Let $n\in\mathbb{N}$. The function $f:[a,b]\rightarrow \mathbb{R}$ is n-times differentiable. Let $x_1,...,x_{n+1} \in [a,b]$ with $x_1<...<x_{n+1}$ and $f(x_i)=0$ for all $1\leq i \leq n+1$.
Prove that there exists a $x_0 \in (a,b)$ for which $f^{(n)}(x_0)=0$ is.
Here's what I've came up with so far:
We are finished if $f(x)= \hat{0}$. Therefore, let $f(x) \neq \hat{0}$. Because $f(x_i)=0$, $i \in \{1,...,n+1\}$, there exists a $U_{\delta}(x_i)$, for which all $x_j\in U_{\delta}(x_i)$, $j \in [a,b]$ and $j\neq i$ are either $j>0$ or $j<0$.
Let $x_0 \in [a,b]$.
Then: $f(x_0) \geq 0$ or $f(x_0) \leq 0$.
If $x_i$ are the local Maxima or Minima and $f^{(1)}(x_i)=...=f^{(n)}(x_i)=0$, we make $f^{(n+1)}(x_i)$ and check for the criteria of local Maxima and Minima and put $x_0 = x_i$.
Are $x_i$ no local extrema, then $x_0$ has to be between the zeros of $f(x)$ and there exists a $x_0$ which $f^{(1)}(x_0)=...=f^{(n)}(x_0)=0$.
Now , I know this prove has lots of potential for improvement, but I would like some feedback as I am quite new to Proves and want to improve.
