Răspuns:
Explicație pas cu pas:
f : R --> R ; f(x) = x/2 + 2 = (x+4)/2
A(0 ; 2) ∈ Gf <=> f(0) = 2
f(0) = (0+4)/2 = 2 => A(0 ; 2) ∈ Gf
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B(3 ; 1) ∈ Gf <=> f(3) = 1
f(3) = (3+4)/2 = 7/2 ≠ 1 => B(3 ; 1) ∉ Gf
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C(-2 ; 0) <=> f(-2) = 0
f(-2) = (-2+4)/2 = 1 ≠ 0 => C(-2 ; 0) ∉ Gf
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D(1 ; 5/2) ∈ Gf <=> f(1) = 5/2
f(1) = (1+4)/2 = 5/2 => D(1 ; 5/2) ∈ Gf
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E(2√2 ; 2+√2) ∈ Gf <=> f(2√2) = 2 +√2
f(2√2) = (2√2+4)/2 = √2+2 => E(2√2 ; 2+√2) ∈ Gf
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F(1/2 ; 9/4) ∈ Gf <=> f(1/2) = 9/4
f(1/2) = (1/2 +4)/2 = 9/2 : 2 = 9/4 => F(1/2 ; 9/4) ∈ Gf
