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Absolute Value

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
solve \(\displaystyle ||x|-2|+|x+1|=3\)
and we find that \(\displaystyle x=0,-2,2,-1\)
I got problem to find the 'case',

Regards,
\(\displaystyle |\rangle\)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Hello MHB,
solve \(\displaystyle ||x|-2|+|x+1|=3\)
and we find that \(\displaystyle x=0,-2,2,-1\)
I got problem to find the 'case',

Regards,
\(\displaystyle |\rangle\)
That doesn't look like the right solution...

Talking about cases, what if x<-2? Or -2<x<-1? And what about -1<x<0? Or perhaps 0<x<2? And x>2?
 

Petrus

Well-known member
Feb 21, 2013
739
That doesn't look like the right solution...

Talking about cases, what if x<-2? Or -2<x<-1? And what about -1<x<0? Or perhaps 0<x<2? And x>2?
I don't mean those are the answer, but those are the point we should check \(\displaystyle \geq\) or \(\displaystyle \leq\), I hope you did understand.
Can I also check this one insted of -1<x<0
-2<x<0?

Regards,
\(\displaystyle |\rangle\)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
I don't mean those are the answer, but those are the point we should check \(\displaystyle \geq\) or \(\displaystyle \leq\), I hope you did understand.
Can I also check this one insted of -1<x<0
-2<x<0?
Sure you can. It's just that |x+1| does something funny at x=-1.
 

Petrus

Well-known member
Feb 21, 2013
739