f(3/2)=5/2
f(7)=0
f(x) =ax+b
3a/2+b=5/2
3a+2b=5
dar -b/a=7 pt ca se anuleaza in 7
deci b=-7a
a tunci
3a+2*(-7a)=5
3a-14a=5
11a=-5
a=-5/11
b=-7a=35/11
f(x)=ax+b=-5x/11+35 /11
verificare
f(7) =-35/11+35/11=0 adevarat
f(3/2) =-15/22+35/11= -15/22+70/22 =55/22=5/2 adevarat
problema e bine rezolvata
