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DLS

  • 3 years ago

Domain of..sqrt{||x-1|-5|-2} ?

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  1. DLS
    • 3 years ago
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    @wio

  2. wio
    • 3 years ago
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    \[ \bigg||x-1|-5\bigg|-2 > 0 \]

  3. DLS
    • 3 years ago
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    Correct,Next?

  4. mathslover
    • 3 years ago
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    \[\sqrt{||x-1|-5|-2}\] better...

  5. wio
    • 3 years ago
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    Add 2 to both sides. Where areyou getting stuck?

  6. DLS
    • 3 years ago
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    I don't know how to deal with mod functions while solving for domain.

  7. wio
    • 3 years ago
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    where is the mod though?

  8. DLS
    • 3 years ago
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    mod/absolute bars

  9. DLS
    • 3 years ago
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    \[\Huge \sqrt{||x-1|-5|-2}\]

  10. wio
    • 3 years ago
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    so when you say 'mod' you mean the absolute value function?

  11. DLS
    • 3 years ago
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    Yes

  12. wio
    • 3 years ago
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    \[ |f(x)| < a\implies -a<f(x)<a\\ |f(x)| > a\implies a<f(x),\quad f(x)<-a\\ \]

  13. wio
    • 3 years ago
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    \[ \bigg||x-1|-5\bigg| >2 \implies |x-1|-5 < -2\quad\quad |x-1|-5 >2 \]

  14. wio
    • 3 years ago
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    The equations will keep splitting up and you'll have to keep track of all of them.

  15. DLS
    • 3 years ago
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    okay,ill try and let you know!

  16. DLS
    • 3 years ago
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    |x-1|<3 |x-1|>7

  17. DLS
    • 3 years ago
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    x-1<3 x-1<-3 x-1>7 x-1>-7 ?

  18. DLS
    • 3 years ago
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    x<4 x<-2 x>8 x>-6 ?

  19. yrelhan4
    • 3 years ago
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    x-1 > -3 *

  20. yrelhan4
    • 3 years ago
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    samajh aaya kyun?

  21. DLS
    • 3 years ago
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    hmm..yeah kinda

  22. yrelhan4
    • 3 years ago
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    x-1< -7 ** okay?

  23. DLS
    • 3 years ago
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    |x-1|<3 |x-1|>7 x-1<3 x-1>-3 x-1>7 x-1<-7

  24. DLS
    • 3 years ago
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    x<4 x>-2 x>8 x<-6

  25. yrelhan4
    • 3 years ago
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    yup. thats right i guess.

  26. DLS
    • 3 years ago
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    |dw:1364716179260:dw|

  27. DLS
    • 3 years ago
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    is it ? :o

  28. DLS
    • 3 years ago
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    ill do after this :/ thanks~

  29. wio
    • 3 years ago
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    looks like you messed up

  30. wio
    • 3 years ago
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    this is a really tedious problem man.

  31. DLS
    • 3 years ago
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    Give me a minute.

  32. yrelhan4
    • 3 years ago
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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?

  33. wio
    • 3 years ago
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    so far, that is right.

  34. DLS
    • 3 years ago
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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.

  35. yrelhan4
    • 3 years ago
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    So, x<4, x<-6 --> x<-6 x>-2, x>8 --> x>8 right? @wio

  36. wio
    • 3 years ago
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    Looking at x<4 and x>-2 is enough to tell you all solutions work.

  37. DLS
    • 3 years ago
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    So drawing the wavy curve graph. |dw:1364716700507:dw|

  38. wio
    • 3 years ago
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    Wait, hold on...

  39. wio
    • 3 years ago
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    you've got: \[ -2<x<4\\ x<-6, \quad x>8 \]What happens at 5?

  40. DLS
    • 3 years ago
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    \[\LARGE (-\infty,-5] U (8,\infty) \]

  41. DLS
    • 3 years ago
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    oh ye to tera answer agaya :P @yrelhan4 lol

  42. yrelhan4
    • 3 years ago
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    ^^^^

  43. DLS
    • 3 years ago
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    but its wrong :/

  44. yrelhan4
    • 3 years ago
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    :/

  45. wio
    • 3 years ago
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    Combining these equations: \[ -2<x<4\\ x<-6, \quad x>8 \]Makes me thing it's \(x\in (-\infty,-6)\cup (-2,4)\cup (8,\infty)\)

  46. DLS
    • 3 years ago
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    #genius^

  47. DLS
    • 3 years ago
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    still wrong :P

  48. wio
    • 3 years ago
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    Hmm, what are you using?

  49. DLS
    • 3 years ago
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    \[x\in (-8,-6)\cup (-2,4)\cup (8,\infty)\] is the answer

  50. yrelhan4
    • 3 years ago
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    Put -10 in the question.. Its >0

  51. wio
    • 3 years ago
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    Yeah, that answer doesn't make sense.

  52. DLS
    • 3 years ago
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    i dont know how -8

  53. DLS
    • 3 years ago
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    i prefer @wio 's answer though,still we might be missing something maybe

  54. wio
    • 3 years ago
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    There would have to be some other thing limiting the domain.

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