A theorem stating that if a curve is continuous, has two x-intercepts, and has a tangent at every point between the intercepts, at least one of these tangents is parallel to the x-axis.
[After Michel Rolle (1652-1719), French mathematician.]
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Rolle's theorem (rôlz, rōlz) ![]() |
[After Michel Rolle (1652-1719), French mathematician.]
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In calculus, a branch of mathematics, Rolle's theorem essentially states that a differentiable function, which attains equal values at two points, must have a point somewhere between them where the slope of the tangent line to the graph of the function is zero.
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If a real-valued function ƒ is continuous on a closed interval [a, b], differentiable on the open interval (a, b), and ƒ(a) = ƒ(b), then there exists a c in the open interval (a, b) such that

This version of Rolle's theorem is used to prove the mean value theorem, of which Rolle's theorem is indeed a special case.
A version of the theorem was first stated by the Indian astronomer Bhāskara II in the 12th century.[1][2] The first known formal proof was offered by Michel Rolle in 1691, which used the methods of differential calculus.
The name "Rolle's theorem" was first used by M.W.Drobisch of Germany in 1834 and by Giusto Bellavitis of Italy in 1846.[3]
For a radius r > 0 consider the function
![f(x)=\sqrt{r^2-x^2},\quad x\in[-r,r].](http://wpcontent.answers.com/math/4/a/c/4ac5d53f9d43cab7da5254aba47b23f8.png)
Its graph is the upper semicircle centered at the origin. This function is continuous on the closed interval [−r,r] and differentiable in the open interval (−r,r), but not differentiable at the endpoints −r and r. Since f(−r) = f(r), Rolle's theorem applies, and indeed, there is a point where the derivative of f is zero.
If differentiability fails at an interior point of the interval, the standard version of Rolle's theorem can fail. Consider the absolute value function
![f(x) = |x|,\qquad x\in[-1,1].](http://wpcontent.answers.com/math/2/d/e/2de0c1532aae33d33bed7fef1a84a1f6.png)
Then f(−1) = f(1), but there is no c between −1 and 1 for which the derivative is zero. This is because that function, although continuous, is not differentiable at x = 0. Note that the derivative of f changes its sign at x = 0, but without attaining the value 0.
The second example illustrates the following generalization of Rolle's theorem:
Consider a real-valued, continuous function f on a closed interval [a,b] with f(a) = f(b). If for every x in the open interval (a,b) the right-hand limit

and the left-hand limit

exist in the extended real line [−∞,∞], then there is some number c in the open interval (a,b) such that one of the two limits

is ≥ 0 and the other one is ≤ 0 (in the extended real line). If the right- and left-hand limit agree for every x, then they agree in particular for c, hence the derivative of f exists at c and is equal to zero.

Since the proof for the standard version of Rolle's theorem and the generalization are very similar, we prove the generalization.
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, say at c, and the function must change from increasing to decreasing (or the other way around) at c. In particular, if the derivative exists, it must be zero at c.
By assumption, f is continuous on [a,b], and by the extreme value theorem attains both its maximum and its minimum in [a,b]. If these are both attained at the endpoints of [a,b], then f is constant on [a,b] and so the derivative of f is zero at every point in (a,b).
Suppose then that the maximum is obtained at an interior point c of (a,b) (the argument for the minimum is very similar, just consider −f ). We shall examine the above right- and left-hand limits separately.
For a real h such that c + h is in [a,b], the value f(c + h) is smaller or equal to f(c) because f attains its maximum at c. Therefore, for every h > 0,

hence

where the limit exists by assumption, it may be minus infinity.
Similarly, for every h < 0, the inequality turns around because the denominator is now negative and we get

hence

where the limit might be plus infinity.
Finally, when the above right- and left-hand limits agree (in particular when f is differentiable), then the derivative of f at c must be zero.
We can also generalize Rolle's theorem by requiring that f has more points with equal values and greater regularity. Specifically, suppose that
Then there is a number c in (a,b) such that the nth derivative of f at c is zero.
Of course, the requirements concerning the nth derivative of f can be weakened as in the generalization above, giving the corresponding (possibly weaker) assertions for the right- and left-hand limits defined above with f (n−1) in place of f.
The proof uses mathematical induction. For n = 1 is simply the standard version of Rolle's theorem. As induction hypothesis, assume the generalization is true for n − 1. We want to prove it for n > 1. By the standard version of Rolle's theorem, for every integer k from 1 to n, there exists a ck in the open interval (ak,bk) such that f' (ck) = 0. Hence the first derivative satisfies the assumptions with the n − 1 closed intervals [c1,c2], . . ., [cn−1,cn]. By the induction hypothesis, there is a c such that the (n − 1)st derivative of f' at c is zero.
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