Determine all strictly increasing functions $ f: R\rightarrow R$ satisfying relationship $ f(x+f(y))=f(x+y)+2005$ for any real values of x and y.
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Tags: function
bbgun34
07.01.2010 19:18
well if you let $ x=0$, then we have
$ f(f(y))=f(y)+2005$, so it follows that one of the functions is $ \boxed{f(x)=x+2005}$. Checking this, both sides are $ x+y+4010$, so it is indeed one.
Syler
07.01.2010 20:15
bbgun34 wrote:
well if you let $ x = 0$, then we have
$ f(f(y)) = f(y) + 2005$, so it follows that one of the functions is $ \boxed{f(x) = x + 2005}$. Checking this, both sides are $ x + y + 4010$, so it is indeed one.
I found only this function too.
x164
07.01.2010 20:43
Since the RHS is symmetric in $ x$ and $ y$, the left hand side must be too. So for all $ x,y \in \mathbb{R}$: $ f(x+f(y)) = f(y+f(x))$. Now, since $ f$ is strictly increasing, $ f$ must be injective. So $ x + f(y) = y + f(x)$ $ f(x) - x = f(y) - y$ So if there is a function that satisfies the equations, it must be of the form $ f(x) = x + c$ for some $ c \in \mathbb{R}$. We'll check if all $ c$ satisfy: $ x+y+2c \stackrel{?}{=} x+y+c+2005$, so $ c=2005$. So the only funtion that satisfies the equation is $ f(x) = x + 2005$.