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I have a function $f(x) = y = x^2(3-2x)$, $0\le x\le 1$, I would like to find the inverse of this function on this restricted domain. ,This seems to be the answer for one of the problems that I have been scratching my head for a day.

I used online calculators, but the problem is I cannot evaluate the inverse in this domain as part of the function is $\sqrt{(y-1)y}$ and y ranges in $0<y<1$. Could someone suggest an inverse of this function restricted to (0,1). Any help will be appreciated.

Michael Hardy
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Satish Ramanathan
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2 Answers2

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Firstly, you should solved $x$ in this equation $$3x^2-2x^3=y.$$

Secondly, you would be to determine the Domain of $f^{-1}$, i.e. the range of $f(x), x\in [0,1]$.

Since $f'(x)=(3x^2-2x^3)'=6x(1-x)\le 0$, $f(1) \ge f(x) \ge f(0), i.e., 1\ge f(x) \ge 0$.

So the domain of $f^{-1}$ is $[0,1]$

Paul
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  • The problem as I had stated in the problem the range for the inverse is (0,1) and when I solve the cubic equation in terms of y, you get sqrt of a negative number and becomes complex as the inverse is found on a wider range of x. Is there a special function that would give x in terms of y where I don't stumble upon the problem that I discussed above. – Satish Ramanathan Oct 29 '14 at 03:16
  • @satishramanathan: it is difficult to give $x$ in terms of $y$ – Paul Oct 29 '14 at 03:24
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According to Mathematica, the solution for $x$ in terms of $y$ is $$x=\frac{1}{4} \left(1+i \sqrt{3}\right) \sqrt[3]{2 \sqrt{y^2-y}+2 y-1}+\frac{1-i \sqrt{3}}{4 \sqrt[3]{2 \sqrt{y^2-y}+2 y-1}}+\frac{1}{2}.$$ Perhaps, more complicated than you expected, particularly since it involves the imaginary unit $i$. However, it is a fundamental fact that cubic equations often require complex numbers to solve completely - even when the solutions are known to be real. This is known as the Casus Irreducibilis and is fully described on its Wikipedia page. That same page indicates how the equation can be solved by hand.

As far as checking whether this is really the "inverse", you can always look at a plot, as inverse functions have graphs that are mirror images of one another about the line $y=x$. I entered the following into WolframAlpha:

plot (3y^2-2y^3, 1/2+(1-I*Sqrt[3])/(4*(-1+2*y+2*Sqrt[-y+y^2])^(1/3))+((1+I*Sqrt[3])*(-1+2*y+2*Sqrt[-y+y^2])^(1/3))/4), y=0..1

to get this:

enter image description here

Mark McClure
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