Let f be a continuous function and its third derivative exists. I need to show that [ -f(x+2h) + 4f(x+h) - 3f(x)] / 2h = f’(x) + O[h^2].

I know how to show that -f(x+2h) + 4f(x+h) - 3f(x) = 2h f’(x) by using the Taylor expansions:

f(x+2h) = f(x) + 2h f’(x) + [(2h)^2] f’’(x) / 2! + O[(2h)^3],

f(x+h) = f(x) + h f’(x) + [h^2] f’’(x) / 2! + O[h^3].

Note that O[h^(n+1)] is the remainder of the Taylor expansion, and O[h^(n+1)] = f^(n+1)(z_n) h^(n+1) / (n+1)!, where z_n is some number between x and c (since h = x-c).

How can I prove that we get O[h^2] in the first equation above? I cannot prove it by directly using the definition above. Please help me answer this question. If you know any links that may help me, please give me.