Thread: Letting X be the subject

1. Letting X be the subject

Suppose I have:

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

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

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

2. 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.

3. That is, of course, a cubic equation for x. As croraf notes, it is NOT "one-to-one". In particular, $y= \frac{1}{2}x(3- x^2)$ which tells us, for example, that $y(0)= y(\sqrt{3})= y(-\sqrt{3})= 0$. We can't solve for x as a single function of y since x(0) cannot equal 0, $\sqrt{3}$, and $-\sqrt{3}$ all at the same time! You could pick any one of three sections, $x\le -1$, $-1\le x\le 1$, or $x\ge 1$, 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.

4. Well then for this case when the density is defined for $0\le x \le 1$. 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?

5. Originally Posted by noob mathematician
Well then for this case when the density is defined for $0\le x \le 1$. 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?
Yes there is. And it is not easy. I doubt you would be expected to apply it. Please post the original question - exact wording - not your version of it.

6. Originally Posted by noob mathematician
Well then for this case when the density is defined for $0\le x \le 1$. 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?
This is the first time you have mentioned "density" or "cumulative function".
I agree with mr fantastic- please tell us what the probem really is!