Răspuns:
a) f:R->R, f(x)=ax³ + b|x|
f -impara ⇔ f(-x)=-f(x) (∀) x∈R
⇔a·(-x)³+b|-x| = -(ax³ + b|x|) (∀) x∈R
⇔-a·(x)³+b|x| = -ax³ - b|x| (∀) x∈R
⇔ b|x| = - b|x| (∀) x∈R
⇔ 2b|x| =0 (∀) x∈R
⇔ b =0, a ∈ R
b) g:R->R, g(x) = {x/2} + {x/3}
g(x+6) = {(x+6)/2} + {(x+6)/3} = {3+x/2} + {2 + x/3} = {x/2} + {x/3} =g(x)
⇒ g -periodica de perioada 6
