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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. 📰 Tn Real Id Appointment 6128473 📰 Hogs Breath Inn Restaurant 6054958 📰 Border Cafe 3632842 📰 Tennisrecord 📰 Sable Offshore Stock 133192 📰 Cuyahoga Falls Weather 7831073 📰 Private Internet Access Download Mac 📰 The Proven Step By Step Guide To Making Outlook Templates Like A Profree Download Inside 7346937 📰 Csf 5346077 📰 Library For Download Java Jdk 8 Fast Start 📰 Rdr2 Walkthrough 📰 Jeff Buckley Grace Album 📰 Shocked Fans Reveal Secret Powers Binding Every Bato In Largest Manga Clue Yet 7415334 📰 Program Guide Spotify App Mac Os Premium Access 📰 Why Is Fortnite Not Letting Me Log In