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For polynom $f(x)=ax^2+bx+c$ equation $f(x)=x$ has no real solutions. Prove that equation $ f(f(x))=x$ also does not have does not have real solutions

Can someone explain solution to me? Why?

If equation $f(x)=x$ has no real solutions than it is either $f(x)>0$ and $a>0$ or it is $f(x)<0$ and $a<0$. In first case $f(f(x))>f(x)>x$ or in second case $f(f(x))<f(x)<x$ for no real number $x$ it can not be $f(f(x))=0$

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  • $\begingroup$ Your assumption about $f$ is wrong: consider $f(x)=(x+2)^2-1$. $\endgroup$ Dec 23, 2018 at 19:59
  • $\begingroup$ The quoted solution is incorrect. $\endgroup$
    – Lance
    Dec 23, 2018 at 20:00
  • $\begingroup$ @John_Wick Your conclusion is wrong. Consider $f(x)=(x+2)^2-1$, then $f(f(-0.9)),f(-0.9)$. $\endgroup$ Dec 23, 2018 at 20:05

3 Answers 3

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The quoted solution should read

If equation $f(x)=x$ has no real solutions than it is either $f(x)>\color{red}x$ and $a>0$ or it is $f(x)<\color{red}x$ and $a<0$. In first case $f(f(x))>f(x)>x$ or in second case $f(f(x))<f(x)<x$ for no real number $x$ it can not be $f(f(x))=\color{red}x$.

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If $a>0$ then $f(x)=ax^2+bx+c>x$ for sufficiently large $x$. And as $f(x)=x$ has no real root then $f(x)>x$ for all $x$ (otherwise there will be a real root). Hence $f(f(x))>f(x)>x$ for all $x$. So, $f(f(x))=x$ has no real root. Similarly for $a<0.$

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Let

$$g(x)=f(x)-x=ax^2+(b-1)x+c$$

no real roots implies that

$$(b-1)^2-4ac<0$$ and $$(\forall x\in \Bbb R)\;\; g(x) \text{ has a constant sign} :$$ $(+ \text{ if } a>0 \text{ and } - \text{ if } a<0)$.

thus $$f(f(x))-x=$$ $$f(f(x))-f(x)+f(x)-x=$$ $$g(f(x))+g(x)$$ will have a constant sign and no root.

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