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DLS

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  • 11 months ago
  • 11 months ago

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  1. DLS
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    \[\LARGE \sqrt{|x-1|+|x-2|+|x-3|-6}\]

    • 11 months ago
  2. DLS
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    I want to do this with graphical method,would be appreciated

    • 11 months ago
  3. TareKKowshiK
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    Domain={R}

    • 11 months ago
  4. DLS
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    no where near,absolutely wrong.

    • 11 months ago
  5. DLS
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    tell me what do you get for x=3?

    • 11 months ago
  6. DLS
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    and x=1 and x=2 too

    • 11 months ago
  7. satellite73
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    http://www.wolframalpha.com/input/?i=|x-1|%2B|x-2|%2B|x-3|-6%3E0

    • 11 months ago
  8. DLS
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    yes that is the correct graph,i need some good explanation..

    • 11 months ago
  9. satellite73
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    apparently it is \((-\infty, 0]\cup [2,\infty)\)

    • 11 months ago
  10. DLS
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    no again

    • 11 months ago
  11. DLS
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    tell me what do u get for x=3 like posted above

    • 11 months ago
  12. satellite73
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    it will take a long time to do you have to solve over different intervals

    • 11 months ago
  13. DLS
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    tell me over one interval

    • 11 months ago
  14. DLS
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    for |x-1|

    • 11 months ago
  15. DLS
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    if x<0 then why is the x-1 thing reversed,first doubt?

    • 11 months ago
  16. satellite73
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    i am sorry my answer was wrong it is \((-\infty, 0]\cup [4,\infty)\)

    • 11 months ago
  17. pursultan
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    taking positive and negative value of piecewise function and solving the square root part greater than equal to 0 you get the following : \[x \geq 0 \& x \geq 4\]

    • 11 months ago
  18. satellite73
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    you have to break it up in to cases if \(x>3\) then \(|x-1|=x-1,,|x-2|=x-2, |x-3|=x-3\)

    • 11 months ago
  19. satellite73
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    so if \(x>3\) you are solving \[x-1+x-2+x-3-6\geq 0\] in which case you get \[3x-12\geq 0\] and so \(x\geq 4\)

    • 11 months ago
  20. pursultan
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    So domain of the function is \[x \geq 4\]

    • 11 months ago
  21. DLS
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    If i have |x-1| then for x<0 it becomes x-1?why?

    • 11 months ago
  22. satellite73
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    if \(x<1\) then \(|x-1|=1-x\)

    • 11 months ago
  23. DLS
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    i wrote that too but why :|

    • 11 months ago
  24. satellite73
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    because if \(x<1\) then \(x-1<0\) and so \(|x-1|=-(x-1)=1-x\)

    • 11 months ago
  25. DLS
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    but modulus function makes it positive

    • 11 months ago
  26. satellite73
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    yes, and if \(x-1\) is negative, then \(1-x\) is positive

    • 11 months ago
  27. DLS
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    why is the sign getting reversed? i thought | | will make it +ve no matter if x>0 or x<0

    • 11 months ago
  28. satellite73
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    lets try it with numbers suppose \(x=-2\) then \(|-2-1|=|-3|=3=1-(-2)\)

    • 11 months ago
  29. DLS
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    seems legit but very non intuitive for me first time :P

    • 11 months ago
  30. satellite73
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    the definition is this \[ |x| = \left\{\begin{array}{rcc} -x & \text{if} & x <0 \\ x & \text{if} & x\geq 0 \end{array} \right. \]

    • 11 months ago
  31. DLS
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    for kind of removing the mod right?

    • 11 months ago
  32. satellite73
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    and so for example \[ |x-1| = \left\{\begin{array}{rcc} 1-x & \text{if} & x <1 \\ x-1& \text{if} & x \geq 1 \end{array} \right. \]

    • 11 months ago
  33. satellite73
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    yes, it seems like absolute value should be easy right? just "make it positive" but in fact it is a very annoying piecewise function and hard to work with because of that fact

    • 11 months ago
  34. DLS
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    hmm..okay..clear enough..thanks for your time :") why do i have 3 medals anyway :P

    • 11 months ago
  35. DLS
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    and the last thing,how do i get the points on the graph? i am supposed to find dy/dx at all these intervals right? If they turn out to be +ve then increasing slope..and vice versa..but what about the points?

    • 11 months ago
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