= Solution
If the bound is to hold with finite $c$, the error functional must vanish on every <polynomial> whose fourth <derivative> is zero. Thus the scheme must be exact for degrees zero through three. Applying it to powers of $t=x+1$ gives
$$
\begin{aligned}
a_{-1}+a_0+a_1+a_2&=0,\\
a_0+2a_1+3a_2&=0,\\
a_0+4a_1+9a_2&=2,\\
a_0+8a_1+27a_2&=0.
\end{aligned}
$$
Solving,
$$
\boxed{a_{-1}=2,
\qquad a_0=-5,
\qquad a_1=4,
\qquad a_2=-1}.
$$
This is the <four-point one-sided second-derivative formula>.
The <Peano kernel theorem> says that if a linear functional $L$ annihilates all <polynomials> of degree below $r$, then for $f\in C^r[a,b]$,
$$
L(f)=\int_a^bK(t)f^{(r)}(t)\,dt,
\qquad
K(t)=L\left(\frac{(x-t)_+^{r-1}}{(r-1)!}\right).
$$
Here $r=4$ and $L=e$. Under the stated nonnegativity assumption,
$$
|e(f)|\leq
\left(\int_{-1}^2K(t)\,dt\right)
\max_{[-1,2]}|f^{(4)}|.
$$
The <integral> equals $e(q)$ for $q(x)=(x+1)^4/24$, since $q^{(4)}=1$. Now $q''(-1)=0$ and
$$
\eta(q)=\frac{-5+4\cdot16-81}{24}=-\frac{11}{12}.
$$
Therefore
$$
\int_{-1}^2K(t)\,dt=e(q)=\frac{11}{12}.
$$
Equality is attained by $q$, so the <sharp Peano-kernel constant for the four-point endpoint second derivative> is
$$
\boxed{c=\frac{11}{12}}.
$$
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