Generalized Rolle's Theorem Confusion This is my first post here (I hope that this hasn't been asked / answered before).  First let me state the Generalized Rolle's Theorem as it is presented to me, then I'll ask my question.
(Generalized Rolle's Theorem)  Assume that $f \in C[a, b]$ and that $f'(x), f''(x), \dotsc, f^{(n)}(x)$ exist over $(a, b)$ and $x_0, x_1, \dotsc, x_n \in [a, b]$. If $f(x_j) = 0$ for $j = 0, 1, \dotsc, n$, then there exists a number $c$, with $c \in (a, b)$, such that $f^{(n)}(c) = 0$.
I was thinking about the simple example $f(x) = x$, where $x \in [0, 1]$. If we let $x_0 = 0 \in [0, 1]$, then the conditions for the Generalized Rolle's Theorem are satisfied, but there is no number $c \in (0, 1)$ such that $f^{(0)}(c) = f(c) = 0$. What am I doing wrong?  Is the answer that $n \geq 1$?  Shouldn't they state this?
 A: Yes, you need $n \geq 1$.  I think you may be getting tripped up by the indexing: maybe it's clearer to say that: for all $n \geq 1$, if there are $a \leq x_1 \leq \ldots \leq x_{n+1} \leq b$ such that $f(x_i) = 0$ for all $i$, and $f$ is continuous on $[a,b]$ and $n$ times differentiable on $(a,b)$ then there is $c \in (a,b)$ with $f^{(n)}(c) = 0$.  That is, the number of zeros needs to be one more than the number of the derivative which is asserted to have a zero.
Do you know/understand the proof?  When the proof is simple enough, that's a good way of checking the accuracy of the statement.  The usual Rolle's Theorem tells you that in each of the $n$ open intervals $(x_i,x_{i+1})$ for $1 \leq i \leq n$ there is a zero $y_1$ of $f'$.  Now you apply Rolle's Theorem on each of the $n-1$ intervals $(y_i,y_{i+1})$ to get $n-2$ zeros of $f''$.  And so forth: each time you pass from one derivative to the next, the number of zeros you can guarantee decreases by $1$.  Since you started with $n+1$ zeros, that's just enough to get one zero of $f^{(n)}$ on $(a,b)$.  (Depending upon your taste, you might want to formalize this argument via induction.)  
A: As you have noticed, what you have written is false. If we generalize Rolle's theorem to higher dimensions, the appropriate conclusion is drawn only regarding the highest ($n$th) derivative.
