f : R ->R, f(x) =3x + 11
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a= ? , a ∈ IR , f(a) + 3 = f(5)
f(a) + 3 = f(5)
(3a+ 11)+ 3= 3·5+ 11
3a+ 11+ 3= 15+ 11
3a+ 14= 26
3a = 26- 14
3a = 12
a = 12: 3
a = 4
probă: f(a) + 3 = f(5)
(3·4+ 11)+ 3= 3·5+ 11
(12+ 11)+ 3= 15+ 11
23+ 3= 26
26=26