The function f (x) whose domain is n + and whose value is also a positive integer: for any n ∈ n +, f (n + 1) > F (n); f (f (n)) = 3N, find f (4), f (5) We know that {f (n)} is a strictly increasing positive integer sequence -- > F (n) ≥ n f(f(1))=3≤f(3)--->f(1)≤3 If f (1) = 1, it is contradictory to f (1)) = 3; if f (1) = 3 --- > F (1)) = f (3) = 3, it is contradictory So: -- > F (1) = 2, f (2) = f (f (1)) = f (2) = 3 f(3)=f(f(2))=6 f(6)=f(f(3))=9 Because {f (n)} is a strictly increasing positive integer sequence -->f(4)=7,f(5)=8 How did you get this step? f(6)=f(f(3))=9 Because {f (n)} is a strictly increasing positive integer sequence

The function f (x) whose domain is n + and whose value is also a positive integer: for any n ∈ n +, f (n + 1) > F (n); f (f (n)) = 3N, find f (4), f (5) We know that {f (n)} is a strictly increasing positive integer sequence -- > F (n) ≥ n f(f(1))=3≤f(3)--->f(1)≤3 If f (1) = 1, it is contradictory to f (1)) = 3; if f (1) = 3 --- > F (1)) = f (3) = 3, it is contradictory So: -- > F (1) = 2, f (2) = f (f (1)) = f (2) = 3 f(3)=f(f(2))=6 f(6)=f(f(3))=9 Because {f (n)} is a strictly increasing positive integer sequence -->f(4)=7,f(5)=8 How did you get this step? f(6)=f(f(3))=9 Because {f (n)} is a strictly increasing positive integer sequence

F (3) = 6, so f (3)) = f (6). And because f (f (n)) = 3N, so f (f (3)) = 9
Because {f (n)} is a strictly increasing sequence of positive integers

If f (x) = xn2 − 3N (n ∈ z) is an even function and y = f (x) is a decreasing function on (0, + ∞), then n=______ ．

∵ y = f (x) is a decreasing function on (0, + ∞) ∵ n2-3n ∵ 0 ∵ n ∵ 3 and ∵ even function ∵ n = 1 or 2, so the answer is: 1 or 2

F (x) = x ^ (n ^ 2-3n) is an even function, and f (x) is a decreasing function on (0, positive infinity)

Because n ^ 2-3n = (n-3) n, the decreasing function shows (n-3) n