a) f:R->R; f(x)=2x+3
g:R->R; g(x)=2+x
Gf∩Gg={A}
=> f(x)=g(x)
2x+3=2+x
2x-x=2-3 => x=-1; il inlocuim pe x
y=f(-1)=g(-1)=-2+3=2-1=1
=> A(-1; 1)
b) f:{0;1;2;3;4} ->R; f(x)=x-4
g:{0;2;4;6}->R; g(x)=-x
Gf∩Gg={B}
=> f(x)=g(x)
x-4=-x
2x=4 => x=2
y=f(2)=g(2)=2-4=-2
=> B(2; -2)
c) f:R->R; f(x)=2x-4
g:R->R; g(x)=2x+1
2x-4=2x+1
2x-2x=1+4
0=5; fals => graficele celor doua functii nu au niciun punct comun; sunt paralele
Gf || Gg; Gf∩Gg=Ф
