Ne folosim de ce stim:
f(1)=1 -> f(1)=a+b+c=1
f(3)=3 -> f(3)=9a+3b+c=3
f(2)=? -> f(2)=2a+2b+c=?
GASIM a
(2) 9a+3b+c=3
(1) a+b+c=1
8a+2b=2 /2
4a+b=1
a=(1-b)/4
Il inlocuim pe a
(1) (1-b)/4+b+c=1
(2) 9(1-b)/4+3(1-b)/4+c=3
(1) 9(1-b)+3(1-b)+4c=12(am adus la acelasi numitor)
(2) 1-b+4b+4c=4(am adus la acelasi numitor)
(1) -12b+4c=0
(2) 3b+4c=3
-15b=-3 => b=1/5
IL gasim pe a
a=(1-b)/4 =>a=1/5
IL gasim pe ce din prima a+b+c=1 1/5+1/5+c=1 => 2/5+c=1 => c=1-2/5 =>
c=3/5
gasim f(2)=2*1/5+2*1/5+3/5=7/5
FINAL