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Differentiating Inverse Functions

  1. Oct 1, 2007 #1
    Please HELP....Differentiating Inverse Functions

    1. The problem statement, all variables and given/known data
    f(x) = x^3 + 2x - 1 when a=2


    2. The attempt at a solution

    I thought you did...
    1/(f '(f-1(x)))
    but I am not sure how to solve for x?

    0=x^3 + 2x - 1
    1=x^3 + 2x -1
    I tried factoring but that did not work either.
     
    Last edited: Oct 2, 2007
  2. jcsd
  3. Oct 1, 2007 #2
    what are you looking for?
     
  4. Oct 1, 2007 #3
    if it is f-1(a) when a = 2...

    set the first equation equal to 2, which will happen when x = 1.

    So, f-1(2) = 1.

    Now if you want f-1'(x), you have:

    = 1 / f'(f-1(x))

    so 1/ f'(1)

    find the derivative of f(X):

    3x^2 + 2

    so,
    answer = 1/(3(1) +2) = 1/5.
     
  5. Oct 1, 2007 #4
    thank you so very much!
     
  6. Oct 2, 2007 #5
    How do you know it is 1? Because thats the only number without an X term?
     
    Last edited: Oct 2, 2007
  7. Oct 2, 2007 #6
    How do you know f-1(2) = 1?
     
  8. Oct 2, 2007 #7

    EnumaElish

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    frasifrasi was not supposed to give away the answer, if that's the answer. That's not how this forum's supposed to work.

    f -1(2) is the answer to question, "at what value of x does x^3 + 2x -1 = 2"?

    frasifrasi assumed f(x) = a, which may or may not be justified. Your statement of the problem does not indicate what a is. If that assumption is right, then f -1(f(x)) = f -1(a), and by the definition of an inverse function, f -1(f(x)) = x. So x = f -1(a). You can verify that when x = 1, f(1) = 2. Therefore 1 = f -1(2).
     
  9. Oct 3, 2007 #8
    Yea I thought it was weird that he just gave the answer. But I am trying to figure out how you figured out that x^3 + 2x - 1 = 2 when x= 1?
     
  10. Oct 4, 2007 #9

    HallsofIvy

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    He solved the equation of course! Cubics can be difficult to solve so I suspect he did what I would: try some easy numbers for x and hope one works. In "real life" that is seldom true but in "made up" exercises it often is.
     
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