 Solution

The finite sum of truncated cubes is $C^2$ by part (b) and is cubic on each knot interval. For $x\le 0$, every term vanishes. For $x\ge 4h$, every truncated power equals its ordinary polynomial, so part (a) applied to $P(x)=x^3$ givesIt is therefore zero on both exterior intervals. Its $C^2$ regularity makes the endpoint second derivatives zero too, proving naturality.

Evaluation at the five knots givesFor example the three interior values in units of $h^3$ are $1$, $8-4=4$, and $27-4\cdot 8+6=1$. At $4h$ the cancellation is $64-4\cdot 27+6\cdot 8-4=0$.

For $h=1$ an explicit graph description isIt is symmetric about $x=2$, increases from zero to its maximum four at the centre, and decreases symmetrically back to zero. It joins the zero tails with zero first and second derivatives.