# If $f(x)$ and $f(x)-x$ have only one real root, then $f(f(x))-x$ has only one real root.

First edition was: Let $f(x)$ be a polynomial such that $f(x)$ and $f(x)-x$ have only one real root. How to prove, without derivatives, that $f(f(x))-x$ also has only one real root?

Second edition: Let $f(x)=a_3 x^3 + a_2 x^2 + a_1 x + a_0$. All $a_i \neq 0$. $f(x)$ and $f(x)-x$ have only one real root. How to prove, without derivatives, that $f(f(x))-x$ also has only one real root?

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It isn't true. Let $f(x)=-x$. –  Jonas Meyer Jan 13 '12 at 6:40
I suspect that the statement is true if we revise the question to require $f(f(x)) - x$ not be the zero function, but I am unable to prove it. –  Austin Mohr Jan 13 '12 at 7:06
One thing that is true is that if $f(x)$ is nondecreasing, then all real roots of $f(f(x)) - x$ are roots of $f(x)-x$. –  Robert Israel Jan 13 '12 at 7:39
Yes! it was my first idea, but i cant proof this. Can you write more detailed proof of this: "if f(x) is nondecreasing, then all real roots of f(f(x))−x are roots of f(x)−x" ? –  Mike Jan 13 '12 at 8:00
If $f(f(r)) = r$ but $f(r) \ne r$, let $f(r) = s$, so $f(s) = r$. Either $r < s$ and $f(r) > f(s)$ or $s < r$ and $f(s) > f(r)$. –  Robert Israel Jan 13 '12 at 15:54

More generally, $f(x)=-x^{2n-1}$ gives a counterexample for each positive integer $n$.

A counterexample for the second edition: $f(x)=-x^3-x^2-x-1$.

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Thanks its really help me. –  Mike Jan 13 '12 at 8:01

As Jonas Meyer has pointed out, the problem is actually not true.

As suggested, let $f(x)= -x$. Then $f(x) -x = -x -x = -2x$. Considering $f(f(x)) -x= f(-x) -x = -(-x) -x = 0$. That is to say, for $f(x)=-x$, we have shown $f(x) -x$ has only one real root, namely $x=0$, while $f(f(x))-x=0$ has infinitely many real roots.

Response to comments: Let $f(x)=x^3 - 2 x^2 + 5 x - 1$. Then, $$f(x) -x = x^3 -2x^2 +5x -1 =x^3 -2x^2 +4x -1$$ Solving for $x$ (by Wolfram Alpha) we have that $x \approx 0.284775$ and that there is only one real root. Considering $f(f(x))-x$ we have the following, $$f(f(x))-x = (x^3 -2x^2 +4x -1)^3 - 2(x^3 -2x^2 +4x -1)^2 +4(x^3 -2x^2 +4x -1) -1$$ which can be "simplifed" to, $$f(f(x))-x = x^9-6 x^8+24 x^7-61 x^6+116 x^5-156 x^4+155 x^3-102 x^2+44 x-8$$ Which has only one real root $x \approx 0.379555$. So, you have shown that there does exist a polynomial, namely $f(x) = x^3 -2x^2 +5x -1$ for which it is true. But this does not show in general that your conjecture is true, since there exists at least one counterexample.

@Jonas: I can't think of another counterexample off the top of my head, I don't want to try to hard at thinking about this.

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I have no idea what I was thinking, after thinking about it for more than 2 seconds it is clear that the intermediate value theorem would not help at all to show that there is only one real root. I was absently thinking that you were trying to show there is at least one real root. –  Samuel Reid Jan 13 '12 at 6:53
Ok. I see. But can you help for example with f(x)=x^3 - 2 x^2 + 5 x - 1 –  Mike Jan 13 '12 at 6:56
@user22867: If your question is actually about a particular polynomial, will you please edit that into your question? –  Jonas Meyer Jan 13 '12 at 6:59
@user22867: If you already know that the degree is odd (or if it is an assumption included in the problem), that information would also be helpful in the problem. –  Jonas Meyer Jan 13 '12 at 7:06
@user22867: If you have any additional information about the problem I suggest you revise your question as I can edit my answer to accommodate. Please always post a full question with all of the information you know when you ask a question. –  Samuel Reid Jan 13 '12 at 7:09
Here's a more general way of making counterexamples. Choose any real numbers $a < b$. Let $g(x)$ be any polynomial of odd degree whose leading coefficient is positive. Take $f(x) = a + b - x - \varepsilon (x-a)(x-b) g(x)$, choosing $\varepsilon \ge 0$ small enough that $f$ is decreasing, and thus $f(x)$ and $f(x) - x$ each have only one zero. But $f(a) = b$ and $f(b)=a$, so $a$ and $b$ are zeros of $f(f(x))-x$.