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Hi I'd like to find the inverse of: $$ y=(1/3)(x^{4} + 4x^{3}) $$

I have learned to do inverses using the following example: $$y=2x-1$$ $$x=2y-1$$ $$x+1=2y$$ $$(x+1)/2=y$$ $$f^{-1}(x)=(x+1)/2$$

Process:

  • Change the x's into y's and the y's into x's
  • Rearrange the equation to get a single y (formerly x) by itself on one side
  • Replace y with $f^{-1}(x)$

But the equation I am working with seems too complicated. I can't get x by itself on one side because the terms are to the power of 2 and 4.

Can anyone suggest a way forward? Thanks.

UPDATE:

As has been pointed out by some people the inverse of the said function is actually quite complicated and it turns out I was over complicating things myself. The main thing I am working on is to show that the roots of the equation $$x^{3}+4x^{2}-3=0$$ can be found (approximately) using iteration formulae which are rearrangements of the equation. After some trial and error I cam up with the following rearrangements which between them cover the three roots. $$x_{n+1}=(3-4x^{2})^{1/2}$$ $$x_{n+1}=(1/4)(3/x - x^{2})$$ $$x_{n+1}=((1/4)(3-x^{3}))^{1/2}$$ That's all I needed to do. Thanks everyone.

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    It ain't pretty: [Wolfram Alpha](http://www.wolframalpha.com/input/?i=y=%281/3%29%28x%5E4%2b4x%5E3%29%20solve%20for%20x)2011-12-22
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    Thanks @BillCook but the solutions produced by wolframalpha are ridiculously complicated. Surely there must be a more simple answer?2011-12-22
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    @NSDigital, why surely?2011-12-22
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    The original function is so simple I just didn't expect the inverse could be so different.2011-12-22
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    But that's how it is. At least there _is_ a closed-form solution; when you move to the fifth power there usually isn't one at all. What do you need the inverse for? Perhaps there's a better way to achieve your eventual goal.2011-12-22
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    I'm using iteration to find the roots of $$x^{3}+4x^{2}-3=0$$ The equation I gave above is a rearrangement of this, and having plotted the rearrangement I think that by symetry of the graphs the inverse equation will provide a better iteration to show there is a root at x=-1.2011-12-22
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    Doesn't plugging in $x=-1$ show that it's a root?2011-12-22
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    @RileyE yes substituting x=-1 shows that it's a root but the idea is investigate the problem using iteration.2011-12-22
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    @NSDigital: Ah. Sorry about that.2011-12-22
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    No worries. I appreciate your input @RileyE.2011-12-22
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    So you have transformed a somewhat difficult problem (your concrete third degree equation) into an extremely difficult problem (inverting a fourth degree polynomial). That does not look like progress to me.2011-12-22
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    Note that inverting a polynomial is a strictly harder problem than finding a root, since you can get a root easily by plugging $0$ into the inverse.2011-12-22
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    you can invert this formula only for ether x>=-3 or x<=3, or part of this http://www.wolframalpha.com/input/?i=%7B1%2F3+%284+x%5E3%2Bx%5E4%29%7D&lk=1&a=ClashPrefs_*Math-2011-12-23

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