Finding inverse of a difficult function 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.

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 came 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. No need for tricky inverses in the end. :0) Thanks everyone.
