Let $f:\left(a,b\right)\to \mathbb{R}$ be twice continuously differentiable and $x$,$h$ be such that $a<x-h<x+h<b$.
Show that there exists $\xi \in (x-h,x+h)$ so that $\dfrac{f(x+h)+f(x-h)-2f(x)}{h^{2}}=f''(\xi)$
My attempt was $\dfrac{f(x+h)+f(x-h)-2f(x)}{h^{2}} \\ = \dfrac{\dfrac{f(x+h)-f(x)}{h}-\dfrac{f(x)-f(x-h)}{h}}{h} \\ =\dfrac{f'(\xi_{1})-f'(\xi_{2})}{h} \\ =(\dfrac{f'(\xi_{1})-f'(\xi_{2})}{\xi_{1}-\xi_{2}})(\dfrac{\xi_{1}-\xi_{2}}{h})\\ =f''({\xi_{3}})\dfrac{\xi_{1}-\xi_{2}}{h}$.
I could not progress I am sorry if something is obvious. I will be pleased if you show me the last idea or write your own. I just applied mean value theorem while finding numbers $\xi_{1}$ and $\xi_{2}$ and $\xi_{3}$.$\xi_{1}$ comes from the mean value theorem applied in $(x,x+h)$. $ \\ \xi_{2}$ comes from the mean value theorem on the interval $(x-h,x)$ and $\xi_{3}$ comes from the mean value theorem applied in $(\xi_{1},\xi_{2})$.