Prove there exists $\xi \in (a,b)$ such that $f(\xi)=f^{(n+1)}(\xi)$. Problem
Let $f(x)$ be $n$-times differentiable over $[a,b]$ and $n+1$-times differentiable over $(a,b)$. $f^{(k)}(a)=f^{(k)}(b)=0$, where $k=0,1,2,\cdots,n$. Prove there exists $\xi \in (a,b)$ such that $f(\xi)=f^{(n+1)}(\xi)$.
Attempt
Consider applying Taylor's formula expanding at $x=a,b$. We have
\begin{align*}
f(x)&amp;=f(a)+f'(a)(x-a)+\cdots+\frac{f^{(n+1)}(\xi_1)}{(n+1)!}(x-a)^{n+1}
=\frac{f^{(n+1)}(\xi_1)}{(n+1)!}(x-a)^{n+1}.
\end{align*}
and
\begin{align*}
f(x)&amp;=f(b)+f'(b)(x-b)+\cdots+\frac{f^{(n+1)}(\xi_2)}{(n+1)!}(x-b)^{n+1}
=\frac{f^{(n+1)}(\xi_2)}{(n+1)!}(x-b)^{n+1}.
\end{align*}
Can we go on from these?
Edit
Later I consulted some reference books and find a similar problem in the book named Problems In Real Analysis: Advanced Calculus On The Real Axis.

Thus, I wonder whether the conclusion holds or not ,if we are only given that   $f(x)$ is $n$-times differentiable over $[a,b]$ and $n+1$-times differentiable over $(a,b)$.
Failed attempt:

Consider the function
$$g(x)=(f(x)+f'(x)+\ldots+f^{(n)}(x))e^{-x} $$
that is continuous(!) over $[a,b]$ and differentiable over $(a,b)$ with derivative
$$\begin{align}g'(x)&amp;=(f'(x)+f''(x)+\ldots+f^{(n+1)}(x))e^{-x}-(f(x)+f'(x)+\ldots+f^{(n)}(x))e^{-x}\\&amp;=(f^{n+1}(x)-f(x))e^{-x}.\end{align}$$
As $g(a)=g(b)=0$, the Rolle theorem tells us that here exists $\xi\in(a,b)$ with $g'(\xi)=0$.

Correction:

Unfortunately, $g$ need not be continuous at $a$ or $b$, so the Rolle theorem cannot be applied. $g$ will only be continuous (and differentiable) over $(a,b)$. In other words, the Rolle theorem holds under somewhat weaker assumptions, but it is usually not presented that way - for the simple reason that Rolle is one of the first theorems about derivatives one encounters.
Here's how to mend the mistake above:

Assume $g$ is injective over $(a,b)$.
Then $g$ must be strictly monotonic over $(a,b)$ (and in particular not constant). Pick $c\in(a,b)$ with $g(c)\ne 0$, wlog. $g(c)&gt;0$ (or else replace $f$ with $-f$). Then either $g(x)&gt;g(c)$ for all $x\in(a,c)$ or for all $x\in (c,b)$; wlog the former (or else replace $f$ with $x\mapsto f(a+b-x)$).
As $e^x$ is bounded over $[a,c]$, it follows that there exist $\delta&gt;0,\epsilon&gt;0$ such that $$\tag1h(x):=e^xg(x)&gt;\epsilon$$ for all $a&lt;x&lt;a+\delta$. But
$ h(x)=f(x)+f'(x)+\ldots+f^{(n)}(x)$ is the derivative of $x\mapsto \int_a^xf(t)\,\mathrm dt+f(x)+\ldots +f^{(n-1)}(x)$ and hence has the Darboux (aka. Inermediate Value) property. In particular, as $h(a)=0$, there must exist $x\in(a,a+\delta)$ with $h(x)=\frac12\epsilon$, contradicting $(1)$.

We conclude that $g$ cannot be injective over $(a,b)$. Hence, there are $u,v$ with $a&lt;u&lt;v&lt;b$ and $g(u)=g(v)$. By Rolle, there exixts $\xi\in(u,v)$ with $g(\xi)=0$. This implies that $f^{(n+1)}(\xi)=f(\xi)$.
First, let's prove the case of $n=1$

You have two contnuous functions. $f(x)$ and $g = f'(x)$. Let's look at the interval $[a, a+ s]$ such that for the first time $f(s)=0$ That means that the derivative has different signs at $a$ and $a+s$. (The case of the function touching the line should be treated carefully, but nothing special there.) So, we have $f-g$ that is positive on the one side of $[a, a+s]$ and negative on the another. By continuousity, there is $0$ of $f-g$ somewhere.

I don't know how to make an indction step, but looks like that's possible.