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9,245,955 20260408

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(2330)-(2330)CMoney

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55.401019370.00 23881353.5$.

20261DRAMSSD2025122026.

(2330)-(2330)CMoney

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But $ d(t) $ agrees with $ t^3 $ at $ t=1,2,3,4 $, yet is a cubic. So define $ p(t) = d(t) - t^3 $. Then $ p(t) $ is a polynomial of degree at most 3 (since $ d(t) $ is cubic, $ t^3 $ is cubic), and $ p(1) = p(2) = p(3) = p(4) = 0 $.
So $ p(t) $ has four roots and is degree ≤ 3 → $ p(t) \equiv 0 $. Hence $ d(t) = t^3 $ is the only possibility.
Therefore, the requirement of a minimum must be interpreted carefully. But the problem says achieves its minimum depth exactly once — so perhaps minimum is a red herring, or we are to assume $ d(t) $ has a unique global minimum, but only one cubic satisfies the four points.

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