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

### Vietnam 2012- Surjective function

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- zadid xcalibured
**Posts:**217**Joined:**Thu Oct 27, 2011 11:04 am**Location:**mymensingh

### 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 $f(x)=4x$