Show that $|E(x)|\le 0.65 (x_1-x_0)^3M$ for the error of interpolation of a 2nd degree polynomial 
Show that on the polynomial interpolation of a function $f \in C^3$ in
3 points $x_0 < x_1 < x_2$ with $x_2-x_1 = x_1 - x_0$, the error of
interpolation verifies $|E(x)| \le 0.65(x_1-x_0)^3M, \forall x$
between $x_0$ and $x_2$, where $M$ is the absolute value of the
maximum of the 3rd derivative of $f$ on the interval $[x_0,x_2]$.

I did
$$\begin{align}
|E(x)| &\le 0.65(h)^3M \\
|E(p_2(x))| &\le \max_{x_2,x_0}|(x-x_0)(x-x_1)(x-x_2)|M/3! = \\
\end{align}$$
$$\begin{align}
 |(x_0+h/2-x_2)(x_0+h/2-x_1)(x_0+h/2-x_2)|M/6&\le 0.65(h)^3M  = \\
|(h/2)(-h/2)(3h/2)|M/6 &\le 0.65h^3M \Leftrightarrow \\
3h^3/8* M/6 &\le 0.65h^3M
\end{align}$$
...q.e.d?
How I got to x being equal to $x_0+h/2$ where h = $x_1-x_0$: I drew some 3rd degree polynomials and "kind of" noticed they were symmetrical and that $x_0$, $x_1$ and $x_2$ were always the zeroes. I also noticed that $f(x_1)$ was always equal to zero and that it stood halfway between the other two, and between each, there was a maximum or minimum. So I chose $x = x_0+h/2$.
Is this correct?
 A: So, you wrote the error formula
$$
|E_2(x)| \leq |(x-x_0)(x-x_2)(x-x_2)| \frac{M}{3!}.
$$
Now you can simply maximize the 3rd degree polynomial... The maximum will be attained at a stationary point, that you can compute explicitly. The stationary points are
$x_0 + (1 \pm \frac{\sqrt{3}}{3})h$, and the corresponding maximum value will be $\frac{2h^3}{3 \sqrt{3}}$. In general, your choice of $x = x_0 +\frac h2$ is wrong and does not yield the maximum.
Using this information,
$$
|E_2(x)|\leq \frac{M h^3}{9 \sqrt{3}}\approx 0.06415 (x_1-x_0)^3 M.
$$
Finally, if this estimate holds, so does the (much worse) proposed estimate.

Regarding the maximization of $|(x-x_0)(x-x_1)(x-x_2)|$ in the interval $[x_0, x_2]$, as noted by Lutz Lehmann, it is equivalent to the maximization of $g(s)=|(s^3-s)h^3|$ for $s \in [-1,1]$. Since $g$ is zero on the boundary and at the branching point for the absolute value, the maximum will occur at a point where $g'(s)=0$, i.e. $s= \pm \frac{\sqrt{3}}{3}$. Substituting back in the function, we conclude that the maximum is $g(\pm \frac{\sqrt{3}}{3}) = \frac{2h^3}{3\sqrt{3}}$.
