Prove there are $b,c$ such that $f(x)=\frac{a}{2}x^2 + bx + c$ 
Let $f:I\rightarrow \mathbb{R}$, a differetiable twice. It is given that $\forall x\in I: f''(x) = a$. Prove there are $b,c$ such that $f(x)=\frac{a}{2}x^2  + bx + c$.  

I was guided to use LMVT twice:
For the first one,   
$$f''(x) = a = \frac{f'(x_1)-f'(x_2)}{x_1-x_2}$$  
$$a(x_1-x_2) + f'(x_2) = f'(x_1)$$
We got an equation of a line. Hence, for $f'$ there is a $b\in\mathbb{R}$ such that: $f'(x) = ax+b$
I'm a little bit confused about how to apply the second LMVT in order to conclude that there is also $c$ such that $f(x) = \frac{a}{2}x^2  + bx + c$. 
I'll be glad for help here.
Thanks in advance! 
 A: I wouldn't approach this problem with the mean value theorem at all, and I'm not sure how one would use the mean value theorem to finish your argument (except in the sense that the fundamental theorem of calculus follows quickly [caution: self-promotion, mildly] from the mean value theorem and this problem is trivial with integration).

Theorem: If $f'(x) = 0$ for all $x$ in $[d,e]$, then $f(x) = c$ on $[d,e]$ for some constant, i.e. $f(x)$ is a constant function there.

Proof:
By the mean value theorem, $\displaystyle \frac{f(x) - f(x_0)}{x - x_0} = f'(\xi) = 0 \implies f(x) - f(x_0) = 0$, so that $f(x) = f(x_0). \;\diamondsuit$

Corollary: If $f'(x) - g'(x) = 0$ for all $x$ in $[d,e]$, then $f(x) = g(x) + c$ on $[d,e]$ for some constant $c$.  

Proof:
From above, $f(x) - g(x)$ is constant, meaning that $f(x) - g(x) = c$ for some constant. $\diamondsuit$

Theorem: If $f''(x) - g''(x) = 0$ for all $x$ in $[d,e]$, then $f(x) = g(x) + bx + c$ on $[d,e]$ for some constants $b,c$.

Proof:
Using the above corollary on the functions $f'(x)$ and $g'(x)$, we know that $f'(x) = g'(x) + b$ for some constant $b$. Let $G(x) = g(x) + bx$. Then $f'(x) - G'(x) = f'(x) - (g'(x) + b) = 0$. By the corollary above, we see that $f(x) = G(x) + c = g(x) + bx + c$ for some constant $c$, which is what we wanted to show. $\diamondsuit$

Claim: If $f''(x) = a$ for all $x \in [d,e]$, then $f(x) = \frac{a}{2}x^2+ bx + c$ on $[d,e]$ for some constants $b,c$.

Notice that $g(x) = \frac{a}{2}x^2 + \beta x + \gamma$ satisfies $g''(x) = a$. Suppose $f$ is another function with $f''(x) = a$. Then as $f''(x) - g''(x) = 0$, we must have that $f''(x) = g''(x) + b'x + c'$ for some constants $b', c'$, which gives us exactly what we wanted to show. $\spadesuit$
In a certain sense, I used the MVT once to show the first Theorem. The corollary follows, implicitly using the MVT once. Then the second theorem uses that corollary twice, and therefore implicitly uses the MVT twice. So in a very weak sense, this does use the MVT twice. 
