Question: An oceanographer models the depth $ d(t) $ (in meters) of a marine layer influenced by tidal forces as a cubic polynomial in time $ t $ (in days). It is known that $ d(1) = 1 $, $ d(2) = 8 $, $ d(3) = 27 $, and $ d(4) = 64 $, and $ d(t) $ achieves its minimum depth exactly once. Find $ d(0) $.

["Deep Insight into Oceanic Tidal Depth via Polynomial Modeling", "Understanding how ocean depth varies over time is crucial for tidal modeling, climate studies, and marine navigation. A recent study models the depth $ d(t) $ of a marine layer—shaped by rhythmic tidal forces—as a cubic polynomial:", "$$\nd(t) = at^3 + bt^2 + ct + d,\n$$\nwhere $ t $ is time in days, and $ d(t) $ is depth in meters.", "We are given:\n- $ d(1) = 1 $\n- $ d(2) = 8 $\n- $ d(3) = 27 $\n- $ d(4) = 64 $\n- $ d(t) $ has exactly one minimum\nWe are to find $ d(0) $.", "---", "### Step 1: Observe the Given Points", "The values $ d(1)=1 $, $ d(2)=8 $, $ d(3)=27 $, $ d(4)=64 $ resemble perfect cubes:\n$$\n1 = 1^3,\quad 8 = 2^3,\quad 27 = 3^3,\quad 64 = 4^3.\n$$\nThis strongly suggests that $ d(t) = t^3 $. But $ t^3 $ is a cubic polynomial and satisfies all four conditions. However, the problem states that $ d(t) $ achieves its minimum exactly once, but $ d(t) = t^3 $ is strictly increasing and has no local minimum—it increases monotonically.", "Thus, $ d(t) <br/>\ne t^3 $, though it agrees with it at $ t = 1,2,3,4 $. Therefore, the cubic polynomial $ d(t) $ agrees with $ t^3 $ at these four points but differs elsewhere, and must have a unique minimum—implying it is not monotonic, so its derivative has exactly one real root (i.e., one critical point, a minimum).", "---", "### Step 2: Assume General Cubic Form", "Let\n$$\nd(t) = at^3 + bt^2 + ct + d.\n$$\nWe apply the known values to form equations:", "1. $ d(1) = a + b + c + d = 1 $\n2. $ d(2) = 8a + 4b + 2c + d = 8 $\n3. $ d(3) = 27a + 9b + 3c + d = 27 $\n4. $ d(4) = 64a + 16b + 4c + d = 64 $", "We now solve this system.", "---", "### Step 3: Solve the System of Equations", "Let’s subtract consecutive equations to eliminate $ d $.", "Equation (2) – Equation (1):\n$$\n(8a + 4b + 2c + d) - (a + b + c + d) = 8 - 1 = 7\n\Rightarrow 7a + 3b + c = 7 \quad \ ext{(Eq A)}\n$$", "Equation (3) – Equation (2):\n$$\n(27a + 9b + 3c + d) - (8a + 4b + 2c + d) = 27 - 8 = 19\n\Rightarrow 19a + 5b + c = 19 \quad \ ext{(Eq B)}\n$$", "Equation (4) – Equation (3):\n$$\n(64a + 16b + 4c + d) - (27a + 9b + 3c + d) = 64 - 27 = 37\n\Rightarrow 37a + 7b + c = 37 \quad \ ext{(Eq C)}\n$$", "Now subtract again:", "(Eq B) – (Eq A):\n$$\n(19a + 5b + c) - (7a + 3b + c) = 19 - 7 = 12\n\Rightarrow 12a + 2b = 12 \Rightarrow 6a + b = 6 \quad \ ext{(Eq D)}\n$$", "(Eq C) – (Eq B):\n$$\n(37a + 7b + c) - (19a + 5b + c) = 37 - 19 = 18\n\Rightarrow 18a + 2b = 18 \Rightarrow 9a + b = 9 \quad \ ext{(Eq E)}\n$$", "Now subtract (Eq E) – (Eq D):\n$$\n(9a + b) - (6a + b) = 9 - 6 \Rightarrow 3a = 3 \Rightarrow a = 1\n$$", "From Eq D: $ 6(1) + b = 6 \Rightarrow b = 0 $", "From Eq A: $ 7(1) + 3(0) + c = 7 \Rightarrow c = 0 $", "Now use Eq (1): $ a + b + c + d = 1 \Rightarrow 1 + 0 + 0 + d = 1 \Rightarrow d = 0 $", "Thus,\n$$\nd(t) = t^3\n$$", "But wait—this again gives $ d(t) = t^3 $, which is strictly increasing, so it has no local minimum. Contradiction with the condition that $ d(t) $ achieves its minimum exactly once.", "This implies our assumption must be re-examined: could $ d(t) $ still be $ t^3 $ at these points but have a different structure? No—because if $ d(t) $ is a cubic polynomial agreeing with $ t^3 $ at four distinct points, and $ t^3 $ is also a cubic, then by the uniqueness of cubic interpolation, $ d(t) = t^3 $ is the only cubic polynomial satisfying those four conditions.", "Hence, the only cubic polynomial satisfying $ d(1)=1, d(2)=8, d(3)=27, d(4)=64 $ is $ d(t) = t^3 $, which is strictly increasing, so it has no minimum—contradicting the given.", "Therefore, the only resolution is that the given values are not incidents of a cubic polynomial matching $ t^3 $ at $ t=1,2,3,4 $—but wait: we derived $ d(t) = t^3 $ from those four points. So unless the model is not cubic, or the data is not exact, contradiction arises.", "But the problem states: “It is known that $ d(t) $ achieves its minimum exactly once” — suggesting $ d(t) $ is not $ t^3 $, yet passes through $ (1,1), (2,8), (3,27), (4,64) $. But with four points, a cubic is uniquely determined. So if $ d(t) $ is cubic and agrees with $ t^3 $ at four points, it must be $ t^3 $. Thus, the only way to resolve this is to interpret the minimum condition as a constraint—perhaps indicating that $ d(t) $ cannot be $ t^3 $, so the values must not be exact, or $ d(t) $ has a different form?", "Wait—reconsider: the problem says $ d(1)=1 $, $ d(2)=8 $, etc.—these are satisfied by $ t^3 $. But $ t^3 $ has derivative $ d'(t) = 3t^2 $, which is zero only at $ t=0 $, and is positive elsewhere. So it has no local minimum—only a global minimum at $ t \ o -\infty $, but in finite domain, strictly increasing.", "But the problem says exactly one minimum, so $ d(t) $ must not be monotonic—hence, its derivative must have exactly one real root (a minimum). That requires $ d'(t) $ to have exactly one real zero, and second derivative changes sign there.", "Thus, contradiction unless our interpolating polynomial is not $ t^3 $, but how?", "Unless: the values $ d(1)=1, d(2)=8, d(3)=27, d(4)=64 $ are not exact? No—the problem states they are.", "Therefore, the only resolution is that the minimum condition is not at a critical point from $ d'(t)=0 $, but wait—every local extremum of a differentiable function occurs where $ d'(t)=0 $. So to have exactly one local minimum, $ d'(t) $ must have exactly one real root (a single critical point, and second derivative positive), and the function increases to the left and right.", "But a cubic derivative is quadratic—so $ d'(t) $ has two roots, one local max, one local min, unless discriminant zero (double root), or one real root (if discriminant negative, but then no local extrema).", "To have exactly one minimum, and no other extrema, $ d'(t) $ must have one real root of multiplicity one, and the other derivative values… no—quadratic derivative always has 0, 1, or 2 real roots.", "- 2 real roots: two critical points — one min, one max → two extrema\n- 1 real root: degenerate (double root), but then inflection, no local min\n- 0 real roots: monotonic", "So a cubic cannot have exactly one local minimum unless it's not generic—because $ d'(t) $ is quadratic, so has\n- two distinct real roots → one min, one max\n- one repeated root → inflection only\n- no real roots → monotonic", "So it always has either one extremum (min or max) or none—and cannot have “exactly one” local minimum among a single local dip—because “exactly one minimum” implies a single global low point, which for a cubic with one min, requires a local max.", "But a cubic with a single local minimum (left-heavy) must have $ d'(t) $ decreasing then increasing—so two critical points: one max, one min.", "Hence, a cubic polynomial can have exactly one global minimum and one local maximum, i.e., two critical points, but only if the middle one is a local min and outer a local max. But then “exactly one minimum” is true.", "Ah—so the phrase “achieves its minimum exactly once” means one global minimum**, not necessarily that there is only one critical point.", "So our goal: find cubic $ d(t) $ such that:\n- $ d(1)=1, d(2)=8, d(3)=27, d(4)=64 $\n- $ d(t) $ has exactly one global minimum (i.e., $ d'(t) $ has one critical point at which $ d''(t) > 0 $)", "But again, $ d(t) $ is determined by four points.", "Let $ d(t) = at^3 + bt^2 + ct + d $", "We found that the only cubic satisfying the four values is $ d(t) = t^3"]