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Homework Help: Calculus question

  1. Apr 23, 2005 #1

    VietDao29

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    Hi,
    The question asks:
    [tex]f(x) = \frac{5x^{2} - 3x - 20}{x^{2} - 2x - 3}[/tex]
    Find:
    [tex]f^{(n)}(x)[/tex]
    I try to find its rule, but I fail:
    I have:
    [tex]f'(x) = \frac{-7x^{2} + 10x - 31}{(x^{2} - 2x - 3)^{2}}[/tex]
    And I have f''(x) is some kind of very very complicated number. What should I do in this kind of problem??
    Any help will be appreciated,
    Viet Dao,
     
    Last edited: Apr 23, 2005
  2. jcsd
  3. Apr 23, 2005 #2

    Zurtex

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    Differentiate a few times simplifying as much as you can. Here is what I got:

    [tex]f(x) = 5 + \frac{4}{x - 3} + \frac{3}{x + 1}[/tex]

    [tex]f'(x) = -\frac{4}{(x - 3)^2} - \frac{3}{(1 + x)^2}[/tex]

    I feel if I go any further I'll completely give the pattern away, carry on differentiating.
     
  4. Apr 23, 2005 #3

    VietDao29

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    So the answer is:
    [tex]f^{(n)} = \frac{(-1)^{n}4(n!)}{(x - 3)^{n + 1}} + \frac{(-1)^{n}3(n!)}{(x + 1)^{n + 1}}[/tex]
    Am I correct??
     
  5. Apr 23, 2005 #4

    dextercioby

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    U can use induction method to prove your formula right...


    Daniel.
     
  6. Apr 23, 2005 #5

    VietDao29

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    So...
    [tex]f^{1}(x) = -\frac{4}{(x - 3)^{2}} - \frac{3}{(1 + x)^2}[/tex]
    So the formula is correct if n = 1.
    Assume it's correct for n = k:
    [tex]f^{(k)}(x) = \frac{(-1)^{k}4(k!)}{(x - 3)^{k + 1}} + \frac{(-1)^{k}3(k!)}{(x + 1)^{k + 1}}[/tex]
    Prove it's correct if n = k + 1:
    [tex]f^{(k + 1)}(x) = 4(k!)(-1)^{k}\frac{-(k + 1)(x - 3)^{k}}{(x - 3)^{2k + 2}} + (-1)^{k}3(k!)\frac{-(k + 1)(x + 1)^{k}}{(x + 1)^{2k + 2}}[/tex]
    [tex]= \frac{4(k + 1)!(-1)^{k + 1}}{(x - 3)^{k + 2}} + \frac{(-1)^{k + 1}3(k + 1)!}{(x + 1)^{k + 2}}[/tex]
    So the formula is true [itex]\forall n \in N*[/itex]
    Am I correct?
    Viet Dao,
     
    Last edited: Apr 23, 2005
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