👤
DaRkBoY
a fost răspuns

Fie functia f:R->R , f(x)=3x+1.
Sa se calculeze fofof...of.(de 2013 ori)
fof = f compus cu f.

Răspuns :

Matepentrutoti
Analizam cateva cazuri particulare:
[tex](fof)(x)=f(f(x))=f(3x+1)=3(3x+1)+1=9x+4=\\ =3^2x+3^1+3^0\\ (fofof)(x)=f(f(f(x)))=f(9x+4)=3(9x+4)+1=27x+13\\ =3^3x+3^2+3^1+3^0\\ ......\\ (fofofofo....of)(x)=3^{2013}x+3^{2012}+3^{2011}+...+3^1+3^0 [/tex]

Alte intrebari