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I want to find all $f(x)$s (if they exist) which $fof(x)=4-x$, I know that $f(x)$ can't be linear because if $$f(x)=ax+b$$ then $$fof(x)=a(ax+b)+b=a^2x+ab+b$$ And $a^2$ can't be -1.

Actually i think $f(x)$ can't be any polynomial but i can't prove it and a combination of trig functions may be an answer.

The answer to my question is a function with that property or a proof that such function doesn't exist. Thanks!

edit:i should be clear that $f$ is $\mathbb R\to \mathbb R $ and it is injective and continuous.

A. H. Z. A.
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    why $f(x) = ix + 4/(1+i)$ isn't good? Do you want it real valued? – Exodd Mar 29 '22 at 15:26
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    might be helpful https://math.stackexchange.com/questions/3409811/prove-that-no-function-satisfies-ffx-x3-if-f-mathbb-z-0%e2%86%92-mathbb?noredirect=1 –  Mar 29 '22 at 15:28
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    Is $f$ meant to be from $\mathbb R\to \mathbb R$? If so, note that $f$ must be injective (why) and therefore, if it is continuous, it must be monotone. Derive a contradiction (again, under the additional requirements I spelled out). – lulu Mar 29 '22 at 15:29
  • yes $f(x)$ is from $\mathbb R \to \mathbb R $ and i want it to be real not containing complex numbers. – A. H. Z. A. Mar 29 '22 at 17:19
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    i also found this function $$f(x)=\sqrt{1-(4-x)^2}+4$$the problem is that $fof(x)$ is $|4-x|$ not $4-x$. – A. H. Z. A. Mar 29 '22 at 17:30

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hint

If we replace $ x $ by $ f\circ f(x)=f^2(x) $, we get

$$f^4(x)=4-f^2(x)=x$$

hamam_Abdallah
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