Page 1 of 1

Ecuatie functionala 3

Posted: Sun Feb 21, 2010 11:31 pm
by Marius Mainea
Sa se determine functiile strict crescatoare \( f:\mathbb{R}\rightarrow\mathbb{R} \) cu proprietatea ca \( f(xf(y))=f(x)f(y) \) oricare ar fi \( x,y\in\mathbb{R} \)

GM

Posted: Mon Feb 22, 2010 2:49 pm
by Claudiu Mindrila
Intai \( f \) e injectiva. Deoarece \( f\left(xf\left(y\right)\right)=f\left(yf\left(x\right)\right)=f\left(x\right)\cdot f\left(y\right)\Longrightarrow xf\left(y\right)=yf\left(x\right)\Longrightarrow\frac{f\left(x\right)}{x}=\frac{f\left(y\right)}{y}=c \), cu \( c \) constant. De aici functiie cautate sunt de forma \( f(x)=cx \)

Posted: Mon Feb 22, 2010 6:34 pm
by Marius Mainea
....