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So I have this question on a homework and I just can't seem to figure it out.

Let $f \in C^4 [0,1]$ and let $p$ be a polynomial of degree $\le 3$ such that $p(0) = f(0)$, $p(1) = f(1)$, $p'(0) = f'(0)$ and $p'(1) = f'(1)$. Show that for all $x \in [0,1]$ there exists $c \in [0,1]$ such that the following holds : $$ f(x) = p(x) + \frac 1{24} x^2(x-1)^2 f^{(4)}(c). $$

Now what I've tried up to now is considering the function $g_x : [0,1] \to \mathbb R$ defined by $$ g_x(t) = f(t) - p(t) - \frac{f(x) - p(x)}{x^2(x-1)^2} (t^2(t-1)^2). $$ Since $g_x(0) = g_x(1) = g'_x(0) = g'_x(1) = 0$, I have a $c \in [0,1]$ such that $g_x^{(3)}(c) = 0$ by applying Rolle's Theorem repeatedly, but to solve my problem I need a zero of $g_x^{(4)}(c)$ and I can't seem to get my way around this thing.

(Note that since $p$ has degree $\le 3$, $p^{(4)}(t)$ is identically zero so rearranging the terms for $g^{(4)}(c)$ gives me what I want.)

Any hints?

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    Reminds me Lagrange remainder for Taylor series: http://en.wikipedia.org/wiki/Taylor's_theorem#Explicit_formulae_for_the_remainder2011-11-01
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    It seems you've mixed up the quantifiers. The way you've written it, $f$ would have to be the sum of two polynomials. I suspect it should say that for all $x\in[0,1]$ there is $c\in[0,1]$ such that ...?2011-11-01
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    Patrick, do you mean $p'(1)=f'(1)$ as the final condition? If so, this is a special case of [Hermite Interpolation](http://en.wikipedia.org/wiki/Hermite_interpolation).2011-11-01
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    http://math.stackexchange.com/questions/73617/existence-and-uniqueness-of-hermite-interpolation-polynomial2011-11-01
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    I noticed that it is a special case of Hermite interpolation, but I still don't know how to do this. =(2011-11-01

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