It states that, for smooth curves, if the function is equal at two points there must be a stationary point somewhere between them. All the assumptions are necessary. For example, if f(x) = |x|, the absolute value of x, then we have that f(-1) = f(1), but there is no x between -1 and 1 for which f '(x) = 0.
Rolle's Theorem is used in proving the mean value theorem, which can be seen as a generalisation of it.
Proof of Rolle's Theorem: The idea of the proof is to argue that if f(a) = f(b) then f must attain either a maximum or a minimum somewhere between a and b, and f ' (x) = 0 at either of these points.
Now, by assumption, f is continuous on [a , b], and by the continuity property is bounded and attains both its maximum and its minimum at points of [a , b]. If these are both attained at endpoints of [a , b] then f is constant on [a , b] and so f ' (x) = 0 at every point of (a , b).
Suppose then that the maximum is obtained at an interior point x of (a , b) ( the argument for the minimum is very similar). We wish to show that f ' (x) = 0. We shall examine the left-hand and right-hand derivatives separately.
For y just below x, ( f(x) - f(y) ) / (x - y) is non-negative, since x is a maximum. Thus the limit limy->x- is non-negative. (Note that we assume that f is differentiable to guarantee that the left-hand and right-hand derivatives exist; it does not follow from the other assumptions).
For y just greater than x, ( f(x) - f(y) ) / (x - y) is non-positive. Thus limy->x+ is non-positive.
Finally, since f is differentiable at x, these two limits must be equal and hence are both 0. This implies that f ' (x) = 0.
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