Let $f\in C^2(\mathbb R)$. I have to prove that there is $c\in [a,b]$ s.t. $f(b)=f(a)+f'(a)(b-a)+\frac{f''(c)}{2}(b-a)^2.$ Let $f\in C^2(\mathbb R)$. I have to prove that there is $c\in [a,b]$ s.t. $$f(b)=f(a)+f'(a)(b-a)+\frac{f''(c)}{2}(b-a)^2.$$
I know that we can apply Rolle's theorem twice with $$g(x):=f(x)-f(a)-f'(a)(x-a)+\frac{f(a)+f'(a)(b-a)-f(b)}{(b-a)^2}(x-a)^2,$$
but set such a function looks so unnatural for me (I would never think to set such a function), I was wondering if there where a more intuitive way to do it.

The aim of the exercise is to prove Taylor theorem, so I can't use Taylor polynomial.
 A: First, we have:
\begin{align*}
f(b) &= f(a) + \int_a^b f'(t)\,dt \\
     &= f(a) + \int_a^b \left( f'(a) + \int_a^t f''(u)\,du \right) \,dt \\
     &= f(a) + (b-a) f'(a) + \int_a^b \int_a^t f''(u)\,du\,dt \\
     &= f(a) + (b-a)f'(a) + \int_a^b \int_u^b f''(u)\,dt\,du \\
     &= f(a) + (b-a)f'(a) + \int_a^b (b-u) f''(u)\,du.
\end{align*}
Now, since $f''$ is continuous on $[a,b]$, it has a minimum value $m$ and a maximum value $M$ on this interval.  Then
$$\frac{1}{2}(b-a)^2 m \le \int_a^b (b-u) f''(u)\,du \le \frac{1}{2}(b-a)^2 M.$$
Therefore, $\frac{2}{(b-a)^2} \int_a^b (b-u) f''(u)$ lies between $m$ and $M$, so by the intermediate value theorem, there is some $c\in [a,b]$ such that $f''(c) = \frac{2}{(b-a)^2} \int_a^b (b-u) f''(u)$.  It then follows that for this $c$,
$$f(b) = f(a) + (b-a) f'(a) + \frac{1}{2} (b-a)^2 f''(c)$$
as desired.
A: Perhaps I can explain what's going on in that auxiliary function.
The way one usually goes about proving the MVT, is that you "slant" your original function by a linear function so that the difference is zero at the end points. This sets you up to use Rolle's Theorem.
It's also useful to notice here that we use a linear polynomial because a linear polynomial is the bare minimum of what we would need to get two desiderata: we want the difference to be $0$ at $a$ and $0$ at $b$.
What's going on in that function is that it's like a "second-order slant". The first tier of this slant is similar to the first application: we want the difference to be zero at the endpoints. This will give us a point in the middle where the first derivative will be zero. But now we want another point where the derivative will be zero so that we can invoke Rolle's Theorem again. Why don't we be easy on ourselves and just stipulate where that other zero will be? Let's construct the slanting function so that the difference will be $0$ at $a$, $0$ at $b$, and so that the derivative will be $0$ at $a$. This is three criteria we want. We can use a quadratic to get these things we want. And the quadratic you have is the one that will do that.
In general, if you want $n+1$ criteria on a polynomial and its subsequent derivatives to satisfy, you can find an $n$ degree polynomial that will do the trick.
This could extend to prove Taylor's Theorem in general. The slanting function would still make the difference $0$ at the endpoints. But you would demand a much higher degree of vanishing for the derivatives at $a$.
If you really want to use induction here, you might try proving this lemma first:

Lemma: Suppose you have $n+1$ real numbers $a_0, a_1, \ldots, a_{n-1}$ and $b_0$. Then there exists a polynomial $p$ of degree $n$ on $[a,b]$ such that
  $$p(a)=a_0,\, p'(a)=a_1,\,\ldots\,,\, p^{(n-1)}(a)=a_{n-1}, \text{ and},\, p(b)=b_0\,.$$

From there you would consider $f-p$ and apply Rolle's theorem to your heart's desire.
A: Consider the taylor expansion of $f(x)$ at $x_0=a$:
$f(x)=f(a)+\frac {f'(a)} {1!}(x-a)^1 +R_2(x)$
,
Applying Lagrange`s form for the remainder we will get :
$R_1=\frac {f''(c_x)(x-a)^2} {2!}$, looking at $x=b$ we get the form needed. ($c_x \in(a,b)$)
Edit: Since you can`t use the taylor expansion, You can define a maybe more intiutive function :
$g(x)=f(x)-(f(a)+f'(a)(x-a))$, This $g(x)$ is actually the remainder function of the 2nd degree expansion of the Taylor expansion.
define $\gamma(t)=(x-a)^2$ , and apply Cauchy`s mean value theorem, to get :
$\exists r\in(x,a): \frac {g(x)-g(a)} {\gamma(x)-\gamma(a)}=\frac {g'(r)} {\gamma'(r)}$, it is easy to see $g(a)=\gamma(a)=0$, so rearranging the terms we get :
$g(x)=\frac {\gamma(x) g'(r)} {\gamma'(x)} = \frac {(x-a)^2(f'(r)-f'(a))} {2(r-a)}$, and applying the mean value again to get $\exists c\in(x,r) : g(x)=\frac{f''(c)(x-a)^2} {2}$.
$\implies \frac{f''(c)(x-a)^2} {2}=f(x)-(f(a)+f'(a)(x-a))$, set $x=b$ and rearrange the terms to get the form needed. 
I still had to use mean-value theorem twice but I hope this answer is any help.
