Let $f\in C^4[0,1]$ and $p$ a polynomial of degree $3$. Suppose: $$f(0)=p(0),\quad f'(0)=p'(0),\quad f(1)=p(1),\quad f'(1)=p'(1)$$ Show that for each $x\in [0,1]$ there exists $\xi\in [0,1]$: $$f(x)-p(x)=x^2(1-x)^2\cdot\frac{f^{(4)}(\xi)}{4!}$$
Work
It was suggested that I use Rolle's theorem. I tried defining $\epsilon(x)=f(x)-p(x):$
By using the information and repeatedly applying the theorem I have show that: $$\epsilon'''(\alpha)=0$$ for some $\alpha\in (0,1)$.
But this doesn't look helpful, so I am thinking that I have not used the theorem in the intended manner. I can't see how to arrive at the $4$th derivative through Rolle's.
Any ideas?
$\forall x\in (0,1)$, let $\displaystyle \phi_{x}(t)=f(t)-p(t)-\frac{f(x)-p(x)}{x^{2}(x-1)^{2}} t^{2}(t-1)^{2}$ where $t\in [0,1]$
Note that $\phi_{x}$ has three roots $t=0,x,1$.
$\exists \alpha_{i}\in (0,1)$ such that $\phi_{x}'(\alpha_{i})=0$ and $\alpha_{i} \notin \{0,x,1\}$ for $i=1,2$
But $\phi_{x}'(0)=\phi_{x}'(1)=0$,
i.e. $\phi_{x}'$ has four distinct roots.
$\exists \beta_{j}\in (0,1)$ such that $\phi_{x}''(\beta_{j})=0$ for $j=1,2,3$
$\exists \gamma_{k}\in (0,1)$ such that $\phi_{x}'''(\gamma_{k})=0$ for $k=1,2$
$\exists \xi\in (0,1)$ such that $\phi_{x}^{(4)}(\xi)=0$
$\therefore \: f^{(4)}(\xi)-p^{(4)}(\xi)-24\left( \frac{f(x)-p(x)}{x^{2}(x-1)^{2}} \right)=0$
The result follows.