Then 9+3(a+ 1)+b=3.

3a+b=-9

b=-9-3a

F(x)≥x is a constant.

That is, X 2+(A+ 1) X+B ≥ X holds.

X 2+ax+b ≥ 0 holds.

δ=a^2-4b<； =0

B=-9-3a is brought into 2-4b.

a^2+36+ 12a<； =0

(a+6)^2<； =0

therefore

a=-6

b=9