Find all function $ f:\mathbb{R}\rightarrow\mathbb{R} $
1)$f$ is surjective
2)For $x>y$, $ f(x)>f(y) $
3)$ f(f(x))=f(x)+12x $
Vietnam 2012- Surjective function
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Nur Muhammad Shafiullah | Mahi
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- zadid xcalibured
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Re: Vietnam 2012- Surjective function
$f(f(0))=f(0)$ and strictly increasing property yields $f(0)=0$.
let $a_n=f^{n}(x)$ and let $a_0=x$
then $a_n=ap^n+bq^n$ where $a$ and $b$ are the solutions of $x^2-x-12=0$
$a_2=f(x)=ap+bq=-3p+4q=-3x+7q$
but $f(0)=0$ implies $q=0$
so $f(x)=-3x$ which is surjective and strictly increasing.
let $a_n=f^{n}(x)$ and let $a_0=x$
then $a_n=ap^n+bq^n$ where $a$ and $b$ are the solutions of $x^2-x-12=0$
$a_2=f(x)=ap+bq=-3p+4q=-3x+7q$
but $f(0)=0$ implies $q=0$
so $f(x)=-3x$ which is surjective and strictly increasing.
Re: Vietnam 2012- Surjective function
$f(x) = -3x$ is strictly decreasing.zadid xcalibured wrote: so $f(x)=-3x$ which is surjective and strictly increasing.
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Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
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Re: Vietnam 2012- Surjective function
The mistake in Zadid vai's solution was perhaps to write $a_n=ap^n+bq^n$. As far as I know the right form should be $a_n=pa^n+qb^n$ and this yields a valid solution $f(x)=4x$
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- zadid xcalibured
- Posts:217
- Joined:Thu Oct 27, 2011 11:04 am
- Location:mymensingh
Re: Vietnam 2012- Surjective function
There are two cases.I missed it.One case gives $a=0$ and $b=x$ another cases gives $b=0$ and $a=x$.One case gives no solution while the other case gives $f(x)=4x$