Răspuns:
Explicație pas cu pas:
f : Z --> Z
f(x) = 2x+1
A(1 , 3) ; B (-2 , 3) ; C(-2 , -3) ; D(2 , 5) ; E(0 , 1) ; F(1/2 ; 2) ∈ Gf ?
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A(1 , 3) ∈ Gf <=> f(1) = 3 => f(1) = 2·1+1 = 3 => A(1 , 3) ∈ Gf
B (-2 , 3) ∈ Gf <=>f(-2) = 3 => f(-2) = 2·(-2) +1 = -4+1= -3 => B (-2 , 3)∉ Gf
C(-2 , -3) ∈ Gf deoarece f(-2) = -3
D(2 ; 5) ∈ Gf <=> f(2) = 5 => f(2) = 2·2+1 = 5 => D(2 ; 5) ∈ Gf
E(0 , 1) ∈ Gf <=> f(0) = 1 => f(0) = 2·0+1 = 1 => E(0 , 1) ∈ Gf
F(1/2 , 2) ∈ Gf <=> f(1/2) = 2 => f(1/2) = 2·1/2 + 1 = 1+1 = 2 => F(1/2 , 2) ∈ Gf
