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Math Help - How to find inverse of polynomial equation?

  1. #1
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    Exclamation How to find inverse of polynomial equation?

    Problem:
    f(x) = x^3 + x + 1

    Question:
    What is f^-1(x)? What is f^-1(0.5)?


    How would we change this equation into it's inverse? Do we solve for x such as this?

    y = x^3 + x + 1
    y - 1 = x(x^2 + 1)
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by florx View Post
    Problem:
    f(x) = x^3 + x + 1

    Question:
    What is f^-1(x)? What is f^-1(0.5)?


    How would we change this equation into it's inverse? Do we solve for x such as this?

    y = x^3 + x + 1
    y - 1 = x(x^2 + 1)
    swap x and y in your function and then convert the equation in terms of y

    that is,

    write  y = x^3+x+1 as

     x = y^3+y+1 .
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  3. #3
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    Quote Originally Posted by harish21 View Post
    swap x and y in your function and then convert the equation in terms of y
    Of course, but how with a cubic? complete the cube?
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by pickslides View Post
    Of course, but how with a cubic? complete the cube?
    Cubic function! now thats troublesome..
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  5. #5
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    Well the answer they got in the back is -0.424 for the inverse of f(x).

    So would we have to factor out that x^3 + x?
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  6. #6
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by florx View Post
    Well the answer they got in the back is -0.424 for the inverse of f(x).

    So would we have to factor out that x^3 + x?
    Check if your question really is y = x^3+x+1. The inverse of this cubic function would look like this:

    inverse
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  7. #7
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    Holy smokes that is a lot of work. Anyways I am pretty sure that is what they are asking for.

    Here is a scan of the question.

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    Quote Originally Posted by florx View Post
    Holy smokes that is a lot of work. Anyways I am pretty sure that is what they are asking for.

    Here is a scan of the question.

    I don't see "What is f^-1(x)?" asked in this attachment. Next time give the actual question, not your (incorrect) interpretation of the question.

    Let f^{-1}(1/2) = a.

    Since (1/2, a) is a point on the graph of y = f^{-1}(x) it follows that (a, 1/2) is a point on y = f(x).

    Therefore \frac{1}{2} = a^3 + a + 1 \Rightarrow a^3 + a + \frac{1}{2} = 0 and you're expected to use technology to solve this equation.
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  9. #9
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    Sorry but I am not familiar with solving cubic polynomials. How would we solve a^3 + a + 0.5 = 0? Do we move the 0.5 to the right side and factor out the a?

    Such as a(a^2 + 1) = -0.5?
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  10. #10
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    Quote Originally Posted by florx View Post
    Sorry but I am not familiar with solving cubic polynomials. How would we solve a^3 + a + 0.5 = 0? Do we move the 0.5 to the right side and factor out the a?

    Such as a(a^2 + 1) = -0.5?
    *Ahem*:

    Quote Originally Posted by mr fantastic View Post
    [snip]
    and you're expected to use technology to solve this equation.
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  11. #11
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    I know that but my teacher is strict about showing work on homework.
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  12. #12
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    Quote Originally Posted by florx View Post
    I know that but my teacher is strict about showing work on homework.
    Then you should ask your teacher how s/he intends you to solve the cubic equation that arises from this question ....
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