DLS
Domain of..sqrt{||x-1|-5|-2} ?
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DLS
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@wio
wio
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\[
\bigg||x-1|-5\bigg|-2 > 0
\]
DLS
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Correct,Next?
mathslover
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\[\sqrt{||x-1|-5|-2}\] better...
wio
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Add 2 to both sides. Where areyou getting stuck?
DLS
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I don't know how to deal with mod functions while solving for domain.
wio
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where is the mod though?
DLS
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mod/absolute bars
DLS
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\[\Huge \sqrt{||x-1|-5|-2}\]
wio
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so when you say 'mod' you mean the absolute value function?
DLS
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Yes
wio
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1
\[
|f(x)| < a\implies -a<f(x)<a\\
|f(x)| > a\implies a<f(x),\quad f(x)<-a\\
\]
wio
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\[
\bigg||x-1|-5\bigg| >2 \implies |x-1|-5 < -2\quad\quad |x-1|-5 >2
\]
wio
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The equations will keep splitting up and you'll have to keep track of all of them.
DLS
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okay,ill try and let you know!
DLS
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|x-1|<3
|x-1|>7
DLS
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x-1<3
x-1<-3
x-1>7
x-1>-7
?
DLS
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x<4
x<-2
x>8
x>-6
?
yrelhan4
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x-1 > -3 *
yrelhan4
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samajh aaya kyun?
DLS
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hmm..yeah kinda
yrelhan4
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x-1< -7 **
okay?
DLS
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|x-1|<3
|x-1|>7
x-1<3
x-1>-3
x-1>7
x-1<-7
DLS
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x<4
x>-2
x>8
x<-6
yrelhan4
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yup. thats right i guess.
DLS
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|dw:1364716179260:dw|
DLS
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is it ? :o
DLS
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ill do after this :/
thanks~
wio
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looks like you messed up
wio
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this is a really tedious problem man.
DLS
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Give me a minute.
yrelhan4
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hey wio.
|x-1|<3
|x-1|>7
x-1<3
x-1>-3
x-1>7
x-1<-7
Thats right no?
wio
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so far, that is right.
DLS
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\[\Huge||x-1|-5\bigg|-2 > 0\]
------->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.
yrelhan4
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So, x<4, x<-6 --> x<-6
x>-2, x>8 --> x>8
right? @wio
wio
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Looking at x<4 and x>-2 is enough to tell you all solutions work.
DLS
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So drawing the wavy curve graph.
|dw:1364716700507:dw|
wio
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Wait, hold on...
wio
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you've got: \[
-2<x<4\\
x<-6, \quad x>8
\]What happens at 5?
DLS
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\[\LARGE (-\infty,-5] U (8,\infty) \]
DLS
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oh ye to tera answer agaya :P
@yrelhan4 lol
yrelhan4
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^^^^
DLS
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but its wrong :/
yrelhan4
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:/
wio
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Combining these equations: \[
-2<x<4\\
x<-6, \quad x>8
\]Makes me thing it's \(x\in (-\infty,-6)\cup (-2,4)\cup (8,\infty)\)
DLS
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#genius^
DLS
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still wrong :P
wio
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Hmm, what are you using?
DLS
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\[x\in (-8,-6)\cup (-2,4)\cup (8,\infty)\]
is the answer
yrelhan4
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Put -10 in the question.. Its >0
wio
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Yeah, that answer doesn't make sense.
DLS
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i dont know how -8
DLS
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i prefer @wio 's answer though,still we might be missing something maybe
wio
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There would have to be some other thing limiting the domain.