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Inverse function + integration help

  1. Jul 17, 2008 #1
    1. The problem statement, all variables and given/known data

    find (f-1)'(0) if f'(x) = root(1+x^4)

    3. The attempt at a solution

    i know (f-1)'(0) = 1 / f'((f-1 (0))

    but to find the value of the inverse at 0, i need to find the inverse, for which i need to find the original function by integrating, and i cannot seem to be able to integrate root (1+x^4)

    i mean..if I set u to 1+x^4, i get du = 4x^3 dx
    but there is nothing other than whats under the root sign in the integral so it doesn't work out , so please guide me :)

    and another question
    f(x) = (ax + b) / (cx + d)
    Determine a, b , c, d for which f = f-1

    so i foudn the inverse to be f(x) = (dx - b) / (a - cx)

    so i set the two functions equal to each other

    (ax + b) / ( cx + d) = (dx - b) / (a - cx)
    I have no idea how to figure out the constants now though...After some algebra i get
    cx^2 (d+a)+ x(d+a)(d-a) - b(d+a) = 0
    so assuming d+ a is not equal to zero, i can simplify to
    cx2 +x(d-a)-b = 0
    but how the hell do i figre out the constants from this point?
  2. jcsd
  3. Jul 17, 2008 #2
    There is a rule that says
    derivative of inverse = 1/derv of the function

    y = x^2
    dy/dx = 2x

    x = sqrt(y)
    dx/dy= -1/(2*sqrt(y))

    oo nvm .. perhaps, it was something else.

    but, you can develop the right thing yourself:

    f(f^-1) = x
    differentiate and use chain rule. Hopefully that would give you an answer

    for b) I think that is much more like common sense question. How you draw an inverse function?!

    Have a graph paper and draw first a function and then its inverse, and then that would answer your question
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