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Linearization of a function

  1. Oct 28, 2013 #1

    Qube

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    1. The problem statement, all variables and given/known data

    http://i3.minus.com/jbt2vueBfwXvWD.jpg [Broken]

    2. Relevant equations

    Linearization: f(x) + f'(x)(dx)

    3. The attempt at a solution

    The derivative of g(x) using the chain rule is (2lnx)/x. x = e, so that simplifies to 2/e.

    Linearization:

    (2/e)x + 1, where 1 is f(e).

    Why is the answer (2/e)x - 1 instead? I thought f(e) is 1, not negative 1?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 28, 2013 #2

    tiny-tim

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    Hi Qube! :smile:

    That works fine for x = 0 … not so much for x = e ! :wink:

    (2/e)x + 1 at x = e is 3

    (2/e)x - 1 at x = e is 1 …

    which do you prefer? o:)

    (2/e)(x - e) + 1 would be ok :smile:
     
  4. Oct 28, 2013 #3

    eumyang

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    I was told that the linearization is defined this way:
    [itex]L(x) = f(a) + f'(a)(x - a)[/itex],
    where f is differentiable at a.
     
  5. Oct 28, 2013 #4

    Qube

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    Yes, I see how working backwards helps, but how can I work to the solution instead of vice versa?
     
  6. Oct 28, 2013 #5

    Qube

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    Yes, and I seem to get 1 + (2/e)x
     
  7. Oct 28, 2013 #6

    tiny-tim

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    f(a) = 1, f'(a) = 2/e, (x - a) = (x - e)

    total: 1 + 2/e (x - e) = (2/e)x - 1 :wink:
     
  8. Oct 28, 2013 #7

    Mark44

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    Show us what you did.

    L(x) = g(x) + g'(a)(x - a)

    For L(e), what do you get? I get (2/e)x - 1.
     
  9. Oct 28, 2013 #8

    Qube

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    Ohh I see. Alright.

    a = e

    f(e) = 1

    f'(e) = 2/e

    1 + (2/e)(x-e) = 1 + (2/e)x - 2 and the 1 - 2 results in the -1. Alright. I think my problem was not sticking closely enough with the definition and instead going with a gut feel / intuition about how linearization works.
     
  10. Oct 28, 2013 #9

    tiny-tim

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    the gut is not linear! :wink:
    250px-Stomach_colon_rectum_diagram.svg.png
     
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