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Se considera functia f:R->R, f(x)=4x-1. Determinati a ∈ R, daca A(a;-9) ∈ Gf.

Răspuns :

A(a,-9)∈Gf⇔f(a)=-9
⇔4a-1=-9
4a=-8
a=-2
A(a;-9) ∈ Gf ⇒ f(a) = -9
                       f(a) = 4a-1

Deci, 4a-1=-9
4a=-9+1
4a=-8
a=-8 : 4
a=-2