TITLE: Mean value theorem for twice differentiable function
QUESTION [6 upvotes]: Let $f:(0,\infty)\to \Bbb R$ be a twice differentiable function. In this answer, it is asserted that the MVT lets one write $$f(x+h)=f(x)+f'(x)h+\frac12 f''(\xi)h^2$$ for some $\xi\in (x,x+h)$.
It is not clear to me why this should be the case. Using the MVT, one can write $f''(\xi)h=f'(x+h)-f'(x)$ for some $\xi\in (x,x+h)$. Using that, the claim rephrases to $$f(x+h)=f(x)+\frac{1}{2}f'(x)h+\frac{1}{2}f'(x+h)h$$ and I don't see why that should hold.
I'm sure I'm being stupid, therefore I much welcome clarification.

REPLY [2 votes]: The basic arguments of the exact proof (which I do not give here) that this works, go as follows: Taylor's theorem says that $$f(x+h)=f(x)+f'(x)h+\underbrace{\frac12f''(x)h^2+ο(h^2)}$$ If we replace $x$ with $ξ$ (where $ξ$ is given by the MVT) in the underlined term we can get rid of the remainder $ο(h^2)$ and achieve an exact calculation of $f(x+h)$: 
$$f(x+h)=f(x)+f'(x)h+\frac12f''(ξ)h^2$$ Why then the approximation in the first place? The MVT says there exists such an $ξ$ but does not give a way to find it, so the approximation in Taylor's theorem is indeed useful.