Rolle's theorem

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 stationary point somewhere between them.

tandard version of the theorem

If a real-valued function "f" is continuous on a closed interval ["a","b"] , differentiable on the open interval ("a","b"), and "f"("a") = "f"("b"), then there is some real number "c" in the open interval ("a","b") such that

:$f\text{'}(c)\; =\; 0,.$

This version of Rolle's theorem is used to prove the mean value theorem, Rolle's theorem is indeed a special case of it.

History

A version of the theorem was first stated by the Indian astronomer Bhāskara II in the 12th century. [ [http://www-history.mcs.st-andrews.ac.uk/history/Projects/Pearce/Chapters/Ch8_5.html 8 V. Bhaskaracharya II ] ] [citation|first=T. A. A.|last=Broadbent|title=Reviewed work(s): "The History of Ancient Indian Mathematics" by C. N. Srinivasiengar|journal=The Mathematical Gazette|volume=52|issue=381|date=October 1968|pages=307-8] The first known formal proof was offered by Michel Rolle in 1691, which used the methods of the differential calculus.

Examples

First example

For a radius "r" > 0 consider the function

$f(x)=sqrt\{r^2-x^2\},quad\; xin\; [-r,r]\; .$

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.

econd example

function

:$f(x)\; =\; |x|,qquad\; xin\; [-a,a]\; .$

Then "f"(−"a") = "f"("a"), but there is no "c" between −"a" and "a" 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.

Generalization

The second example illustrate 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

:$f\text{'}(x+):=lim\_\{hsearrow0\}frac\{f(x+h)-f(x)\}\{h\}$

and the left-hand limit

:$f\text{'}(x-):=lim\_\{h\; earrow0\}frac\{f(x+h)-f(x)\}\{h\}$

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

:$f\text{'}(c+)quad\; ext\{and\}quad\; f\text{'}(c-)$

is ≥ 0 and the other one is ≤ 0. 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.

Remarks

#If "f" is convex or concave, then the right- and left-hand derivatives exist at every inner point, hence the above limits exist and are real numbers.
#This generalized version of the theorem is sufficient to prove convexity when the one-sided derivatives are monotonically increasing: [cite book
last = Artin
first = Emil
authorlink = Emil Artin
others = trans. Michael Butler
title = The Gamma Function
origyear = 1931
year = 1964
publisher = Holt, Rinehart and Winston
pages = pp. 3-4 ]

::

Proof of the generalized version

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 is maximum at "c". Therefore, for every "h" > 0,

:$frac\{f(c+h)-f(c)\}\{h\}le0,$

hence

:$f\text{'}(c+):=lim\_\{hsearrow0\}frac\{f(c+h)-f(c)\}\{h\}le0,$

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

:$frac\{f(c+h)-f(c)\}\{h\}ge0,$

hence

:$f\text{'}(c-):=lim\_\{h\; earrow0\}frac\{f(c+h)-f(c)\}\{h\}ge0,$

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.

Generalization to higher derivatives

We can also generalize Rolle's theorem by requiring that "f" has more points with equal values and greater regularity. Specifically, suppose that
* the function "f" is "n" − 1 times continuously differentiable on the closed interval ["a","b"] and the "n"^th derivative exists on the open interval ("a","b"), and
* there are "n" intervals given by "a"₁ < "b"₁ &le; "a"₂ < "b"₂ &le; . . .&le; "a_n" < "b_n" in ["a","b"] such that "f"("a_k") = "f"("b_k") for every "k" from 1 to "n".Then there is a number "c" in ("a","b") such that the "n"^th derivative of "f" at "c" is zero.

Of course, the requirements concerning the "n"^th 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".

Proof

The proof uses mathematical induction. For "n" = 1 it is just 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 "c_k" in the open interval ("a_k","b_k") such that "f' "("c_k") = 0. Hence the first derivative satisfies the assumptions with the "n" − 1 closed intervals ["c"₁,"c"₂] , . . ., ["c"_"n"−1,"c_n"] . By the induction hypothesis, there is a "c" such that the ("n" − 1)^st derivative of "f' " at "c" is zero.

See also

*Mean value theorem
*Extreme value theorem
*Intermediate value theorem
*Linear interpolation

Footnotes

External links

* [http://www.cut-the-knot.org/Curriculum/Calculus/MVT.shtml Rolle's and Mean Value Theorems] at cut-the-knot