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Math Help - find the inverse function

  1. #1
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    find the inverse function

    The question asks me to find the inverse of a function and find inverse of (3) and f(inverse of 2)

    f(t)=t^5 + t^3 + t
    I tried going by the steps, but it seemed weird.
    y=t^5 + t^3 + t
    t=y^5 + y^3 + y
    So inverse of f(t) = t^5 + t^3 + t since I can't simplify if any further?

    Then for inverse of (3), it would be 3^5 + 3^3 + 3 = 273?
    For f(inverse of 2), it would be f(2^5 + 2^3 + 2) = f(42) = 42^5 + 42^3 + 42? Woah...but that seems kind of high...

    PLEASE HELP
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: find the inverse function

    Quote Originally Posted by Taurus3 View Post
    The question asks me to find the inverse of a function and find inverse of (3) and f(inverse of 2)

    f(t)=t^5 + t^3 + t
    I tried going by the steps, but it seemed weird.
    y=t^5 + t^3 + t
    t=y^5 + y^3 + y
    So inverse of f(t) = t^5 + t^3 + t since I can't simplify if any further?

    Then for inverse of (3), it would be 3^5 + 3^3 + 3 = 273?
    For f(inverse of 2), it would be f(2^5 + 2^3 + 2) = f(42) = 42^5 + 42^3 + 42? Woah...but that seems kind of high...

    PLEASE HELP
    The problem to find the inverse of a function of the type...

    y=f(t)= c_{1}\ t + c_{2}\ t^{2}+ c_{3}\ t^{3} + ...\ ;\ c_{1}\ne 0

    ... today has been analysed here...

    http://www.mathhelpforum.com/math-he...on-188961.html

    Kind regards

    \chi \sigma
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  3. #3
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    Re: find the inverse function

    ? Ouch. I still don't get it. Must you solve this as a complex function?
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: find the inverse function

    Quote Originally Posted by Taurus3 View Post
    ? Ouch. I still don't get it. Must you solve this as a complex function?
    Real and complex functions in this case are treated in the same fashion. The only You have to know are the basic derivation rules...

    The alternative is to solve a quintic equation, for which a solving formula doesn't exist...

    Kind regards

    \chi \sigma
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  5. #5
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    Re: find the inverse function

    Quote Originally Posted by Taurus3 View Post
    The question asks me to find the inverse of a function and find inverse of (3) and f(inverse of 2)

    f(t)=t^5 + t^3 + t
    I tried going by the steps, but it seemed weird.
    y=t^5 + t^3 + t
    t=y^5 + y^3 + y
    So inverse of f(t) = t^5 + t^3 + t since I can't simplify if any further?

    Then for inverse of (3), it would be 3^5 + 3^3 + 3 = 273?
    For f(inverse of 2), it would be f(2^5 + 2^3 + 2) = f(42) = 42^5 + 42^3 + 42? Woah...but that seems kind of high...

    PLEASE HELP
    I don't think you need the inverse function per se to answer the two questions,

    for f(t) = t^5 + t^3 + t , f(1) = 3 , so f^{-1}(3) = 1


    for f(t) = 2

    0 = t^5 + t^2 + t - 2

    you may need to solve this w/ a calculator ... say you get x = a , then f^{-1}(2) = a

    edit: Since I've discovered that this post was originally in the calculus forum, I have to ask ... was this a calculus problem looking for the value of a derivative for an inverse function?
    Last edited by skeeter; September 27th 2011 at 10:56 AM.
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