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Function Homework Help

  1. Aug 17, 2005 #1
    Q) Given F(X) = 1/x and (f/g)(x) = (x+1)/(x^2-x). find the function g.

    A: (x+1)/x(x-1) = (1/x) . (x+1)/(x-1)

    Hence, the function G is (x+1)/(x-1)


    Is this answer correct?
     
  2. jcsd
  3. Aug 17, 2005 #2

    HallsofIvy

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    Almost! You have shown that [itex]\frac{x+1}{x^2- x}= \frac{x+1}{x(x-1)}= \frac{1}{x}\left(\frac{x+1}{x-1}\right)= f(x)\left(\frac{x+1}{x-1}\right)[/itex] correctly.

    But you want f/g, not f*g! How do you divide by a fraction?
     
  4. Aug 20, 2005 #3
    So that would be (1/x) / (x+1)(x-1) = (x-1) / ( x^2+x)

    Is that the correct answer?


    thanks
     
  5. Aug 20, 2005 #4

    VietDao29

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    Nope, that's still wrong. You can recheck your calculation.
    g(x) = f(x) / ((f / g)(x))
    By the way, at your first post, it reads F(x) = 1 / x. Do you really mean that or f(x) = 1 / x?
    Viet Dao,
     
  6. Aug 20, 2005 #5

    HallsofIvy

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    What I wrote before was
    [tex]\frac{x+1}{x^2- x}= \frac{x+1}{x(x-1)}= \frac{1}{x}\left(\frac{x+1}{x-1}\right)= f(x)\left(\frac{x+1}{x-1}\right)= \frac{f(x)}{g(x)}[/tex]

    So [tex]\frac{1}{g(x)}= \frac{x+1}{x-1}[/tex].

    Now do you see what g(x) is?

    Once again, how do you divide by a fraction?
     
    Last edited: Aug 20, 2005
  7. Aug 22, 2005 #6
    damn.... So does g(x) = (x-1)/(x+1) ?



    I mean, f(x)=1/x
     
    Last edited: Aug 22, 2005
  8. Aug 22, 2005 #7

    VietDao29

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    Yup. That's correct.
    Viet Dao,
     
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