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Finding delta Algebraically

  1. Sep 14, 2009 #1
    Given f(x) = x2, L = 4, xo = -2, e = 0.5 find delta.

    -0.5 < x2 - 4 < 0.5

    3.5 < x2 < 4.5

    (3.5)1/2 - (-2) < x - (-2) < (4.5)1/2 - (-2)

    =>|x - xo| < (3.5)1/2 - (-2) ~ 3.87

    My text says the answer is 0.12 ?

    I was convinced that I have been doing these right. Am I?
     
  2. jcsd
  3. Sep 14, 2009 #2

    lanedance

    User Avatar
    Homework Helper

    you have a skipped an important step here, when you take the square root, do you take the positive or negative squareroot.... thoughts?

    ie. assuming
    a2 < x2 < b2
    you need to be careful about whether you treat a & b as positive or negative numbers


    As a test try using you delta & see what e it gives you?

    As an aside I would tray & start a little more explicit:
    so there exists d>0 such that for all |x-x0|< d, |f(x)-f(x0)|<e

    Now
    |f(x)-f(x0)|= |x^2-x0^2|= |(x+x0)(x-x0)| < e

    Any ideas how to re-write this in terms of delta?
     
    Last edited: Sep 15, 2009
  4. Sep 15, 2009 #3
    You did a mistake.

    It should be:

    √3.5 < |x| < √4.5
     
  5. Sep 15, 2009 #4
    Okay then! :smile:

    I am not really sure what you are doing here? Could you please explain?

    Thanks :)
     
  6. Sep 15, 2009 #5

    Mark44

    Staff: Mentor

    Starting with your inequality 3.5 < x2 < 4.5, it must be either that
    1) [itex]\sqrt{3.5}[/itex] < x < [itex]\sqrt{4.5}[/itex], OR
    2) [itex]-\sqrt{4.5}[/itex] < x < [itex]-\sqrt{3.5}[/itex]

    Since x0 in your problem is -2, which of the inequalities above is the one that should be used for this problem? For your limit, x should be reasonably close to x0.
     
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