Question

F:R->R f(x)=3x+2 si fn(x)=fofof...of (de n ori)
Rezolvati inecuatia fn(x)<=x-1

Answer 1

(f°f) (x)=3(3x+2)+2=9x+6+2=9x+8
(f°f°f)(x)=3(9x+8)+2=27x+26
(f°f°f°f)(x) =3(27x+26)+2=81x+80
presupunem ca
(f°f...°f)(x) compusade n ori=3^n *x +3^n-1
este verificat deja pt n=1;2;3;4 (nu era nevoie dar m'am "prins" pe la n=2 si am fost sigur la n=3)

 Pn adevarat
P(n+1) =(f°f...°f)(x) compusade (n +1) ori= f°((f°f...°f)(x) compusade n ori
3(3^n *x +3^n-1)+2= 3^(n+1)x +3^(n+1)-3+2=3^(n+1)x +3^(n+1)-1
deci Pn⇒P(n+1) egalitatea este demo.perin inductie completa

atuci avem de rezolvat inecuatia
3^n *x +3^n-1≤x-1
3^n *x +3^n≤x
3^n *x-x≤-3^n
x(3^n-1)≤-3^n
x≤(-3^n)/(3^n-1)
are sens pt n≥1 si va avea mereu o valoare negativa
x∈Q-

(absolut OK, pt ca, pt n=0 nu ar exista f)