Then we know from the conditions that there is (3x 2+a) (2x+b) ≥ 0 in the interval.

Then draw f' (x) = 3x 2+a, which is a parabola with a vertex of (0, a) and an upward opening.

And draw g'(x)=2x+b, which is a straight line.

Are you sure the topic is not bigger than A or B? If not, the topic will be much more complicated.

It is discussed in two situations. Let's assume that b is greater than a, then the interval is (a, b). According to the image, we can know that the intersection of the straight line and the X axis is (-b/2,0). If b is greater than 0, then b is greater than -b/2, so it is in the interval (-b/2,0), and g'(x).

When b is not greater than 0, and the intersection (-b/2,0) is on the right side of the Y axis, or on the Y axis (b=0), then g'(x) is always less than or equal to 0 in the interval (a, b), that is to say, f'(x) is always less than or equal to 0 in (a, b). It can be found from the image that A is greater than or equal to-1/3. So the range of a is-1/3,0), and the range of b is (a,0), so the maximum value of |a-b| is1/3.

When b is less than a, then b is directly less than 0. As shown above, a is greater than or equal to-1/3, and b is greater than or equal to -√-a/3, and the result cannot be solved.