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The derivation below was taken from a book on Classical Differential Geometry. It uses Rolle's theorem to find the characteristic line of a family of planes, but I don't see how it applies.

Given is a family of planes,

$\mathbf{x} \cdot \mathbf{a}(u) + p(u) = 0$

with normal vector $\mathbf{a}$ and "distance to origin" $p$ both depending on a parameter $u$ (the planes are supposed not to be parallel). The objective is to find the equation of the characteristic line, which (in my understanding) is the line of intersection of two planes whose $u$'s differ only infinitesimally. Now, the book says

The planes determined by the parameters $u_1$ and $u_2$ (with $u_1 < u_2$) intersect in a straight line, which also lies in the plane $ \mathbf{x} \cdot (\mathbf{a}(u_1)- \mathbf{a}(u_2)) + p(u_1) - p(u_2) = 0$

or, applying Rolle's theorem,

x_1a'_1(v_1) + x_2a'_2(v_2) + x_3a'_3(v_3) + p'(w) = 0

with $u_1 \le v_i \le u_2$ and $u_1 \le w \le u_2$.

I don't understand how Rolle's theorem is used here. The theorem says that, for a function $f$ that is continuous on an interval $[a, b]$ with $f(a) = f(b)$, the interval contains a point $q$ at which f'(q) = 0.

The interval seems to be $[u_1, u_2]$ in the proof above, but where is $f$ for which $f(a)=f(b)$?

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    @koletenbert: Sorry, my fault -- you're quite right. This is supposed to hold for general $\mathbf x$ on the entire plane, so the mean values have to be taken on functions independent of $\mathbf x$. My other comments were off the mark.2011-04-26

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A variation of Rolle's theorem is the following: If $f$ is realvalued, continuous on $[a,b]$ and differentiable on $]a,b[$, there exists $c \in ]a,b[$ such that f'(c)= \frac{f(b)-f(a)}{b-a}.

This is a consequence of Rolle's theorem by considering the function $g : x \mapsto f(x) - \frac{f(b)-f(a)}{b-a} (x-a)$, which satisfies $g(a)=g(b)=f(a)$ and g'(x) = f'(x) - \frac{f(b)-f(a)}{b-a} for all $x \in ]a,b[$.

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    OK, so it's the mean value theorem then... that makes sense, thank you!2011-04-26