= Solution
The <Peano kernel theorem> says that if a continuous linear error functional $L$ annihilates all polynomials of degree less than $r$ and permits the required interchange with integration, then for $f\in C^r[a,b]$,
$$
L(f)=\int_a^bK(\theta)f^{(r)}(\theta)\,d\theta,\qquad
K(\theta)=L\left[\frac{(x-\theta)_+^{r-1}}{(r-1)!}\right].
$$
It follows by applying $L$ to the integral-remainder <Taylor formula with integral remainder>. Here choose $r=2$ and
$$
L(f)=f(1)-\frac13f(0)-f(2)+\frac13f(3).
$$
This functional annihilates constants and linear functions (indeed quadratics too), and its <Peano kernel> is
$$
K(\theta)=(1-\theta)_+-(2-\theta)_++\frac13(3-\theta)_+
=\begin{cases}
-\theta/3,&0\leq\theta\leq1,\\
2\theta/3-1,&1\leq\theta\leq2,\\
1-\theta/3,&2\leq\theta\leq3.
\end{cases}
$$
Thus $\int_0^3|K|\,d\theta=1/2$ and $\max|K|=1/3$. The integral representation gives
$$
\boxed{\alpha_{\min}=\frac12,\qquad\beta_{\min}=\frac13.}
$$
For the first inequality, continuous functions $f''$ with $\|f''\|_\infty\leq1$ can approximate $\operatorname{sgn}K$ away from its zeros, making $L(f)$ arbitrarily close to $\int|K|$. Such derivatives integrate twice to admissible $C^2$ functions, proving no smaller $\alpha$ works. For the second, use continuous $f''$ of $L^1$ norm one concentrated near $\theta=1$ or $2$, where $|K|=1/3$, with the sign of $K$ there. This makes $|L(f)|$ approach $1/3$, proving optimality of $\beta$.
For $f=x^3$, $L(f)=2$, $\max|f''|=18$ and $\|f''\|_1=27$. Both right-hand bounds equal nine, so \b[both inequalities hold].
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