simple algebra question

[Semidevil]

given |x| + |x + 1|.

for x <-1, this implies |x| = -x and |x + 1| = -x -1
for -1<x<0, this implies |x| = -x and |x + 1| = x + 1
for x > 0, this implies |x| = x and |x| = x + 1.

I completl forgot how they did those case analysis.....

how did they get all those implications??

 

[DrPizza]

Are you saying that you don't understand that if x = -4 or x = -5 or anything else, x<-1, that the absolute value of x is -x?
In other words, if x = -4, then the absolute value of x does this: -x.
-(-4) = 4

and, probably to complete it, then abs(x) + abs(x+1) = (depending on the part of the domain)
-2x -1 or 1 or 2x + 1

 

[Semidevil]

so what if x is between -1 and 0?? how do they do that?

 

[snoopdoug1]

they probably specified it was for integers only

 

[tikwanleap]

They got all the "critical points" of x for |x| and |x+1|

The "critical point" of x for |x| is 0. The is the point where |x| changes from behaving like -x to behaving like x.

The "critical point" of x for |x+1| is -1. The is the point where |x + 1| changes from behaving like -x - 1 to behaving like x + 1.

Then all the ranges are:

x < -1 : This is when |x| behaves like -x and |x+1| behaves like -x - 1
-1 < x < 0 : This is when |x| behaves like -x and |x+1| behaves like x + 1
0 < x : This is when |x| behaves like x and |x+1| behaves like x + 1

 

[Semidevil]

I know where they got the critical points, but I dont know how the implications work.....

 

[Semidevil]

morning bump

 

[aux]

Originally posted by: Semidevil
I know where they got the critical points, but I dont know how the implications work.....

See DrPizza's and tikwanleap's explanations.

Here is my attempt for an explanation:
for any a
|a| = a (if a >=0) or -a (if a<0).

Therefore, replacing a by x:
|x| = x (if x >=0) or -x (if x<0)
and replacing a by x+1
|x+1| = x+1 (if x+1 >=0, that is x>=-1) or -(x+1) (if x+1<0, that is x<-1)

Then apply the above for the 3 segments, separated by the critical points x=-1 and x=0.


Disclaimer: typos possible

 

[SpecialEd]

In thie first one, x is negative and x+1 is negative so taking absolute value reverses the sign of the variable

In the second one, x is negative and x+1 is positve, so abs of x is -x and abs of x+1 changes nothing.

In the third one, x and x+1 are both positive, so absolute value doesn't effect anything.

Abs(x)=x+1 is only true for x=-.5, so you thrid statement is false, but you may have meant abs(x+1) = x+1.

 

[DrPizza]

Originally posted by: Semidevil
so what if x is between -1 and 0?? how do they do that?

isn't that the third line of your original post??
-1<x<0. . .

 

[tikwanleap]

Originally posted by: Semidevil
I know where they got the critical points, but I dont know how the implications work.....

What do you mean by implications?