Absolute Value

[Petrus]

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]

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]

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]

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]

Sure you can. It's just that |x+1| does something funny at x=-1.

Thanks, got the correct answer now
Regards,
\(\displaystyle |\rangle\)