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One-to-One Intervals

  1. Dec 12, 2009 #1
    1. The problem statement, all variables and given/known data
    I'm trying to do problem #5 on this worksheet
    http://ugnotesonline.com/attachments/008_t1.5pst.pdf [Broken]

    2. Relevant equations

    3. The attempt at a solution
    (a) I tried to take the derivitive of the function in the intergral so I could see where it was increasing and decreasing and prove that it was one-to-one but that didn't work.

    (b) I don't even know what this part is trying to ask.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Dec 12, 2009 #2
    Try again and show us your work because that is as far as I can see the easiest approach. Remember that if f is everywhere differentiable and the derivative is strictly positive, then the function is strictly increasing.

    For b, the notation [itex]f^{-1}(x)[/itex] denotes the inverse function. That is the function such that
    [tex]f^{-1}(f(x)) = f(f^{-1}(x)) = x[/tex]
    for all x. (its existence is guaranteed by part a). To solve this part use the chain rule as follows:
    [tex]\frac{d(f \circ f^{-1})(x)}{df^{-1}} \times \frac{df^{-1}(x)}{dx} = \frac{d(f^{-1} \circ f)(x)}{dx}[/tex]
    You should be able to evaluate everything here except [tex]\frac{df^{-1}(x)}{dx}[/tex] so you can solve the problem.

    [tex]\frac{d(f \circ f^{-1})(x)}{df^{-1}} = \frac{df(x)}{dx}[/tex]

    Also in case you didn't know [tex]g|_{x=0}[/tex] just means g(0), so in b you're just asked to evaluate the derivate at 0.
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