f(x)=(x+2)/2, daca x<=0 si f(x)=x+1, daca x>0.
daca x,y<=0: f(x)=f(y) <=> (x+2)/2=(y+2)/2 <=> x=y.
daca x<=0, y>0: f(x)=f(y) <=> (x+2)/2=y+1 <=> x=2y, imposibil! => f(x) diferit de f(y).
daca x,y>0: f(x)=f(y) <=> x+1=y+1 <=> x=y.
Din cele 3 cazuri deducem ca f este injectiva.
