[tex]Conditia \ de \ existenta: \ f(x)\cdot f(y) \neq -1\\ \\ x=y=0\\ \\ f(0+0)=\frac{f(0)+f(0)}{1+f(0)\cdot f(0)}\\ \\ \Rightarrow f(0)=\frac{2\cdot f(0)}{1+f(0)^2}, \ in \ acest \ caz \ f(x)\cdot f(y)=f(0)^2 \geq 0, \ Deci \neq -1\\ \\ \Rightarrow f(0)+f(0)^3=2\cdot f(0)\\ \\ \Rightarrow f(0)^3-f(0)=0\\ \\ \Rightarrow f(0)[f(0)^2-1]=0\\ \\ \Rightarrow f(0)\cdot [f(0)+1]\cdot [f(0)-1]=0\\ \\ Cazul \ I:\\ \\ f(0)=0, \ nu \ verifica \ conditia \ 1) \ din \ enunt. \ In \ acest \ caz, \ | \ f(0) \ |=0[/tex]
[tex] Cazul \ II:\\ \\ f(0)+1=0 \Rightarrow f(0)=-1, \ verifica \ conditia \ 1)\\ \\ Cazul \ III:\\ \\ f(0)-1=0 \Rightarrow f(0)=1, \ verifica \ conditia \ 1)\\ \\ \\ \\ x=0 \Rightarrow f(0+y)=\frac{f(0)+f(y)}{1+f(0)\cdot f(y)} \Rightarrow f(y)=\frac{f(0)+f(y)}{1+f(0)\cdot f(y)}; \ f(0)\cdot f(y) \neq -1\\ \\ \Rightarrow f(y)+f(0)\cdot f(y)^2=f(0)+f(y) \Rightarrow f(0)\cdot f(y)^2-f(0)=0\\ \\ Cazul \ 1: \ f(0)=-1[/tex]
[tex] f(y)^2\cdot (-1)-(-1)=0\\ \\ f(y)^2=1 \Rightarrow f(y)=-1\\ \\ (nu \ poate \ fi \ 1, \ numitorul \ ar \ fi \ 0)\\ \\ Cazul \ 2: \ f(0)=1\\ \\ f(y)^2\cdot 1-1=0\\ \\ f(y)^2=1 \Rightarrow f(y)=1\\ \\ (nu \ poate \ fi \ -1, \ numitorul \ ar \ fi \ 0)\\ \\ \\ \\ Deci \ f(x)=1 \ sau \ f(x)=-1, \ avem \ doua \ functii[/tex]
