Răspuns:
Explicație pas cu pas:
a) f(2) =f(1+1) = 2*1+4-f(2) => f(2)+f(2) = 6 => 2f(2)=6 => f(2)=3
b) f(x)=ax+b => f(x+1) = a(x+1)+b = 2x+4-3 => ax+a+b=2x+1 =>
a=2 si a+b=1 => b= -1 => f(x) =2x-1
c) A(1,1) ∈ Gf <=> f(1)=1 => 2*1-1 = 1 => A∈Gf
B(5,9) ∈ Gf <=> f(5)=9 => 2*5-1 = 10-1=9 => B ∈Gf
C(-1,3) ∈ Gf <=> f(-1)=3 => 2*(-1)-1 = -2-1 = -3 ≠ 3 => C∉ Gf
