Wait — unless our initial assumption that $ d(t) = t^3 $ is forced is correct, but it violates the minimum condition. So perhaps $ d(t) $ is not exactly $ t^3 $, but fits it at four points. But a cubic is uniquely determined by four points, so $ d(t) = t^3 $ is the only cubic polynomial satisfying the conditions.

Sat, 02 May 2026 22:13:55 +0000

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