f[f(x)]=f(ax+b)=a(ax+b)+b=a^2 x+a b+ b = 9x+8

So a 2 = 9, ab+b = 8, the solution is, a=3, b=2 or a=-3, b=-4.

F(x)=3x+2 or f(x)=-3x-4.

(2) let f (x) = ax 2+bx+c (a is not equal to 0)

f(x+ 1)+f(x- 1)=a(x+ 1)^2+b(x+ 1)+c+a(x- 1)^2+b(x- 1)+c=2ax^2+2bx+2a+2c

Since f (x+1)+f (x-1) = 2x2-4x+4,

So 2a=2, 2b=-4, 2a+2c=4, while a= 1, b=-2, c= 1, so f (x) = x 2-2x+ 1.

Note: All the above are the mean square of 2 ~

Does this question have anything to do with the tick function ~