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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.
But perhaps we misread: the problem says d depicts the depth... as a cubic polynomial and achieves its minimum depth exactly once. So the polynomial must have a unique critical point where $ d'(t) = 0 $ and $ d''(t) > 0 $.
But $ d(t) = t^3 $ has $ d'(t) = 3t^2 \geq 0 $, zero only at $ t=0 $, but second derivative $ 6t $, so $ d''(0)=0 $ — not a strict minimum.

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