|x-1|>=|x+1|

[ramone]

hello, i'm having some problems with this:
|x-1|>=|x+1|
after thinking a while i found what the catch is, x is less than 0, that's the only way in wich |x-1| will be less than |x+1|, but i don't know what is the procedure to solve it, can you help me with this please?

thank you

 

[Matt]

Consider three cases:

(1) x <= -1
(2) -1 <= x <= 1
(3) x <= 1

In each case you can remove the absolute value symbols from the equation.

In the first case, for example, \(\displaystyle x-1 < 0,\) and therefore \(\displaystyle |x-1|=-(x-1).\) Likewise, \(\displaystyle x+1 < 0,\) and therefore \(\displaystyle |x+1|=-(x+1).\)

 

[soroban]

Hello, ramone!

How about a graphic approach?

\(\displaystyle |x\,-\,1|\:\geq\:|x\,+\,1|\)

The graph of: \(\displaystyle y\;=\;|x|\,\) is a "V" with its vertex at the origin.

The graph of: \(\displaystyle y\:=\:|x - 1|\,\) is the same graph moved 1 unit to the right.

The graph of: \(\displaystyle y\:=\;|x\,+\,1|\,\) is moved 1 unit to the left.

              \     |     /           /
    \           \   |   /           /
      \           \ | /           /
        \           *           /
          \       / | \       /
  y = |x+1| \   /   |   \   / y = |x-1|
    - - - - - * - - + - - * - - - - -
             -1     |     1

Now you can see where \(\displaystyle \,|x\,-\,1|\,\) is greater than or equal to \(\displaystyle \,|x\,+\,1|\)