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What reasoning error am I making here?

  1. Jun 7, 2010 #1
    I apologize for a misclick double post:

    Mentor action: double posted thread merged. No harm, no foul.
    Last edited by a moderator: Jun 7, 2010
  2. jcsd
  3. Jun 7, 2010 #2
    Apparently I cannot read. I figure this is the wrong thread. Please move or kill as needed :(

    You need to find M where

    [tex]\left|\frac{x+2}{x}-5\right| \le M, x \in (1, 4)[/tex]

    To do this, I put 4 for the top x, 1 for the bottom x, to give the greatest quotient possible, plus |-5|, which finally gives 11. The correct answer is to combine 5 into the fraction with x's, and that gives a different answer.

    What did I overlook by NOT combining them?

    You can view the entry on google books preview here:

    http://books.google.com/books?id=10... by first writing the given function"&f=false

    Gratitude for enlightenment.
    Last edited: Jun 7, 2010
  4. Jun 7, 2010 #3


    Staff: Mentor

    Look at the range of values for (x + 2)/x for x in [1, 4]. This function is defined at all points in this interval, and is decreasing on this interval. The largest value of (x + 2)/x on this interval comes when x = 1.
  5. Jun 7, 2010 #4
    This I understand. If you substitute x with 1, you can get | 3 | + | 5 | = 8 as an upper bound.

    The solution as linked actually has M = 18. What makes 18 a better upper bound than 8? Or, is the book wrong?

    Thank you
  6. Jun 7, 2010 #5
    It shouldn't be a better upper bound. I don't know how the book got 18, but it is a M that works, so it's technically correct as well.
  7. Jun 7, 2010 #6
    The book got it by merging the -5 into the fraction, giving

    [tex]\left| \frac{-4x + 2}{x} \right| [/tex]

    Then it maximizes this fraction within the allowed values for x by using 4 for the top x, 1 for the bottom x, and applying absolute value to each term since |-4x + 2| <= |4x| + |2|

    This gives you 18. What makes this a good method?

    It's really confusing, because by turning 5 into 5x/x, and plugging DIFFERENT values into the top x and bottom x, it greatly magnifies the result, apparently without good reason.
  8. Jun 7, 2010 #7
    Magnifying the result is a good enough reason, since you're looking for any upper bound. But your way is another perfectly fine way to find an upper bound.
  9. Jun 10, 2010 #8
    Makes sense, thank you.
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