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\[ \bigg||x-1|-5\bigg|-2 > 0 \]
Add 2 to both sides. Where areyou getting stuck?
I don't know how to deal with mod functions while solving for domain.
where is the mod though?
so when you say 'mod' you mean the absolute value function?
\[ |f(x)| < a\implies -a
\[ \bigg||x-1|-5\bigg| >2 \implies |x-1|-5 < -2\quad\quad |x-1|-5 >2 \]
The equations will keep splitting up and you'll have to keep track of all of them.
okay,ill try and let you know!
x-1<3 x-1<-3 x-1>7 x-1>-7 ?
x<4 x<-2 x>8 x>-6 ?
x-1 > -3 *
samajh aaya kyun?
x-1< -7 ** okay?
|x-1|<3 |x-1|>7 x-1<3 x-1>-3 x-1>7 x-1<-7
x<4 x>-2 x>8 x<-6
yup. thats right i guess.
is it ? :o
ill do after this :/ thanks~
looks like you messed up
this is a really tedious problem man.
Give me a minute.
hey wio. |x-1|<3 |x-1|>7 x-1<3 x-1>-3 x-1>7 x-1<-7 Thats right no?
so far, that is right.
\[\Huge||x-1|-5\bigg|-2 > 0\] ------->x<4 So lets take x as 3. Substituting, Okay positive. x>-2 So lets take x as -1. Substituting, Okay positive. x>8 So lets take x as 9. Substituting, Okay positive. x<-6 So lets take x as -7. Substituting, Okay positive. So that means all the solutions are acceptable.
So, x<4, x<-6 --> x<-6 x>-2, x>8 --> x>8 right? @wio
Looking at x<4 and x>-2 is enough to tell you all solutions work.
So drawing the wavy curve graph. |dw:1364716700507:dw|
Wait, hold on...
you've got: \[ -2
\]What happens at 5?
\[\LARGE (-\infty,-5] U (8,\infty) \]
oh ye to tera answer agaya :P @yrelhan4 lol
but its wrong :/
Combining these equations: \[ -2
\]Makes me thing it's \(x\in (-\infty,-6)\cup (-2,4)\cup (8,\infty)\)
still wrong :P
Hmm, what are you using?
\[x\in (-8,-6)\cup (-2,4)\cup (8,\infty)\] is the answer
Put -10 in the question.. Its >0
Yeah, that answer doesn't make sense.
i dont know how -8
i prefer @wio 's answer though,still we might be missing something maybe
There would have to be some other thing limiting the domain.