1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculus question

  1. Apr 23, 2005 #1

    VietDao29

    User Avatar
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

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


    Daniel.
     
  6. Apr 23, 2005 #5

    VietDao29

    User Avatar
    Homework Helper

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Calculus question
  1. Calculus question (Replies: 5)

  2. Calculus question (Replies: 4)

  3. Calculus question (Replies: 6)

  4. Calculus Question (Replies: 13)

  5. Calculus Question (Replies: 10)

Loading...