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Math Help - Inverse Functions of polynomials

  1. #1
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    Sep 2008
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    Inverse Functions of polynomials

    Hey I was hoping you guys could help me out.

    I have to find the inverse of

    f(x)= 3 + x^2 + tan((y x pi)/2))


    I know how to do inverses, but I cant figure out how to do them when its with polynomials. Anouther one I have that I can't figure out is:

    f(x)= (e^x)/(1+2e^x)

    and

    f(x)=x^5+2x^3+3x+1


    If anyone could help me with just a few of them, or give me an example of how to do one like this, Im sure i could figure the rest out.

    Thanks for your time and help
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  2. #2
    MHF Contributor
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    If you know how to do inverses, then you know that the first step is replacing all instances of x with y and all instances of y with x. I will run through this problem with you: y = \frac{e^x}{1+2e^x}

    First, switch x and y: x = \frac{e^y}{1 + 2e^y}

    Then multiply both sides by 1 + 2e^y: x(1 + 2e^y) = e^y

    Distribute: x + 2xe^y = e^y

    Subtract 2xe^y from both sides: x = e^y - 2xe^y

    Factor: x = e^y(1 - 2x)

    Divide both sides by 1 - 2x: \frac{x}{1-2x} = e^y

    Take the natural logarithm of both sides: \ln{\frac{x}{1-2x}} = y

    And that's it.
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  3. #3
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    Quote Originally Posted by icemanfan View Post
    If you know how to do inverses, then you know that the first step is replacing all instances of x with y and all instances of y with x. I will run through this problem with you: y = \frac{e^x}{1+2e^x}

    First, switch x and y: x = \frac{e^y}{1 + 2e^y}

    Then multiply both sides by 1 + 2e^y: x(1 + 2e^y) = e^y

    Distribute: x + 2xe^y = e^y

    Subtract 2xe^y from both sides: x = e^y - 2xe^y

    Factor: x = e^y(1 - 2x)

    Divide both sides by 1 - 2x: \frac{x}{1-2x} = e^y

    Take the natural logarithm of both sides: \ln{\frac{x}{1-2x}} = y

    And that's it.
    Thanks a lot I get it now, its pretty simple, but it wasn't before I saw how to do it lol.

    Thanks again
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  4. #4
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    Jan 2010
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    Ulsan
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    Inverse polynomial function

    Hi
    I have read your response to finding an inverse function and it looks reasonably straight forward but was wondering whether you might be able to assist me in finding an inverse function for the following;

    Y = aX + bX^c

    Clearly it is not possible to find 'X' in terms of 'Y' so how does one approach the finding of the inverse function? I also have to find the integral to the inverse function but just cannot get past the first step using the examples already provided in the forum. Any helpful tips would be greatly appreciated.

    Thanks!
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  5. #5
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    Jan 2010
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    Ulsan
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    9

    Addendum

    Hi,
    If anybody is interested in following up on the question I posed in January referring to the implicit function above, then please have a look in the 'University Calculus' section of this forum. After reviewing the comments by the administrator in terms of what constitutes "Pre-University Math Help" I felt that posting the question there was more appropriate to the level of the forum and proceeded to add a little more detail by way of explanation of my problem under the thread heading "Finding the inverse of an implicit polynomial function". A 'super user' by the handle "Shawsend" provided a most eloquent reply which I have now managed to verify against empirical data by integrating the final equation numerically using MatLab. It works out very well and I have even used the method provided by "Shawsend" to verify some explicit functions which can ordinarily be solved in the good old-fashioned analytical way.
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