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

J

joshuaG

Member
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
 
A

AH

Member
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.
 
T

TedS

Member
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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