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Shown is the graph of $y=f(x)$,a polynomial function of degree $10$ whose domain is restricted to $[1,5]$.Function $f$ is symmetric about $x=3$.Compute the number of solutions to the equation $f(x)=f(f(x))$.

enter image description here

My effort

Considering the cases where $f(a)=f(b)$ where $a \ge 1 $ and $b \le 5 $ ,I must have that $a=b$ or $b=6-a$.

In the first case I have that $f(x)=f(f(x))$ iff $f(x)=x$ but since $\deg f(x) =10$ I haven't solutions .

In the latter case $f(x)=f(f(x))$ iff $f(x)=6-x$ so when $6-x=f(6-x) $ which hasn't solutions since if we take $x=6-c$ we have $c=f(c)$ which is just the first case ,which hasn't solutions

My conclusion is that there are not solutions,but I guess there are some mistakes in my argument and if so I would like to know the reason for them .

Mr. Y
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  • So $f(f(x))=f(x)$ only holds if $f(x)\in[1,5]$? –  Feb 04 '16 at 08:58
  • @vrugtehagel I guess the range isn't restricted as long as the domain is respected (if I am understanding what you're asking). – Mr. Y Feb 04 '16 at 08:59
  • I think you proved that on $f(x)\in[1,5]$, we must have $f(x)=x$. So having $f(p)=1$ and $f(q)=5$, we look at $g(x)=f(x)-x$, so that $g(x)=0$ for $x\in[p,q]$. Thus, $g(x)=0$. So $f(x)=x$, contradiction. –  Feb 04 '16 at 09:06
  • Looks like the polynomial could be $f(x) = -(x-3)^{10} + 824$ – vnd Feb 04 '16 at 09:12
  • $f(x)=x$ has solutions about $x\sim 1.04$ and $x\sim 4.96$, so these are two solutions of the original problem, too. It's easy to prove by inequalities that the intersection of the curve above with $y=x$ has to exist for $x=1+\epsilon$ and $x=5-\epsilon$. When we add solutions to $f(x)=6-x$ which is also sufficient, here are 4 solutions in total, about 1.043, 1.044, and 6-these two values. But the latter two, from $f(x)=6-x$, don't work. – Luboš Motl Feb 04 '16 at 09:19
  • Can't I state that there $4$ solutions since $f(x)=x$ has two solutions (suppose I proved that) and $6-x=f(6-x) \implies c=f(c)$ which is the first case and so it implies $2$ other solutions ? – Mr. Y Feb 04 '16 at 09:52
  • Probably yes, I forgot to add "6-" at one place, and both pairs of solutions seem to be OK. – Luboš Motl Feb 04 '16 at 10:02

1 Answers1

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Your problem has at least $4$ solutions.

You can prove it by drawing the graph of $f(f(x))$ over the graph of $f(x)$: you have a zero $x_0$ near $x=1$, as your function is a polyomial it is continuous so you can assume the function takes all the values between $-200$ and $824$, around the zero we will, then, have $f(x)=1$, $f(x)=3$ and $f(x)=5$ you can then draw $f(f(x))$ as a small parabolic shaped curve at the very right of $x_0$.

This parabola will intersect $f(x)$ in two points.

As you can make the same arguments for the other zero, you should expect two other intersections at the very left of the zero near $x=5$.

So the solutions we found are $4$. As the polynomial is of 10th degree it is very hard to know from your graph if we should expect any other intersection. If you know that $x=3$ is the only stationary point for your function you can say that the solutions are exactly $4$.

N74
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  • How would I prove rigorously that ? – Mr. Y Feb 04 '16 at 09:36
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    Sorry, I'm an engineer... rigorously is out of my vocabulary :) – N74 Feb 04 '16 at 09:46
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    @N74: to prove it, one really wants to say (and see from the graph, whatever it is, or from the explicit form of the function with $824-(x-3)^{10}$) that the function $f(x)-x$ is monotonic around the points $x=1+\epsilon$ (e.g. in the interval 0.5-1.5) and $x=5-\epsilon$ (in 4.5-5.5), so by monotonicity and continuity, one finds exactly 1 solution of $f(x)=x$ in each interval. Similarly, one finds a pair of solutions to $f(x)=6-x$ in these two intervals which is also OK. The monotonicity follows from the graph visually and from monotonicity of $y=x$. – Luboš Motl Feb 04 '16 at 10:05
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    One may also prove that $f(x)=x$ and $f(x)=6-x$ have no solutions outside these two intervals, and all these four solutions are different from each other because $f(x)=x$ and $f(x)=6-x$ could only be obeyed by the same $x$ if $x=6-x$ i.e. $x=3$ but that's clearly not a solution. – Luboš Motl Feb 04 '16 at 10:07