Absolute Value

Engi44

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Apr 2, 2010
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I'm having some trouble understanding this.

If x<-2, then |x+2|/x+2= -x+2/x+2= -x/x= -1, right? But what if x<-1? Could someone show me how this breaks down when x is less than negative 1?

The problem is written like this:

|x+2|/x+2, x< -2

What I'm wondering is how it breaks down if:

|x+2|/x+2, x< -1


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You could help me out further by breaking down this:

|x-1|/ x-1, x> 0
 
|x+2|/x+2, x< -2

You should do it for cases, which I believe comes up at times when solving absolute values. Ckeck

x+2/x+2 and -(x+2)/x+2 and 0/x+2.

These are the cases that result if x is bigger than -2, equal to -2, or less than -2. All cases must be investigated to simplify through an absolute value sign, and all cases are present. I will not say for now which expressions stands for which case; I will leave that for you to figure out.

After you have figured that out, look at the problem's condition on x. Now, which equation is the only equation that applies to your situation?
 
Engi44 said:
I'm having some trouble understanding this.

If x<-2, then |x+2|/x+2= -x+2/x+2= -x/x= -1, right? <----you got the right answer, but for the WRONG reason.....

If x < -2, then x + 2 < 0. By the definition of absolute value, if a < 0, | a | = -a. So, if (x + 2) < 0, then | x + 2 | = - (x + 2).....NOT -x + 2 as you've indicated. And you cannot simplify (-x + 2) / (x + 2) by dividing out the 2s to get -x/x.........

If x < -2, | x + 2 | / (x + 2) = -(x + 2) / (x + 2), which is -1 (because (x + 2)/(x + 2) = 1)



But what if x<-1? Could someone show me how this breaks down when x is less than negative 1?

The problem is written like this:

|x+2|/x+2, x< -2

What I'm wondering is how it breaks down if:

|x+2|/x+2, x< -1
If x < -1, you have three possibilities to consider...
If x = -2 (which could happen if x < -1), then the fraction is undefined because the denominator is 0.
If -2 < x < -1, (x + 2) is non-negative. When a > 0, | a | = a, so when x + 2 is non-negative, | x + 2 | = x + 2, and your fraction becomes (x + 2)/(x + 2), or 1.
If x < -2, (which would be part of the domain x < -1), as we've seen above, |x + 2| / (x + 2) = -1



____________________________________________
You could help me out further by breaking down this:

|x-1|/ x-1, x> 0
This is similar to your second question above...you need to consider three possibilities....

If 0 < x < 1, (x - 1) is negative....so |x - 1| = -(x - 1) and you should be able to figure out the value of the fraction |x - 1| / (x - 1)

If x = 1, the fraction is undefined (because the denominator will have a value of 0).

If x > 1, (x - 1) is positive...and |x - 1| = x - 1. Again, you should be able to figure out the value of the fraction in this case.
 
pitoten said:
|x+2|/x+2, x< -2

You should do it for cases, which I believe comes up at times when solving absolute values. Ckeck

x+2/x+2 and -(x+2)/x+2 and 0/x+2.

These are the cases that result if x is bigger than -2, equal to -2, or less than -2. All cases must be investigated to simplify through an absolute value sign, and all cases are present. I will not say for now which expressions stands for which case; I will leave that for you to figure out.

After you have figured that out, look at the problem's condition on x. Now, which equation is the only equation that applies to your situation?

x+2/x+2 if x is bigger than -2. -(x+2)/x+2 if x is less than -2. 0/x+2 if x is equal to -2.

Thank you, it took a little mulling over to help me realise that this just asks you whether the result of |x+2|/x+2 if x<-2 is positive, negative, or zero. I couldn't stop thinking I was trying to solve for x.

So this means in the problem |x+2|/x+2 if x<-1, it will reduce to +-1 or it will be undefined.

Thank you both for the help.


Oh yes, "If x<-2, then |x+2|/x+2= -x+2/x+2= -x/x= -1, right?"
Thank you mrspi for correcting me. If x<-2, then |x+2|/x+2 will reduce to -(x+2)/(x+2) and from there we get -1.
 
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