(1) let f (x) >: 0, then f [f (x)] >; 0, does not meet the meaning of the question.

(2) If f(x)=0 has a unique real root, then b 2-4c = 0.

F[f(x)]=0 has a unique real root, that is, f(x)=-b/2 has a unique real root.

At this time, b=c=0.

(3) Let f(x)=0 have two real roots, and let f(x)=(x-x 1)(x-x2).

Then f [f (x)] = [f (x)-x1] [f (x)-x2]

Ensure that f[f(x)]=0 has a unique real root, that is, f(x)-x 1=0 has a real root and f(x)-x2=0 has no real root.

X 1=minf(x), x2.