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What is Rolle's theorem

  1. Jul 23, 2014 #1
    Definition/Summary

    Rolle's theorem states that between any two points at which a sufficiently smooth function has equal values, there is a point at which it has zero derivative.

    To be precise: if a real-valued function [itex]f[/itex] is continuous on the closed interval [itex][a,b][/itex] and differentiable on the open interval [itex](a,b)[/itex], and if [itex]f(a)\ =\ f(b)[/itex], then there is a point [itex]c[/itex] in [itex](a,b)[/itex] such that [itex]f'(c) \ =\ 0[/itex]

    Rolle's theorem is a special case of the mean value theorem: if a real-valued function [itex]f[/itex] is continuous on the closed interval [itex][a,b][/itex] and differentiable on the open interval [itex](a,b)[/itex], then there is a point [itex]c[/itex] in that interval such that [itex]f'(c) \ =\ (f(b) - f(a))/(b - a)[/itex].

    Rolle's theorem may be proved using the extreme value theorem: a continuous real-valued function on a closed interval attains its extreme values (its minimum and maximum).

    Equations



    Extended explanation

    Dedekind-completeness theorem:

    A bounded real-valued function has a least upper bound and a greatest lower bound.

    Bolzano-Weierstrass theorem:

    Every bounded real-valued sequence has a convergent sub-sequence.

    Boundedness theorem:

    A continuous function on the closed interval [itex][a,b][/itex] is bounded on that interval.

    This may be proved using the Bolzano-Weierstrass theorem.

    Extreme value theorem:

    A continuous function on the closed interval [itex][a,b][/itex] attains both extreme values (a maximum and minimum) on that interval.

    To be precise: if a real-valued function [itex]f[/itex] is continuous on the closed interval [itex][a,b][/itex], then there are points [itex]c\text{ and }d[/itex] in that interval such that [itex]f(c)\text{ and }f(d)[/itex] are the extreme values (the minimum and maximum) of [itex]f[/itex] in that interval.

    This may be proved using the boundedness theorem the Dedekind-completeness theorem and the Bolzano-Weierstrass theorem: the function is bounded, therefore has a least upper bound, therefore there is a sequence on which the function converges to that least upper bound, therefore there is a subsequence which converges, and the function attains that least upper bound at that point of convergence: in other words, the function attains its maximum value at that point. Similarly, the function attains its minimum value (its greatest lower bound).


    Extreme value derivative theorem:

    If a function has a derivative at an extreme value (a maximum or minimum) in an open interval, then that derivative is zero.

    This can be proved directly from the definition of derivative, by showing that a non-zero derivative cannot be at an extreme value.

    Proof of Rolle's Theorem:

    From the extreme value theorem, the function attains its extreme values on [a,b]. If it attains them both at a and b, then the function is constant, and so has zero derivative everywhere. If it attains either of them at an interior point, then by the extreme value derivative theorem the derivative at that point is zero.

    Corollary:

    A function with positive derivative everywhere on (a,b) is strictly increasing, and a function with negative derivative everywhere on (a,b) is strictly decreasing.

    Generalisaton of Rolle's theorem to higher derivatives:

    Between any two points between which there are n non-overlapping intervals at the endpoints of each of which a sufficiently smooth function has equal values, there is a point at which it has zero nth derivative.

    To be precise: if a real-valued function [itex]f[/itex] is continuously differentiable n-1 times on the closed interval [itex][a,b][/itex] and differentiable n times on the open interval [itex](a,b)[/itex], and if [itex]a_1<b_1\leq a_2<b_2\leq\cdots a_n<b_n[/itex] in [itex][a,b][/itex], and if [itex]f(a_i)\ =\ f(b_i)\,\forall 1\leq i\leq n[/itex], then there is a point [itex]c[/itex] in [itex](a,b)[/itex] such that [itex]f^{(n)}(c) \ =\ 0[/itex]

    This may be proved by induction on the value of n.

    * This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
     
  2. jcsd
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