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Let $p(x)$ such that $\deg(p) \le 2n+1$ and $$p(x_i) = f(x_i) \\ p'(x_i) = f'(x_i)$$ $i=0,\ldots, n$.
Show that $p(x)$ is the only polynomial satisfying those demands for all polynomials with a degree $\le 2n+1$.
Assumption: for every $j\ne k$: $x_j \ne x_k$.

So let's assume by contradiction that there's another polynomial, $q(x)$ satisfying those constraints.

I'm not sure why but I think we would like to assume WLOG that $p(x),q(x)$ are monic.

Anyways, let's look at $h(x) = p(x) - q(x)$.

We notice that for $x_i$:$$h(x_i) = p(x_i) - q(x_i) = f(x_i) - f(x_i) = 0$$

So $h(x)$ has at least $n$ roots, and therefore, $h'(x)$ has at least $n-1$ roots .

My goal is to show that $h(x)\equiv 0$ by showing that there're more than $2n$ roots (I think $\deg(h(x)$ should be $\le 2n$ since we looked at monic polynomials)

How should I proceed?

EDIT
Basically, the same applies for the derivative:

$$h'(x_i) = p'(x_i) - q'(x_i) = 0$$

but aren't those the same roots as before?

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    You may need additional constraints for the claim to hold. For example, the $x_i$ better be distinct. And, no, you cannot assume monic – 2017-02-23
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    Alright. I'll add the the distinct constraint to my question (maybe it was omitted but was the intention of the author) – 2017-02-23
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    WLOG assume $x_1$h(x_1)=0, h(x_2)=0$ so Rolle's theorem tells us that $h'(x)=0$ for some $x_1$n+1+n=2n+1$ roots to $h'(x)=0$. So $h$ would have degree $2n+2$ ot higher. Contradiction. – 2017-02-23
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    @daruma, how did you get $2n+1$? I'm missing it.. – 2017-02-23
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    $n+1$ from $x_0, x_1, ....,x_n$ and then $n$ from the Rolle's theorem – 2017-02-23
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    right.... ! Cheers :) – 2017-02-23
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    Between each interval $(x_i, x_{i+1})$ , there is a root to $h'(x)=0$ by Rolle. – 2017-02-23

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Wlog. $x_0