First Order Lagrange Remainder Using Ordinary MVT I would like to show my Calc I class that $f(x)=f(a)+f'(a)(x-a)+(f''(c)/2)(x-a)^2$ for some $c$ in $(a,x)$ (for $f$ smooth). Just from the form of the statement, it seems as though this should be possible using only the ordinary MVT, with no integrals. However, I can't seem to get the symbols to behave properly; all my attempts have ended with remainder term $f''(c)(d-a)(x-a)$ for some $d$ in $(a,x)$, which is also correct but not the formula that I want. Help me out or let me know if what I seek is unreasonable. Thanks.
 A: Let $$g(x)=f(b)-f(x)-(b-x)f'(x)-\left(\frac{b-x}{b-a}\right)^2[f(b)-f(a)-(b-a)f'(a)]$$
Then $$g'(x)=\frac{2(b-x)}{(b-a)^2}[f(b)-f(a)-(b-a)f'(a)-\frac{1}{2}(b-a)^2f''(x)]$$
(Differentiate, then factor out $\frac{2(b-x)}{(b-a)^2}$.)
Applying the mean value theorem, $g(b)-g(a)=g'(c)(b-a)$ for some $a<c<b$. $g(a)=g(b)=0$, so $g'(c)$ must equal zero for some $a<c<b$. Therefore, $$f(b)=f(a)+(b-a)f'(a)+\frac{1}{2}(b-a)^2f''(c)$$ for some $a<c<b$.
This proof is from Hardy's "A Course of Pure Mathematics".
A: Another way to solve the problem: Let
$$F(x)=f(x)-f(a)-f'(a)(x-a)-k(x-a)^2$$
where $k$ is a constant determined later. Then $F(a)=0$. Let $F(x)=0$; then we have
$$ k=\frac{f(x)-f(a)-f'(a)(x-a)}{(x-a)^2}. $$
Since $F(a)=F(x)=0$ and $F$ is differentiable in $(a,x)$, by Rolle's Theorem, there is $c\in(a,x)$ such that $F'(c)=0$, namely
$$ f'(c)-f'(a)-2(c-a)k=0. $$
By Langrange's MVT, there is $\bar{c}\in(a,c)$ such that
$$ f'(c)-f'(a)=f''(\bar{c})(c-a). $$
Thus we have
$$ f''(\bar{c})(c-a)-2(c-a)k=0 $$
or $k=\frac{1}{2}f''(\bar{c})$. This completes the proof.
