Inverse function of y = -2x^3 + 3x^2

Question

I want to get the inverse of aforementioned function. I know it is not one-to-one normally, but my $x$ is restricted to $[0,1]$ where we have no problem. I didn't really know how to inverse such a complicated function so I tried to get an answer online.

I visited 2 different websites, wolframalpha and emathhelp. My problem is, these 2 "calculators" give me a slightly different result and I can't find out if it is the same or not, so I don't know what the correct inverse is.

I post 2 pictures with the 2 results. The first is from wolfram and the second for emath.

I've noticed that if I insert the $-1$ in the cubic root by writing it as $\sqrt[3]{-1}$ the signs of $2x$ and $1$ change accordingly, but the sign of the square root is staying the same.

Perhaps I somehow transform the $\sqrt[3]{-1}$ into a complex number involving $i$ and that affects the sign of the square root ? Not sure.

Any help would be greatly appreciated.

EDIT: My question seems to be the same with another and thank you for pointing it out to me. However people in the comments pointed out that for $ x$ in $[0,1]$ the term inside the square root becomes negative. Is that a problem? Can I accept this solution as the inverse function in the restricted domain $[0,1]$? I would still like a more detailed answer about the actual inverse of the function if this isn't the correct one, if of course someone can offer it. Thanks a lot .

Answer 1

If by [0,1] you mean the numbers 1 and 0 then the square root is of no interest to you and can be left out, in both cases its result is 0, its just that wolfram and emathhelp tried to solve for a general inverse.

Answer 2

In fact, it is a bit more complex. You face a cubic equation in $x$ $$2x^3-3x^2+y=0$$

Solving for $x$, you should get for the branch you consider $$x=\frac 12+\frac{\left(1+i \sqrt{3}\right)}{4} \sqrt[3]{2 \sqrt{y^2-y}+2 y-1}+\frac{(1-i \sqrt{3})}{4 \sqrt[3]{2 \sqrt{y^2-y}+2 y-1}}$$ which is real.

If you use the trigonometric method, you will see (after simplifications) that $$x=\frac{1}{2}-\sin \left(\frac{1}{3} \sin ^{-1}(1-2 y)\right)$$