Rolle's theorem - Wikipedia

Rolle's theorem is a mathematical theorem; developed by Rolle, and published in 1691. This concept is, perhaps, best described with a parabola (see the below image); the theorem argues that if a continually differentiable curve has two algebraic points, with equal y-values; then, at some point between the first two points, the curve has a slope of zero.

For a real valued function of one variable, f(x); if P and Q are two points on the graph of said function, and these points are equidistant from the x-axis (ie; have the same y-value); then, there is a point, R, on the graph, between P and Q; such that, the tangent to the graph ( through the point R) is both horizontal, and parallel, to the x-axis.

Assume that a function, f(x), is continuous within the closed interval [a,b], differentiable within the open interval ]a,b[, and that f(a) = f(b). If these clauses are satisfied; there exists a point, c (within the open interval ]a,b[ ); such that, the value of f '(c) = 0.

Or symbolically:

<math>f(x) \in [a,b], f'(x) \in ]a,b[, f(a) = f(b) \Rightarrow \exists c, f'(c) = 0</math>

Or:

Let f : [a , b] -> R be continuous, on the interval [a , b], differentiable on (a , b), and suppose that f(a) = f(b); then, there exists some x (in (a , b) ); such that, f ' (x) = 0.

Table of contents showTocToggle("show","hide")

1 Example

2 Note on Diferentiability

3 See Also

4 References

Example

Consider the graph of x²...For any two points (on this graph), having equivalent y-values, there is the point (0,0), where the slope is zero.

Note on Diferentiability

Should one remove the clause that a function be differentiable, upon the open interval ]a,b[; then, f(x) will have a critical point at R; although, it might not have a horizontal tangent.

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 lim_y->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 lim_y->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.

Generalization: The theorem is usually stated in the form above, but it is actually valid in a slightly more general setting: We only need to assume that f : [a , b] -> R is continuous on [a , b], that f(a) = f(b), and that for every x in (a , b) the limit lim_h->0 (f(x+h)-f(x))/h exists or is equal to +/- infinity.

References

KarlsCalculus.Org (http://www.karlscalculus.org/calc5_3.html)