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