Lagrange's Theorem exercise Be $u : \mathbb{R} \rightarrow \mathbb{R}$ a $C^2$ function. Proof that exists a $x \in \mathbb{R}$ with $0<x<2$ for which $u(2)-2u(1)+u(0)=u''(x)$
Applying Lagrange's Theorem I showed:
$u(2)-u(1)=u'(x_1)$ with $0<x_1<2$
$u'(x_1) - u'(0) = u''(x_2) x_1$ with $0<x_2<x_1$
$u(2)-u(1)-u'(0) = u''(x_2)x_1$
How i can finish?
Thanks

If $f \in C^2(a,b)$ then for every $x\in (a,b)$ and suitable $h>0$
$$
f(x+h)=f(x)+hf^{\prime}(x) + \frac{h^2}{2}f^{\prime\prime}(\xi_+), \qquad \xi_+ \in (x,x+h)
$$
and 
$$
f(x-h)=f(x)-hf^{\prime}(x) + \frac{h^2}{2}f^{\prime\prime}(\xi_-), \qquad \xi_- \in (x-h,x).
$$
Summing we get 
$$
f(x+h)+f(x-h) - 2f(x) = h^2\frac{f^{\prime\prime}(\xi_+) + f^{\prime\prime}(\xi_-)}{2}
$$
and by intermediate value theorem for $f^{\prime\prime}$ there exists $\xi \in (x-h,x+h)$ such that 
$$
f(x+h)+f(x-h) - 2f(x) = h^2 f^{\prime\prime}(\xi)
$$
Now just plug into the last equation $f=u$, $(a,b)=(0,2)$, $x=1$ and $h=1$. 
