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How to show that If $y= f(x)$ satisfying $y^3 -3y + x = 0$ then $f$ is an odd function for $x \in(-2,2)$

NOTE: Given f(x) is continuous and $f(0)=0.$

My Approach: I have been trying to factorise y get an equation of the sort $y=f(x)$ and then do it however I am not able to succeed in doing so. How should I got about to check the same? Desmos does show me a neat graph for the same but I am not able to do it mathematically on paper.

Thomas Andrews
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marks_404
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    The phrasing is strange. The equation is an odd function?!? Could you perhaps try to clarify exactly what you mean? – Hans Lundmark Jun 03 '21 at 19:56
  • @HansLundmark Ok so I want to be able to know whether f(1) = f(-1) or f(1) = -f(1) or if f(1) is not equal to f(-1) whatsoever given the relation between f(x) and x. I added the odd function thing but that might be the wrong use of the term by me. The implicit equation of the function has been given – marks_404 Jun 03 '21 at 20:04
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    @ThomasAndrews it is mentioned that y=f(x) is an implicit function which satisfies f(0)=0 and the given equation – marks_404 Jun 03 '21 at 20:05
  • @ThomasAndrews Right now I have an implicit equation of the function. I was wondering if somehow I could write y interms or x directly then I might be able to know whether f(1)=-f(-1) but perhaps that might not be the way – marks_404 Jun 03 '21 at 20:07
  • Never mind, didn’t notice the $=0$ in the question. – Thomas Andrews Jun 03 '21 at 20:09
  • @ThomasAndrews sure I will. However when I plot the equation $y^{3} - 3y + x = 0$ desmos shows me that y at x=1 is negative of y at x=-1 and have the same value. This is what I meant to be able to prove basically. Probably used the wrong terminologies – marks_404 Jun 03 '21 at 20:11
  • The information that $f(0)=0$ should be included in the question! And that $f$ is continuous, otherwise it's false, since the equation doesn't define $y$ uniquely as a function of $x$. But there is a branch of the curve that is singled out by those conditions, as you can easily see if you just plot the curve (you can plot $x$ as a function of $y$ easily). – Hans Lundmark Jun 03 '21 at 20:11
  • OK, so you've already plotted it (your last comment appeared as I was typing). So can't you just argue using the fact that $x$ is an odd function of $y$? – Hans Lundmark Jun 03 '21 at 20:13
  • I'm fairly sure you simply want to check that $G(-x,-y)=0$ whenever $G(x,y)=0$. Here $G(x,y)=y^3-3y+x$ is the two variable function that you want to use to define $y$ as an implicit function of $x$. But checking that is easy. Mind you, you won't get a unique $y$, when $x\in(-2,2)$. That is, unless you also constrain $y$ to the interval $[-1,1]$. – Jyrki Lahtonen Jun 03 '21 at 20:18
  • @HansLundmark yeah I understand what you are saying that makes sense! Never did it the other way round. – marks_404 Jun 03 '21 at 20:18
  • One thing I had a doubt on though. Why does plotting $y^3 - 3y + x$ give me a unique curve. Isnt it the same thing aswell. But according you all there would be multiple such functions that should exist. – marks_404 Jun 03 '21 at 20:21
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    The equation gives you three functions $y_1(x)\in[-2,-1]$, $y_2(x)\in[-1,1]$ and $y_3(x)\in[1,2]$ when $x\in[-2,2]$. Only $y=y_2(x)$ is an odd function. We do have $y_3(-x)=-y_1(x)$ also. – Jyrki Lahtonen Jun 03 '21 at 20:29

1 Answers1

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Let $g(u)=3u-u^3.$ Then show $g(u)$ is strictly on on $u\in[-1,1]$ and decreasing outside that intervals. Since $g(-1)=2, g(1)=2.$ This means that for each $x\in(-2,2)$ there is a unique $y\in (-1,1)$ so that $g(y)+x=0.$

We want a continuous function $f$ such that $g(f(x))=x$ and $f(0)=0.$

Show that continuity of $f$ and $f(0)=0$ and $g(f(x))=x$ means that $f(x)\in(-1,1)$ for $x\in(-2,2).$ (1)

Then show if $g(y)=x$ then $g(-y)+(-x)=0,$ and, since $y\in(-2,2),$ $-y\in (-2,2).$

The only hard part is $(1).$ If $f(x_0)=y<-1$ or $f(x_0)>1$ then the function $g(u)=u^2-3u$ takes duplicate values for some $u_1,u_2)$ between $f(0)=0$ and $f(x_0),$ and since the values must be in the image of $f,$ we must have $u_i=f(x_i)$ for some $x_1,x_2,$ by the intermediate value theorem. But if $g(f(x_1))=g(f(x_2)))$ then $$x_1=g(f(x_1))=-g(f(x_2))=x_2!.$$

So it is not possible for $g$ to take duplicate values in the range of $f.$

The more general result: If $g$ is continuous and, for some $a<b<c<d,$ $g$ strictly decreasing on $[a,b]$ and $[c,d]$ and strictly increasing on $[b,c]$ and there is a continuous $f$ with $g(f(x))=x,$ and $f(x_0)\in(b,c).$ then $f(x)\in [b,c]$ for all $x.$ This is also true, of course, if we swap “increasing” and “decreasing.”

We can make the domain of $f$ be $[g(b),g(c)]$ by defining:

$$f(x)=u\text{ where } u\in [b,c], f(u)=x$$

We know that $f$ is well-defined because $f$ is continuous and strictly increasing, so there must be a $u$ and it must be unique.

This must be our $f,$ because the intermediate value theorem and the fact that if $f$ has a range including $f(x_0)$ means, if the range had some point outside $[a,b],$ there would be a solution to $g(u_1)=g(u_2)=x_2$ for distinct values $u_i$ in the range of $f,$ which is not possible.

Thomas Andrews
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