Because f (x) >; 0, so f(0)= 1.

(2) f(x)=f？ (x/2)≥0

Let x1; 0, so that the

f(x2-x 1) f(x 1)=f(x2)

Get 0

So f (x2) < f(x 1)

F(x) is a decreasing function on r.

(3) f(-x？ +6x- 1+y)= 1, then -x? +6x- 1+y=0

That is A={(x, y)|y=x? -6x+ 1}

Because a ∩ b = φ, that is, y=x? -6x+ 1 does not intersect with y=a,

Make a≤(x? -6x+ 1) Minimum value =-8