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Math Help - Inverse function of a funtion

  1. #1
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    Inverse function of a funtion

    Hello, I'm trying to find the an expression to define the inverse of a function of a function (i think that's what it is anyway) h(x) = f(\frac{x}{b}) and I was wondering if my method of multiplying both sides by b and then inversing to get h^{-1}(x) = f^{-1}(x*b) is the correct one or if I've done something right, because there seems to be something a little off to that method... or at least i think so.
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  2. #2
    MHF Contributor chisigma's Avatar
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    If y=f (\frac{x}{b}), that means that \frac{x}{b}= f^{-1}(y) \rightarrow x=b\cdot f^{-1} (y). Your formula would be correct in the case in which f^{-1} (b\cdot y)= b\cdot f^{-1} (y), but in general that isnt

    Kind regards

    \chi \sigma
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  3. #3
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    i'm sorry chisigma.... but that really doesn't make a lot of sense to me
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  4. #4
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    Quote Originally Posted by mattty View Post
    i'm sorry chisigma.... but that really doesn't make a lot of sense to me
    Part of the difficulty may be that we're not quite sure what you mean...?

    Please reply with the exact text of the exercise, along with a clear listing of your thoughts and efforts so far.

    Thank you!
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  5. #5
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    Suppose f is a one-to-one function.

    Find an expression for the inverse of , where
    sorry if that wasn't clear the question really confuses me so i guess it should make sense that it confuses other people more when i word it differently.

    Ok, so what i've done from here is multiply both sides by b so that
    h(x*b) = f(x) and then instead of swapping the x and y terms like I would usually do, i swapped the f and h terms and let the expression for the inverse be h^{-1}(x) = f^{-1}(x*b)
    Hopefully that clears up the question and my thought process (wrong though it probably is) a bit to help anyone else help me.
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  6. #6
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    I think you might be needing the result illustrated in this article; namely:

    . . . . . \mbox{For }\, h(x)\, =\, f(g(x))\mbox{, we have }\, h^{-1}(x)\, =\, g^{-1}(f^{-1}(x))

    If we let g(x) = x/a, then f(g(x)) = f(x/a). We already know that f is one-to-one (and thus invertible), and obviously the linear function g = x/a is one-to-one. So the inverses are well-defined.

    So find the expression for g^(-1), and write the composition statement. I think this is all they're looking for.
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  7. #7
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    thanks a heap for that stapel i might not understand it to the best degree possible, but using the stuff around the edges of the composition of functions and inverse function rules I've worked it out to be h^{-1}=f^{-1}(\frac{x}{b}) I'll just hope that that is acceptable.
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