Proving the Maximum and Minimum Values of Absolute Value Functions

[silvashadow]

How do I prove that: http://img177.imageshack.us/img177/9184/eqnnq3.th.jpg ?

 

[sutupidmath]

How do I prove that: http://img177.imageshack.us/img177/9184/eqnnq3.th.jpg ?

well here it is what i would do, consider first the values of x,y<-1 and you will get
I x-1 I= -(x-1) and I y+1 I= -(y+1), in which case you would get -3+2 which is smaller than 5.
then -1<x,y<1, in which case you would get
Ix-1I= -(x-1) and for Iy+1I= y+1, see what happens now when you plug in?
and the other case would be x,y>1
In which case you would ged
Ix-1I=x-1 and Iy+1I=y+1 when you plug in see what u get.

Note; this is not exactly as i am saying but i did it in purporse, this is the idea, but just that you have to do it in a better order, like taking the values of x smaller/greater than sth first, and then those of y, etc

P.s.Show some work of yours first, and then maybe you will get some more help!

 

[nicksauce]

What is the maximum value of (x-1) / |x-1| ?

 

[sutupidmath]

Since the OP hasn't shown up yet, i will give him/her some more hints.
Indeed, i will just pick up at what nicksauce suggested, it is shorter and nicer, look teh maximum value of \frac{(x-1)}{ abs{(x-1)}}, well the maximum here is 3.
Now look at the other part what is the minimum value of
2\frac{(y+1)}{abs(y+1)}, obviously it is -2, or you might want to look for the maximum of the:
-2\frac{(y+1)}{abs(y+1)}, which would be 2,
now going back to what you have to prove, why does that inequality hold?
Can you figure it out now, do u understand how it goes?