Răspuns:
a)n:3=c rest q q≥0 q<3 q∈{0,1,2}=>
Im f={0,1,2}
b) 0≤2 2≤2
Deci Im f∈[0,2]=> f marginita superior si inferior
peroiodicitatea
f(3k)=3k;3=k rest 0
f(3k+1)=3k+1):3= k rest 1
f(3k+2)=(3k+2):3=k rest 2
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f(3k+3)=3k+3):3=k+1 rest 0 ciclul se repeta=> perioada T= 3
monotonia
f(3k)=0
f(3k+1)=1
f(3k+2)=2
Observi ca pt 3k<3k+1<3k+2 f(3k)<f(3k+1)<f(3k+2)=>
f monoton crescatoare
pe perioada T=3
Explicație pas cu pas:
