If 3(f(x))2+2af(x)+b=0, then there are two f(x) that make the equation hold.

X 1=f(x 1), assuming that X2 > x 1 and x 1=f(x 1), then x1is the maximum point.

The following schematic images:

At this point, there are three intersections,

On the other hand, if x2 < X 1, x 1=f(x 1), then X 1 is the minimum point.

So choose a.