the key values here are x = 1 & x= 2
if x < 1 then the problem is -(x-1) - [ -(x-2) ] = 1.{none}
if 1 x < 2 then it is (x-1) -[-(x-2) ] = 1 { none}
if 2 x the it is (x-1) - (x-2) = 1 { all}
 
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