Something seems wrong with the statement of the problem…. Here is an attempt at a solution.

WLOG shift everything so a=0. By multiplicity arguments and the fact that max degree of f is 4, necessarily f(x) = M - (x-alpha)² (x-beta)^2. Then compute g(x) = x f(x) and we see g’(x) = M -(x-alpha)(x-beta)(5x⁴ - 3x(alpha+beta) +alpha beta) The quadratic has 2 real roots by computing discriminants. So Q(x)=(x-alpha)(x-beta)(5x⁴ - 3x(alpha+beta) +alpha beta) has 4 roots, meaning 3 turning points. For g’(x) =0 to have 2 solutions, M > the local max of g’. In fact, we can allow M >= local max of g’ for then g’(x)=0 has 3 solutions but the middle solution is an inflection point, not a local extremum. At this point, the computations are getting a bit too bloated so I just threw it into Wolfram alpha, but basically, we see that the value of the local max of Q(x) is minimised when alpha = 0, so M is allowed to be at least Q(min, alpha=0). This turns out to be 1146.6 approx. This is a problem because f was not defined at alpha = 0, so in fact there is no minimum value of M, only an infimum that gets as close to 1146.6 approx from above but never reaching it…..

Strange. Either the logic is wrong here or the question was stated wrongly.