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Suppose we have a cubic polynomial function $f:\mathbb{R}\to\mathbb{R}$ given by $$y=f(x)=ax^3+bx^2+cx+d,$$ where $a,b,c,d\in\mathbb{R}$ and $a\neq 0$. As far as I know there are two ways to find the inverse function of $f$, namely the Cardano method (where we use the depressed cubic equation) and the one with the hyperbolic functions.

Both methods have particular conditions for which they hold. For example, in the Cardano method with $t^3+tx+q$, we need $4p^3+27q^2 > 0$ where $p$ and $q$ are given by $\frac{3ac-b^2}{3a^2}$ and $\frac{2b^3-9abc-27a^2d}{27a^3}$, respectively.

Now, what if we have the polynomial function $y=-0.06x^3+0.9x^2+3x$ where it fails to be a bijection on its domain? Can we restrict the domain and still get an inverse function? For example, we can define $g:[0,5]\to[0,30]$ with $g(x)=-0.06x^3+0.9x^2+3x$, where $g$ is a bijection on its domain $[0,5]$.

I have tried using the Cardano method and the hyperbolic functions method to find the inverse function but both fail.

But from the graph below (the red line is $g$ and the green $h$), it is obvious that $g$ should have an inverse. So, I tried the following: reflect $g$ about $y=x$ and get $h(y)=-0.06y^3+0.9y^2+3y$. Checking now $$h(g(x))=x, \quad \text{and}\quad g(h(y))=y$$ should give the values of $x$ and $y$ such that $h$ is the inverse of $g$, right?

The issue here is that I don't know if I can force $g:[0,5]\to[0,30]$ and $h:[0,30]\to[0,5]$ with $g(x)=y\iff h(y)=x$.

Any hints or help would be appreciated.

enter image description here

johnny09
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  • Short answer, yes, but what type of explicit formulation will depend on the domain. The methods you mentioned as having failed should in fact work, you just need to correctly interpret the formulas you get and place appropriate restrictions on the variables (often a very difficult task). On the other hand, if you restrict the function enough, you can get simple formulas. You can use a constant function if you restrict the domain to a single point, you can use a first-degree algebraic function if you restrict the domain to two points, etc. – Dave L. Renfro Nov 23 '19 at 18:36
  • @DaveL.Renfro the conditions depend on the coefficients, so I'm not sure how I can change those. When you say restrict the function, do you mean the domain? Again, I'm not sure I follow you because both the Cardano method and the hyperbolic method depend on the coefficients of the cubic polynomial. – johnny09 Nov 24 '19 at 02:39
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    If you want an inverse that incorporates as parameters the coefficients of the cubic, then it's really going to be messy, because you'll have to develop (or find formulas somewhere) for the turning points and such in terms of the coefficients, which will likely wind up bifurcating into many cases. Regarding "find formulas somewhere", maybe look through some of the many dozens of theory of equations books published from the early 1800s to the 1960s, of which Burnside/Panton's book is one of the most complete. – Dave L. Renfro Nov 24 '19 at 09:37
  • Not sure if relevant to you, but I found your question while looking for an answer to a similar problem and found this article, which answered at least my question: https://iquilezles.org/articles/ismoothstep/ – fweth Oct 01 '23 at 11:33

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