Suppose I have:

$\displaystyle y=\frac{3}{2}(x-\frac{x^3}{3})$

How can I make $\displaystyle x$ the subject of the equation?

Actually I am trying to find the inverse of a function $\displaystyle G(x)= \frac{3}{2}(x-\frac{x^3}{3})$

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- Oct 29th 2010, 02:57 AMnoob mathematicianLetting X be the subject
Suppose I have:

$\displaystyle y=\frac{3}{2}(x-\frac{x^3}{3})$

How can I make $\displaystyle x$ the subject of the equation?

Actually I am trying to find the inverse of a function $\displaystyle G(x)= \frac{3}{2}(x-\frac{x^3}{3})$ - Oct 29th 2010, 03:24 AMcroraf
x = 3/2 (y-y^3/3)

Wolfram alpha gives this: http://www.wolframalpha.com/input/?i=inverse++(+3/2(x-x^3/3)+)&asynchronous=false&equal=Submit

How to derive this I'm not sure, but because the function is not bijection actually it does not have the inverse. To find the inverse of the bijective part i'll try to find out. - Oct 29th 2010, 05:29 AMHallsofIvy
That is, of course, a cubic equation for x. As croraf notes, it is NOT "one-to-one". In particular, $\displaystyle y= \frac{1}{2}x(3- x^2)$ which tells us, for example, that $\displaystyle y(0)= y(\sqrt{3})= y(-\sqrt{3})= 0$. We

**can't**solve for x as a single function of y since x(0) cannot equal 0, $\displaystyle \sqrt{3}$, and $\displaystyle -\sqrt{3}$ all at the same time! You could pick any one of three sections, $\displaystyle x\le -1$, $\displaystyle -1\le x\le 1$, or $\displaystyle x\ge 1$, on which y is a one-to-one function of x but even then I think you would have to use Cardano's cubic formula, The "Cubic Formula", which is very complicated. - Oct 29th 2010, 05:38 AMnoob mathematician
Well then for this case when the density is defined for $\displaystyle 0\le x \le 1$. And all cumulative distribution is non-negative and non-decreasing, there should be an inverse function right?

Is there any other way to obtain the inverse then? - Oct 29th 2010, 01:05 PMmr fantastic
- Oct 29th 2010, 01:45 PMHallsofIvy