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new years eve outfit - GetMeFoodie

📅 May 1, 2026 👤 scraface

May 01, 2026

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📌 But $ d(t) = t^3 $ is strictly convex and has no minimum (it decreases on $ (-\infty, 0) $, increases on $ (0, \infty) $), so it does not achieve a minimum. However, the problem states that the **minimum depth is achieved exactly once**, which implies $ d(t) $ has a unique global minimum. But $ t^3 $ has no such minimum. Contradiction?

📌 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.

📌 But $ d(t) = t^3 $ has no minimum. Unless minimum depth refers to **local minimum**? But still, $ t^3 $ has no local minimum.

📌 Hence, the only resolution is that the model is not $ d(t) = t^3 $, so perhaps there's a typo? But wait — reconsider: the values $ d(1)=1, d(2)=8, d(3)=27, d(4)=64 $ **force** $ d(t) = t^3 $, due to interpolation.

📌 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.

📌 So $ d(t) = t^3 $ fails the minimum condition.

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