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Inverse function

  1. Feb 8, 2005 #1
    I have no idea how to do this question, can anyone provide some help?

    Suppose f is differentiable with derivative f '(x) = (1+x^3) ^(1/2).
    If g = f^-1, show that g ''(x) = 3/2 g(x)^2.
     
  2. jcsd
  3. Feb 8, 2005 #2
    These sort of problems are fun, inverse calculus.

    Write f(x) = int[0,x, (1+t^3) ^(1/2),dt]

    Set the integral equal to u so that you have:

    f(x) = u

    Take the inverse of both sides:

    x = g(u)

    Differentiate wrt x, using the chain rule:

    (dg/du)(du/dx) = 1

    Using FTC, du/dx = (1+x^3) ^(1/2)

    Then dg/du = (1+x^3) ^(-1/2)

    but remember x =g(u)

    So g' = (1+g^3) ^(-1/2)

    differentiate again (remember to use the chain rule):

    g'' = (-1/2)(3g)(g')(1+g^3) ^(-3/2)

    g'' = (-1/2)(3g)(1+g^3) ^(-1/2))(1+g^3) ^(-3/2)

    Which simplifies to your expression. I am sorry to have deprived you of solving this wonderful problem, but it is pretty difficult to get started on a problem like this if you do not know where to begin.

    I invented this method to to derive an expression for the integral int[0,x, 1/t, dt] .

    :smile:
     
  4. Feb 8, 2005 #3
    wow, thank you so much, you are amazing
     
  5. Feb 8, 2005 #4

    mathwonk

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    if f '(x) = dy/dx = (1+x^3) ^(1/2). where x = g(y).
    If g = f^-1, show that g ''(x) = 3/2 g(x)^2. \\

    this seems incorrect, as shown by the previous post, which also seems however slightly incorrect.

    i.e. dx/dy = 1/(dy/dx). so dx/dy = (1+x^3)^(-1/2). so d^2x/dy^2, by chain rule,

    = (-1/2)(1+x^3)^(-3/2) (3x^2) dx/dy = (-1/2)(1+x^3)^(-3/2) (3x^2)(1+x^3)^(-1/2)

    = (-1/2) (1+x^3)^(-2) (3x^2)

    = (-3/2)[1+g(y)^3)^(-2)] (g(y)^2). (the previous post omitted the power 2.)

    in particular all the integration in the previous post is entirely superfluous.

    now i may easily be missing something here, but my problem is that now I do not see how this simplifies to (3/2) [g(y)]^2, since it does not appear to me that

    [1+g(y)^3)^(-2)] = -1, i.e. that


    1 + x^3 = -1, i.e.that x = cuberoot(-2), for all x.

    what am i missing?
     
  6. Feb 9, 2005 #5

    matt grime

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    You are correct, Mathwonk, as far as I'm concerned: I independently reached exactly the same answer as you.
     
  7. Feb 9, 2005 #6

    mathwonk

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    thanks, Matt; I guess you are kindly giving me credit for what I meant to write, even if one of my braces is misplaced in the last line.

    I probably meant [1+g(y)^3)]^(-2) = -1, so the left side is a square while the right side is negative.

    the point is the problem was probably miscopied from the source.
     
  8. Feb 9, 2005 #7

    matt grime

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    we had the same issue, and it roughly stated all x satisfied some polynomial relation, i didn't check the details - perhaps i should stop using the word exactly when i don't mean it...
     
  9. Feb 9, 2005 #8

    mathwonk

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    I assumed you quit reading my details as soon as you saw we raised the same point and found the same result.
     
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