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Do you know how I could compute the inverse function of the following polynomial on $[0,1]$?

$f(x)=\alpha x^3-2\alpha x^2+(\alpha+1) x$

(where $\alpha$ is within $]0,3[$)

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What is the domain? How do we know if is injective? We cannot find the inverse unless we know the domain. –  smanoos Jan 4 '12 at 17:03
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Cubics have three roots. You haven't given enough information for us to be able to pick which of the three are needed. –  J. M. Jan 4 '12 at 17:16
    
The domain is $[0,1]$. $\alpha$ is smaller than 3, which ensure the bijection on $[0,1]$. –  julien Jan 4 '12 at 17:33
    
Is $\alpha$ nonnegative? –  savick01 Jan 4 '12 at 18:04
    
$\alpha$ is within $]0,3[$ –  julien Jan 4 '12 at 18:18
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2 Answers

up vote 3 down vote accepted

J.M. is right - in general there is no inverse function, because your function is not injective.

I assume that the domain is real numbers.

However, we can check when it is not injective. If it is monotonic, then it is injective (since it is a polynomial so it does not have a constant fragments unless it is a constant function - a polynomial of degree 0). And if a polynomial is monotonic, then its derivative is nonpositive or nonnegative. On the other hand if its derivative takes values of both signs, then it is not injective (it first go down from some value, then there is minimum, and then it goes up - a U-shaped function is obviously not injective; the same for first going up and then down).

So we calculate the derivative $3\alpha x^2 - 4 \alpha x + \alpha + 1$. It takes values of both signs if it has two distinct roots, so when the $\Delta$ is positive. $$\Delta = 16\alpha^2 - 4 (\alpha+1)3\alpha = 16\alpha^2 - 12\alpha^2 - 12\alpha = 4\alpha^2 - 12\alpha= 4\alpha(\alpha-3)$$ so it is positive for all $\alpha \in [-\infty,0)\cup(3,\infty]$.

So the task have sense only for $\alpha \in [0,3]$ (assuming $x$ is real, otherwise [in the case of $\mathbb{C}$] the function is not injective for $\alpha \neq 0$).

When you want to calculate $f^{-1}(y)$ you want to find $x$ such that $y=f(x)$, so you're solving the equation: $$\alpha x^3-2\alpha x^2+(\alpha+1) x - y =0.$$ There are different formulas for that on wikipedia. You can ask wolframalpha for help, too.

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Thanks! wolframalpha formula looks great. –  julien Jan 4 '12 at 18:22
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Indeed, $0 < \alpha \le 3$ ensures injectivity by computing the derivative of $f$ (which gives us a quadratic) and we obtain the condition by making sure that the discriminant of the derivative is negative.

Now to get the inverse, you know that $$ (f^{-1})' = \frac{1}{f'} $$ hence $$ (f^{-1})' = \frac{1}{3\alpha x^2 - 4\alpha x + (\alpha + 1)} $$ and since $f(0) = 0$, $f^{-1}(0) = 0$, hence $$ f^{-1}(x) = \int_0^x \frac{1}{3\alpha t^2 - 4 \alpha t + \alpha + 1} \, dt. $$ I don't feel like computing the integral though, it looks like a pain.

Hope that helps,

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thanks patrick. –  julien Jan 5 '12 at 11:27
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Patrick actually provided the most relevant answer. To evaluate the integral, find the roots of the quadratic and then use partial fraction expansion. –  user49309 Nov 13 '12 at 21:05
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