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Inverse of a Cubic Function

  1. Dec 4, 2007 #1
    I am working on a paper to find the equation of the inverse function of a cubic function. The function is

    f(x) = x^3 + 6x^2 +12x +7

    I have already graphed the function and its inverse. I have found the inflection point of (-2, -1). I found the 3 roots (1 real & 2 complex). The real root is x=-1, so I am left with:

    (x+1)(x^2 + 5x + 7)

    which leads to the complex roots of:

    (5+i(3)^.5)/2 and (5 - i(3)^.5)/2

    At this point though I am totally stuck. I know that I am supposed to switch the placement of the x & y variables, but I can't figure out what the next step should be. So I now have:

    x = y^3 + 6y^2 +12y +7,​

    but I have no idea what to do with it! I know from above that it can be rewritten as:

    x = (y+1)(y^2 + 5y + 7).​

    How do I get the y variable all by itself though? Looking for some direction on what I should do at this point to come up with the equation of the inverse.

    Thanks for your help in advance!
  2. jcsd
  3. Dec 4, 2007 #2


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    You can Google for Cardano's formula... but why do you want an explicit inverse?
  4. Dec 4, 2007 #3
    The Cardano's formula will give me the values for x which I already have. I am not sure how that formula will help me get an explicit inverse other than giving me the values for x.

    The person who assigned this paper asked us to "find the inverse function of the given function",. He also stated "the purpose of theis research assignment is to test your ability to apply the concept of an inverse function and other algebraic concepts to a problem."

    So I am assuming that he wants an actual equation of the inverse. We tried to ask for further details, but we were told that it was a research paper and he would not give us any help on it. Perhaps I am misreading what he is requesting? If you have any thoughts on what he may actually be asking for, it would be terrific! As mentioned, I have already graphed it, found the inflection point & the 3 roots.
    Thanks again!
    Last edited: Dec 4, 2007
  5. Dec 4, 2007 #4


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    Well, there's always Galois theory, perhaps that could give some insight.
  6. Dec 4, 2007 #5
    I will take a look at it in a few minutes, but I should have mentioned this is for an intermediate Algebra class, not a high level course. Thanks for the suggestion!
  7. Dec 4, 2007 #6
    Okay...I took a quick peek at the Galois Theory...but it appears to be way beyond the Algebra 2 level. Any other thoughts on this problem?
  8. Dec 4, 2007 #7
    Generally speaking, cubics often do NOT have nice, closed form inverses. You are trying to find [tex]f^{-1}(x)[/tex] as a nice formula in terms of x, however this form implicitly assumes that [tex]f^{-1}(x)[/tex] is a function. Generally, it is not. Here's how to see this geometrically.

    If you have the graph of a function f, you can easily obtain the graph of the "inverse" of by reflecting the graph of f across the line y=x. Draw a general cubic, and reflect its graph across y=x. You can find a cubic so that its inverse graph does not satisfy the vertical line test.
  9. Dec 4, 2007 #8

    Ben Niehoff

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    If you have

    [tex]x = y^3 + 6y^2 +12y +7[/tex]

    then write

    [tex]y^3 + 6y^2 +12y + (7-x) = 0[/tex]

    and solve using Cardano's method.
  10. Dec 4, 2007 #9
    Or you could try "completing the cube", as an analogy to completing the square? If I am not mistaken, this should yield a nice algebraic expression without too much computation?
  11. Dec 4, 2007 #10

    Ben Niehoff

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    "Completing the cube" doesn't work, unfortunately. For a quadratic, all you have to do to complete the square is add a constant. But a cubic has too many variables to do that. Try it.
  12. Dec 4, 2007 #11

    Ben Niehoff

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    But wait! This particular cubic can be completed by adding a constant! Well, then.

    To the OP: Remember that

    [tex](a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3[/tex]

    Now, write

    [tex]y^3 + 6y^2 + 12y = x - 7[/tex]

    and see what you can do.
  13. Dec 4, 2007 #12
    In general yes, but x^3 + 6x^2 + 12x + 7 = (x+2)^3-1. So one would think that the coefficients 6 and 12 were not chosen arbitrarily?
    Last edited: Dec 4, 2007
  14. Dec 4, 2007 #13
    Thank you ALL so much...you have definitely sent me in the correct direction particularly Big-T and Ben Niehoff! I will finish my paper right now! Thanks again...Nan
    Last edited: Dec 4, 2007
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