# Can an inverse function of a special cubic function be found?

• I
phymath7
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There's a simple method for finding an inverse function of a quadratic function.But I wonder whether an inverse function of a special cubic function can be found.
Like this one: ##f(x)=3x^3 -18x^2 +36x##.Is there any way to find the inverse of this function without using the general solution to cubic equation?

Staff Emeritus
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I would be surprised if there was for that function in particular.

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$$f(x)=3x^3 -18x^2 +36x$$
complete the cube
$$f(x)=3(x-2)^3 +24$$
There will be three inverses
For real values other than two, two preimages are complex

• Infrared, phymath7, Office_Shredder and 1 other person
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Is there any way to find the inverse of this function without using the general solution to cubic equation?
Not sure what you mean by the inverse. These polynomials don't have an inverse. If you mean to find their zeros, i.e. solve ##f(x)=0## for possible values of ##x##, then it depends on the case. Some are easy, but there is no general way to solve them other by using the formulas for degrees ##2,3,## and ##4.##

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^I think the question is if the particular example is easier than the general case. This example is easier, it is a sum of cubes. I don't know what you mean by polynomials don't have an inverse, as they clearly do. The inverse is not a function so we need to choose branches. In this example we have a real branch and two complex branches. If we are only concerned with the real branch the inverse is unique.

• FactChecker
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I don't know what you mean by polynomials don't have an inverse, as they clearly do.
No. they do not. E.g., ##f(x)=x^2-16## cannot have an inverse since two values in the domain map to one value in the rage: ##f(4)=f(-4)=0.## What would ##f^{-1}(0)## be? A parabola isn't injective, ergo not bijective, ergo no inverse.

Even if you meant the reciprocal ##1/f(x)## we get into trouble as soon as there are zeros of ##f(x)## as in my example.

The inverse is not a function ...
... in which case we cannot call it inverse ...
... so we need to choose branches. In this example we have a real branch and two complex branches. If we are only concerned with the real branch the inverse is unique.
You make it artificially injective. That is a different function then!

• dextercioby
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$$x^2-16$$ has an inverse. The complete inverse is multivalued, so you can chose branches if you like. That is like saying sine or exp don't have inverses, my pocket calculator says they do.
I would say
$$f^{-1}(0)=\{-4,4\}$$
If want a single valued inverse pick a branch.

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$$x^2-16$$ has an inverse. The complete inverse is multivalued, so you can chose branches if you like. That is like saying sine or exp don't have inverses, my pocket calculator says they do.
I would say
$$f^{-1}(0)=\{-4,4\}$$
If want a single valued inverse pick a branch.
Again, nobody calls this an inverse.

An inverse of ##f## is a function ##g## such that ##f\circ g = \operatorname{id}## and ##g\circ f = \operatorname{id}.## And if you do not define domain and range precisely to mean something else than the maximal possible, then there is no inverse. A polynomial is defined on the entire real line or on the entire complex plane, but neither has an inverse as soon as it is not linear anymore.

• dextercioby
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No. they do not.
Over the Reals, some do and some don't. @lurflurf has shown that this one does and is fairly easy to give an equation for it.

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Over the Reals, some do and some don't. @lurflurf has shown that this one does and is fairly easy to give an equation for it.
Sorry, but this is serious misinformation. A parabola does not have an inverse. It is not injective.

Where are we that some do and some don't?

Edit: In case ##f(x)=x^2-16## then ##f^{-1}(\{0\})=\{\pm 4\}## is called pre-image of zero, not inverse! Pre-image is a set-theoretical term, inverse refers to functions.

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• SammyS and dextercioby
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Sorry, but this is serious misinformation. A parabola does not have an inverse. It is not injective.
Sorry, but my statement is absolutely correct. Over the field of Reals, some polynomials have an inverse. The fact that you have examples that do not have inverses does not contradict what I said.

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• fresh_42
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Again, nobody calls this an inverse.

An inverse of ##f## is a function ##g## such that ##f\circ g = \operatorname{id}## and ##g\circ f = \operatorname{id}.## And if you do not define domain and range precisely to mean something else than the maximal possible, then there is no inverse. A polynomial is defined on the entire real line or on the entire complex plane, but neither has an inverse as soon as it is not linear anymore.
Yes they do, in particular they say
given two sets A and B the relation R is a subset of their Cartesian product
$$R\subset A\times B$$
the inverse of R is defined by
$$R^{-1}=\{(b,a)|(a,b)\in R\}$$

If you are more interested in functions
it is true that not all functions have full inverses.
It is odd to say they have no inverse when it is very useful to restrict the domain and consider partial inverses.
I would say $x^2$ has a partial inverse or that is does not have a total inverse, not that it has no inverse.

Polynomials over reals do not need to be linear to have inverses consider the given example as a counter example as it has a total inverse as a real function.

• fresh_42
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Sorry, my bad. Polynomials ##x^{2n+1}## can be inversed. Nevertheless, these are the exceptions: no local extrema, and of odd degree.

• FactChecker
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You can always do the check of the derivative to see if it's monotone, so that it's 1-1. And for odd powers, limits at ##+\infty=+\infty##, at ##-\infty=-\infty##, so they will be onto.

The really hairy question is for polys into ##\mathbb R^p ; p>1##. Deciding if they're or onto.

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Yes they do, in particular they say
given two sets A and B the relation R is a subset of their Cartesian product
$$R\subset A\times B$$
the inverse of R is defined by
$$R^{-1}=\{(b,a)|(a,b)\in R\}$$

If you are more interested in functions
it is true that not all functions have full inverses.
It is odd to say they have no inverse when it is very useful to restrict the domain and consider partial inverses.
I would say $x^2$ has a partial inverse or that is does not have a total inverse, not that it has no inverse.

Polynomials over reals do not need to be linear to have inverses consider the given example as a counter example as it has a total inverse as a real function.
Yes, ##x^3## itself has inverse ##x^{1/3}##. I guess having a monotonic derivative is sufficient. But not sure if it's necessary.

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^positive derivative suffices
I found an interesting paper
https://www.researchgate.net/publication/292674112_Fast_and_flexible_methods_for_monotone_polynomial_fitting

that states the general monotonic polynomial is

$$a+b\int_0^x \bigg(p_1(u)^2+p_2(u)^2\bigg) du$$

and that it may be proved using the theory of Tchebycheff systems, the theory of
canonical moments, or directly.

• WWGD