NationMaster - Encyclopedia: Rolle's theorem

In calculus, Rolle's theorem states that if a function f is continuous on a closed interval $left[a,bright]$ and differentiable on the open interval $left(a,bright)$ , and $fleft(aright) = fleft(bright)$ then there is some number c in the open interval $left(a,bright)$ such that Image File history File links Rolle's_theorem. ... For other uses, see Calculus (disambiguation). ... In mathematics, a continuous function is a function for which, intuitively, small changes in the input result in small changes in the output. ... For a non-technical overview of the subject, see Calculus. ...

$f'left(cright) = 0$ .

Intuitively, this means that if a smooth curve is equal at two points then there must be a stationary point somewhere between them. Just continuity is not sufficient. (However, differentiability is not quite necessary; see note below.) For example, if

$fleft(xright) = |x|$

the absolute value of x, then we have that In mathematics, the absolute value (or modulus[1]) of a real number is its numerical value without regard to its sign. ...

$fleft(-1right) = fleft(1right)$

but there is no x between -1 and 1 for which $f'left(xright) = 0$ . This is because that function, although continuous, is not differentiable at x=0.

A version of the theorem was first stated by Indian astronomer Bhaskara in the 12th century.[1] A proof of the theorem had to wait until centuries later when Michel Rolle in 1691 used the methods of differential calculus. Bhaskara (1114 â€“ 1185), also known as Bhaskara II and Bhaskara AchÄrya (Bhaskara the teacher), was an Indian mathematician and astronomer. ... Michel Rolle (April 21, 1652 - November 8, 1719) was a French mathematician. ... Events March 5 - French troops under Marshal Louis-Francois de Boufflers besiege the Spanish-held town of Mons March 20 - Leislers Rebellion - New governor arrives in New York - Jacob Leisler surrenders after standoff of several hours March 29 - Siege of Mons ends to the cityâ€™s surrender May 6...

1 Proof
2 Relaxed assumptions
3 Generalizations
4 See also
5 External links
6 Footnotes

Proof

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 extreme value theorem attains both its maximum and its minimum in $[a,b]$ . 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 $(a,b)$ . A continuous function in a closed interval has a minimum (blue) and a maximum (red). ...

Suppose then that the maximum is obtained at an interior point $x in (a,b)$ (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 $x_1$ less than $x$ , the quantity $f(x)-f(x_1) over {x-x_1}$ is non-negative since $x$ is a maximum. Thus the limit as $x 1$ approaches $x$ from below is non-negative. (Note that we assume that $f$ is differentiable to guarantee that the left-hand and right-hand derivatives exist). Wikibooks Calculus has a page on the topic of Limits In mathematics, the concept of a limit is used to describe the behavior of a function as its argument either gets close to some point, or as it becomes arbitrarily large; or the behavior of a sequences elements as...

For $x_2$ greater than $x$ , the quantity $f(x)-f(x_2) over {x-x_2}$ is non-positive. Thus the limit as $x 2$ approaches $x$ from above is non-positive.

Finally, since $f$ is differentiable at $x$ , these two limits must be equal and hence are both 0. This implies that $f'(x) = 0$ .

Relaxed assumptions

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

$f : left[a,bright] rightarrow mathbb{R}$

is continuous on $left[a,bright]$ , that In mathematics, a continuous function is a function for which, intuitively, small changes in the input result in small changes in the output. ...

$fleft(aright)=fleft(bright)$

and that

$forall x in left(a, bright) ,; lim_{h rightarrow 0}frac{fleft(x + hright) - fleft(xright)}{h} in left[ -infty,+infty right].$

An example of such a function on $[ - 1,1]$ is

$f(x)=begin{cases} sqrt{x}(1-x)&text{for }xin[0,1], -sqrt{-x}(1+x)&text{for }xin[-1,0], end{cases}$

which has infinite slope at the origin and satisfies $f (x) = - f ( - x)$ . Look up Slope in Wiktionary, the free dictionary. ...

Generalizations

The mean value theorem gives a similar statement but for functions that do not have the same value at the end points; that is, $f(a) neq f(b)$ . The conclusion is that there is a point of the domain where the instantaneous slope equals the mean slope. Rolle's theorem can be used to prove the mean value theorem and vice versa. In calculus, the mean value theorem states, roughly, that given a section of a smooth curve, there is a point on that section at which the derivative (slope) of the curve is equal to the average derivative of the section. ...

We can also generalize Rolle's theorem by requiring that f has more zeros and greater regularity. Specifically, suppose the following

the function f is n times differentiable: $displaystyle fin C^{n-1}[a,b]$ , and $displaystyle f^{(n)}(x)$ exists on $displaystyle (a,b)$ , and
the function f has n+1 roots: $displaystyle f(x_i)=0$ for distinct points $x_0, ldots, x_n$ , in $displaystyle [a,b]$ .

Then there is a $cin (a,b)$ such that $displaystyle f^{(n)}(c)=0$ .

The theorem may be generalized to functions that only have one-sided derivatives: If $f left( x right)$ is continuous on $left[ a, b right]$ , and has one-sided derivatives on $left( a, b right)$ , and $f left( a right) = f left( b right)$ then $exists c in left( a, b right)$ such that one of $f'left( c+ right)$ and $f'left( c- right)$ is $ge 0$ and the other is $le 0$ . This would cover the counterexample of $fleft(xright) = |x|$ given earlier. This version of the theorem is sufficient to prove convexity when the one-sided derivatives are monotonically increasing: $f' left( x_0- right) le f' left( x_0+ right) le f' left( x_1- right), x_0 < x_1$ . ^[1] In mathematics, convex function is a real-valued function f defined on an interval (or on any convex subset C of some vector space), if for any two points x and y in its domain C and any t in [0,1], we have Convex function on an interval. ...

External links

Rolle's and Mean Value Theorems at cut-the-knot

cut-the-knot is an educational website maintained by Alexander Bogomolny and devoted to popular exposition of a great variety of topics in mathematics. ...

Footnotes

^ Artin, Emil [1931] (1964). The Gamma Function, trans. Michael Butler, Holt, Rinehart and Winston, pp. 3-4.

Categories: Calculus | Mathematical theorems | Real analysis | Articles containing proofs Emil Artin (March 3, 1898-December 20, 1962) was an Austrian mathematician born in Vienna who spent his career in Germany (mainly in Hamburg) until the Nazi threat when he emigrated to the USA in 1937 where he was at Indiana University 1938-1946, and Princeton University 1946-1958. ... Holt, Rinehart and Winston, somethimes abbreviated as HRW or referred to as Holt, is an Austin, Texas based publishing company, that specializes in textbooks for use in secondary schools. ...

Results from FactBites:

Rolle's theorem : Rolles theorem (607 words)
	Rolle's theorem is a mathematical theorem; developed by Rolle, and published in 1691.
	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.

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Contents

Proof GA_googleFillSlot("encyclopedia_square");

Relaxed assumptions

Generalizations

See also

External links

Footnotes

Proof