A course problem asks me, assuming that $f$ is $C^3$ on $\mathbb{R}$ (and $f'''$ is bounded and continuous on $\mathbb{R}$), to show that
$$\left| \frac{-3f(x) + 4f(x+h) - f(x + 2h)}{2h} - f'(x) \right| \le \frac{1}{3} |\max_{x \in \mathbb{R}}(f'''(x))||h^2|.$$
I rewrote the expression as $$\left|\left[\frac{f(x+h) - f(x)}{h} - f'(x) \right] - \left[\frac{-f(x) + 2f(x+h) - f(x + 2h)}{2h} \right] \right|$$
and then since $f(x+h) = f(x) + f'(x)h + \frac{f''(x)}{2}h^2 + \frac{f'''(\xi)}{6}h^3$, the left bracketed term is $\frac{f''(x)}{2}h + \frac{f'''(\xi)}{6}h^2$; and since $f(x+2h) = f(x) + 2f'(x)h + 2f''(x)h^2 + 4\frac{f'''(\zeta)}{3}h^3$, the right bracketed term simplifies as $\frac{f''(x)}{2}h + \frac{2f'''(\zeta)}{3}h^2 - \frac{f'''(\xi)}{6}h^2$. Thus, the absolute value of their difference is $$\left|\left[\frac{f''(x)}{2}h + \frac{f'''(\xi)}{6}h^2 \right] - \left[\frac{f''(x)}{2}h + \frac{2f'''(\zeta)}{3}h^2 - \frac{f'''(\xi)}{6}h^2 \right] \right|$$ $$ = \left|\frac{f'''(\xi)}{3}h^2 - \frac{2f'''(\zeta)}{3}h^2\right|.$$
I can't see any way from here of obtaining a remainder as small as the one the problem asks for. Assuming the problem is well posed, how can I get that remainder?
Sorry, guys. The instructor said it was a typo. There's no better bound.