The condition that a cubic achieves a minimum must be compatible. But $ t^3 $ has no minimum. Therefore, the only possibility is that the model is still $ d(t) = t^3 $, and the phrase minimum depth refers to the **lowest recorded value in a physical window**, but mathematically, over $ \mathbb{R} $, $ t^3 $ has no minimum.

Tue, 05 May 2026 00:54:29 +0000

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