O functie este injectiva daca si numai daca pt x1≠x2 f(x1)≠f(x2)
a( fie x1 x2 astfel incat f(x1)=f(x2)
2x1+5=2x2+5 => 2x1=2x2 => x1=x2 deci f(x1)=f(x2) =>x1=x2 f injectiva
b)x1 x2 a.i. f(x1)=f(x2)
-5x1+2=-5x2+2 =>
-5x1=-5x2 => x1=x2 f injectiva
c)fie x1 x2≥0 a.i.f(x1)=f(x2)
2x1+3=2x2+3 =>2x1=2x2 =>x1=x2 => f injectiva
d) x1 x2 a.i. f(x1)=f(x2)
(1-x1)/(1+x1)=(1-x2)/(1+x2)aduci la acelasi numitor fractiile
(1-x1)((1+x2)=(1-x2)(1+x1)
1-x1+x2-x1X2=1-x2+x1-x1*x2
x2-x1=- x2+x1
2x2=2x1 => x1=x2 f injevctiva