I started reading "The Mathematics of nonlinear Programming - AL peressini" recently.
Theorem : Suppose that $f(x) , f'(x) ,f''(x)$ exist on the closed interval $[a,b]$. If $x^* ,x $ are two different points of $[a,b]$ , then there exists a point $z$ strictly between $x^*$ and $x$ such that
$$f(x)=f(x^*)+f'(x^*)(x-x^*)+\frac{f''(z)}{2}(x-x^*)^2$$
I tried to prove this theorem using The Fundamental theorem of calculus. But I can not stop it at the second derivative . Do I have to use mean value theorem ? or is there a simple proof for this ?
Let $$F(t)=f(t)-f(x^*)-f'(x^*)(t-x^*)-K(t-x^*)^2$$
where let's set K such that $$0=f(x)-f(x^*)-f'(x^*)(x-x^*)-K(x-x^*)^2$$
So we know that $F(x)=0$ and $F(x^*)=0$
By Rolle's rule, there is a $c1 \in (x, x^*)$ such that $F'(c1)=0$
Now look at the derivative $$F'(t) = f'(t) - f'(x^*)-K \cdot 2(t-x^*)$$
We know $F'(x^*)=0$ and $F'(c1)=0$, so again applying Rolle's theorem, there is a $c2 \in (c1, x^*) \subset (x, x^*)$ such that $F''(c2)=0$
$$F''(t) = f''(t) -2K$$ Therefore $$F''(c2)=0 \rightarrow f''(c2) -2K=0$$
This gives $K=f''(c2)/2$ and you can complete the proof.