Find all functions $f : (0, +\infty) \to \mathbb{R}$ satisfying the following conditions: $(i)$ $f(x) + f(\frac{1}{x}) = 1$ for all $x> 0$; $(ii)$ $f(xy + x + y) = f(x)f(y)$ for all $x, y> 0$.
Problem
Source: 239 2019 S6
Tags: algebra, functional equation
01.08.2020 12:28
SinaQane wrote: Find all functions $f : (0, +\infty) \to \mathbb{R}$ satisfying the following conditions: $(i)$ $f(x) + f(\frac{1}{x}) = 1$ for all $x> 0$; $(ii)$ $f(xy + x + y) = f(x)f(y)$ for all $x, y> 0$. Second equation easily gives solutions $f(x)=e^{g(\ln(x+1))}$ whatever if $g(x)$ additive. Plugging this back in first equation we get $g(x)=-x$ and $\boxed{f(x)=\frac 1{x+1}\quad\forall x>0}$ Which indeed is a solution.
04.08.2020 12:16
mmh, could you explain how you solved the second equation, I kind of dont get it.
04.08.2020 13:35
Lukas8r20 wrote: mmh, could you explain how you solved the second equation, I kind of dont get it. Let $h(x)$ from $(1,+\infty)\to\mathbb R$ defined as $h(x)=f(x-1)$ Second equation if $h((x+1)(y+1))=h(x+1)h(y+1)$ $\forall x,y> 0$ And so $h(xy)=h(x)h(y)$ $\forall x,y>1$ This implies $h(x)\ge 0$ and so $h(x)>0$ (here, I use the fact that $f\equiv 0$ is discarded thru first equation) So we can define $g(x)$ from $(0,+\infty)\to\mathbb R$ as $g(x)=\ln(h(e^x))$ and equation becomes $g(x+y)=g(x)+g(y)$ $\forall x,y>0$ So $g(x)$ is additive and $f(x)=e^{g(\ln(x+1))}$, as claimed