Therefore, the only way is if $ d(t) $ is not $ t^3 $, but a different cubic that fits the four points and has a unique minimum. But four points force $ d(t) = t^3 $. - Parker Core Knowledge

Understanding the Context

Why $ t^3 $ Might Seem Unavoidable

Let’s suppose we are fitting a cubic function through four distinct points $ (t_i, d_i) $. A general cubic polynomial has four coefficients, so in principle, specifying four values uniquely determines a cubic. The standard cubic $ d(t) = t^3 $ passes through the origin and has a single inflection point—not a local minimum. However, if our data demands a smooth monotonic or unimodal behavior with a unique critical point, the function must deform $ t^3 $.

But here’s the catch: the cubic $ d(t) = t^3 $ has no local minimum at a point; it has only an inflection point at $ t = 0 $. To fulfill the requirement of a unique minimum (a strict local minimum), the function must concave upward near that point—something $ t^3 $ fails to do. So even though $ t^3 $ fits the interpolation for many point sets, it does not satisfy the critical requirement of having a local minimum.

The Hidden Flexibility in Cubic Functions

Key Insights

Cubic polynomials are far more versatile than we often assume—especially when shape and symmetry are not enforced. While $ d(t) = t^3 $ is symmetric about the origin, a shifted and scaled cubic can introduce a unique minimum while still agreeing with four given points.

For instance, consider a cubic of the form:
$$ d(t) = a(t - h)^3 + c $$
This has only one critical point at $ t = h $, where the derivative $ d'(t) = 3a(t - h)^2 $ vanishes. If $ a > 0 $, this critical point is a local (and global) minimum. The values of $ a $ and $ h $ can be solved to match a specified cubic curve through four points.

With four points, we generate a system of equations. While $ d(t) = t^3 $ fixes $ a = 1 $, $ h = 0 $, it ignores the degrees of freedom in $ a $ and $ h $. In reality, a unique cubic passing through four points with a distinct local minimum exists if and only if the data permits a non-zero curvature and sign of derivative changes at some point, which is perfectly compatible with cubic form.

When Is $ d(t) = t^3 $ the Only Cubic?

Interestingly, $ d(t) = t^3 $ is not the only cubic that can pass through four points. Without additional constraints (such as zero at origin or specific symmetry), infinitely many cubics fit four points—some smooth, some with extrema. But if the goal is a unique minimum, then the cubic must be chosen to have a negative leading coefficient and a shape matching curvature changes.

Final Thoughts

In summary:

• $ t^3 $ fits four points only if they align with its natural symmetry.
  
• It lacks a local minimum.
  
• A cubic that satisfies interpolation and contains a unique minimum must differ from $ t^3 $, leveraging shifts and scalings to shape the curve appropriately.

Conclusion

Therefore, the assertion that “the only way is if $ d(t) $ is not $ t^3 $, but a different cubic with a unique minimum” is absolutely correct. While $ t^3 $ frequently appears as a cubic interpolant, it fails the essential criterion of having a local minimum. A true cubic with a unique minimum exists only when the function is redefined—through parameter choices that prioritize curvature and extremum—without abandoning cubic form.

So next time you’re fitting a cubic through data with a mini-depth requirement, don’t assume $ t^3 $ is your only choice. Instead, recognize the greater flexibility—and the unique path—to a cubic that both fits and behaves.

Keywords: cubic polynomial, $ d(t) $, local minimum, interpolation, non-monotonic cubic, unique extremum, $ t^3 $, optimization, function fitting, calculus, mathematical modeling, critical points.