Editing Rolle's theorem

{{Short description|On stationary points between two equal values of a function}}
{{Calculus}}
[[File:RTCalc.svg|thumb|300 px|right|If a [[real number|real]]-valued function {{mvar|f}} is [[continuous function|continuous]] on a [[closed interval]] {{closed-closed|''a'', ''b''}}, [[derivative|differentiable]] on the [[open interval]] {{open-open|''a'', ''b''}}, and {{math|1=''f&thinsp;''(''a'') = ''f&thinsp;''(''b'')}}, then there exists a {{mvar|c}} in the open interval {{open-open|''a'', ''b''}} such that {{math|1=''f&thinsp;''′(''c'') = 0}}.]]

In [[calculus]], '''Rolle's theorem''' or '''Rolle's lemma''' essentially states that any real-valued [[differentiable function]] that attains equal values at two distinct points must have at least one point, somewhere between them, at which the slope of the tangent line is zero. Such a point is known as a [[stationary point]]. It is a point at which the first derivative of the function is zero. The theorem is named after [[Michel Rolle]].

== Standard version of the theorem ==

If  a [[real number|real]]-valued [[Function (mathematics)|function]] {{mvar|f}} is [[continuous function|continuous]] on a proper [[closed interval]] {{closed-closed|''a'',&nbsp;''b''}}, [[Differentiable function|differentiable]] on the [[open interval]] {{open-open|''a'', ''b''}}, and {{math|1=''f&thinsp;''(''a'') = ''f&thinsp;''(''b'')}}, then there exists at least one {{mvar|c}} in the open interval {{open-open|''a'', ''b''}} such that <math display="block">f'(c) = 0.</math>

This version of Rolle's theorem is used to prove the [[mean value theorem]], of which Rolle's theorem is indeed a special case. It is also the basis for the proof of [[Taylor's theorem]].

== History ==
Although the theorem is named after [[Michel Rolle]], Rolle's 1691 proof covered only the case of polynomial functions. His proof did not use the methods of [[differential calculus]], which at that point in his life he considered to be fallacious. The theorem was first proved by [[Augustin-Louis Cauchy|Cauchy]] in 1823 as a corollary of a proof of the [[mean value theorem]].<ref>{{cite web |first=A. |last=Besenyei |title=A brief history of the mean value theorem |date=September 17, 2012 |url=https://abesenyei.web.elte.hu/publications/meanvalue.pdf }}</ref> The name "Rolle's theorem" was first used by [[Moritz Wilhelm Drobisch]] of Germany in 1834 and by [[Giusto Bellavitis]] of Italy in 1846.<ref>See {{cite book |first=Florian |last=Cajori |author-link=Florian Cajori |title=A History of Mathematics |year=1999 |page=224 |publisher=American Mathematical Soc. |isbn=9780821821022 |url=https://books.google.com/books?id=mGJRjIC9fZgC&pg=RA1-PA119-IA5 }}</ref>

== Examples ==

===Half circle===
[[File:semicircle.svg|thumb|300px|A semicircle of radius {{mvar|r}}]]
For a radius {{math|''r'' > 0}}, consider the function
<math display="block">f(x)=\sqrt{r^2 - x^2},\quad x \in [-r, r].</math>

Its [[graph of a function|graph]] is the upper [[semicircle]] centered at the origin. This function is continuous on the closed interval {{closed-closed|−''r'', ''r''}} and differentiable in the open interval {{open-open|−''r'', ''r''}}, but not differentiable at the endpoints {{math|−''r''}} and {{mvar|r}}. Since {{math|1=''f&thinsp;''(−''r'') = ''f&thinsp;''(''r'')}}, Rolle's theorem applies, and indeed, there is a point where the derivative of {{mvar|f}} is zero. The theorem applies even when the function cannot be differentiated at the endpoints because it only requires the function to be differentiable in the open interval.

===Absolute value===
[[File:Absolute value.svg|thumb|300px|The graph of the absolute value function]]
If differentiability fails at an interior point of the interval, the conclusion of Rolle's theorem may not hold. Consider the [[absolute value]] function
<math display="block">f(x) = |x|,\quad x \in [-1, 1].</math>

Then {{math|1=''f&thinsp;''(−1) = ''f&thinsp;''(1)}}, but there is no {{mvar|c}} between −1 and 1 for which the {{math|''f&thinsp;''′(''c'')}} is zero. This is because that function, although continuous, is not differentiable at {{math|1=''x'' = 0}}. The derivative of {{mvar|f}} changes its sign at {{math|1=''x'' = 0}}, but without attaining the value 0. The theorem cannot be applied to this function because it does not satisfy the condition that the function must be differentiable for every {{mvar|x}} in the open interval. However, when the differentiability requirement is dropped from Rolle's theorem, {{mvar|f}} will still have a [[critical number]] in the open interval {{open-open|''a'', ''b''}}, but it may not yield a horizontal tangent (as in the case of the absolute value represented in the graph).

===Functions with zero derivative===

Rolle's theorem implies that a [[differentiable function]] whose derivative is {{tmath|0}} in an interval is constant in this interval. 

Indeed, if {{mvar|a}} and {{mvar|b}} are two points in an interval where a function {{mvar|f}} is differentiable, then the function
<math display=block>g(x)=f(x)-f(a)-\frac{f(b)-f(a)}{b-a}(x-a)</math>
satisfies the hypotheses of Rolle's theorem on the interval {{tmath|[a,b]}}.

If the derivative of {{tmath|f}} is zero everywhere, the derivative of {{tmath|g}} is
<math display=block>g'(x)=\frac{f(b)-f(a)}{b-a},</math>
and Rolle's theorem implies that there is {{tmath|c\in (a,b)}} such that
<math display=block>0=g'(c)=\frac{f(b)-f(a)}{b-a}.</math>

Hence, {{tmath|1=f(a)=f(b)}} for every {{tmath|a}} and {{tmath|b}}, and the function {{tmath|f}} is constant.

== Generalization ==

The second example illustrates the following generalization of Rolle's theorem:

Consider a real-valued, continuous function {{mvar|f}} on a closed interval {{closed-closed|''a'', ''b''}} with {{math|1=''f&thinsp;''(''a'') = ''f&thinsp;''(''b'')}}. If for every {{mvar|x}} in the open interval {{open-open|''a'', ''b''}} the [[One-sided limit|right-hand limit]]
<math display="block">f'(x^+):=\lim_{h \to 0^+}\frac{f(x+h)-f(x)}{h}</math>
and the left-hand limit
<math display="block">f'(x^-):=\lim_{h \to 0^-}\frac{f(x+h)-f(x)}{h}</math><!-- The notation "lim as h tends to 0 minus" means h is negative. See example 2 at [[One-sided limit#Examples]] -->

exist in the [[extended real line]] {{closed-closed|−∞, ∞}}, then there is some number {{mvar|c}} in the open interval {{open-open|''a'', ''b''}} such that one of the two limits
<math display="block">f'(c^+)\quad\text{and}\quad f'(c^-)</math>
is {{math|≥ 0}} and the other one is {{math|≤ 0}} (in the extended real line). If the right- and left-hand limits agree for every {{mvar|x}}, then they agree in particular for {{mvar|c}}, hence the derivative of {{mvar|f}} exists at {{mvar|c}} and is equal to zero.

===Remarks===
* If {{mvar|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 [[Convex function|convexity]] when the one-sided derivatives are [[monotonically increasing]]:<ref>{{citation |last=Artin |first=Emil |author-link=Emil Artin |translator-first= Michael |translator-last= Butler |title=The Gamma Function |orig-year=1931 |year=1964 |publisher=[[Henry Holt and Company|Holt, Rinehart and Winston]] |pages=3–4}}.</ref> <math display="block">f'(x^-) \le f'(x^+) \le f'(y^-),\quad x < y.</math>

== 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 {{math|1=''f&thinsp;''(''a'') = ''f&thinsp;''(''b'')}}, then {{mvar|f}} must attain either [[maxima and minima|a maximum or a minimum]] somewhere between {{mvar|a}} and {{mvar|b}}, say at {{mvar|c}}, and the function must change from increasing to decreasing (or the other way around) at {{mvar|c}}. In particular, if the derivative exists, it must be zero at {{mvar|c}}.

By assumption, {{mvar|f}} is continuous on {{closed-closed|''a'', ''b''}}, and by the [[extreme value theorem]] attains both its maximum and its minimum in {{closed-closed|''a'', ''b''}}. If these are both attained at the endpoints of {{closed-closed|''a'', ''b''}}, then {{mvar|f}} is [[constant function|constant]] on {{closed-closed|''a'', ''b''}} and so the derivative of {{mvar|f}} is zero at every point in {{open-open|''a'', ''b''}}.

Suppose then that the maximum is obtained at an [[interior point]] {{mvar|c}} of {{open-open|''a'', ''b''}} (the argument for the minimum is very similar, just consider {{math|−''f&thinsp;''}}). We shall examine the above right- and left-hand limits separately.

For a real {{mvar|h}} such that {{math|''c'' + ''h''}} is in {{closed-closed|''a'', ''b''}}, the value {{math|''f&thinsp;''(''c'' + ''h'')}} is smaller or equal to {{math|''f&thinsp;''(''c'')}} because {{mvar|f}} attains its maximum at {{mvar|c}}. Therefore, for every {{math|''h'' > 0}},
<math display="block">\frac{f(c+h)-f(c)}{h}\le0,</math>
hence
<math display="block">f'(c^+):=\lim_{h \to 0^+}\frac{f(c+h)-f(c)}{h}\le0,</math>
where the limit exists by assumption, it may be minus infinity.

Similarly, for every {{math|''h'' < 0}}, the inequality turns around because the denominator is now negative and we get
<math display="block">\frac{f(c+h)-f(c)}{h}\ge0,</math>
hence
<math display="block">f'(c^-):=\lim_{h \to 0^-}\frac{f(c+h)-f(c)}{h}\ge0,</math>
where the limit might be plus infinity.

Finally, when the above right- and left-hand limits agree (in particular when {{mvar|f}} is differentiable), then the derivative of {{mvar|f}} at {{mvar|c}} must be zero.

(Alternatively, we can apply [[Fermat's theorem (stationary points)|Fermat's stationary point theorem]] directly.)

== Generalization to higher derivatives ==

We can also generalize Rolle's theorem by requiring that {{mvar|f}} has more points with equal values and greater regularity.  Specifically, suppose that
* the function {{mvar|f}} is {{math|''n'' − 1}} times [[Smoothness#Differentiability_classes|continuously differentiable]] on the closed interval {{closed-closed|''a'', ''b''}} and the {{mvar|n}}th derivative exists on the open interval {{open-open|''a'', ''b''}}, and
* there are {{mvar|n}} intervals given by {{math|''a''<sub>1</sub> < ''b''<sub>1</sub> ≤ ''a''<sub>2</sub> < ''b''<sub>2</sub> ≤ ⋯ ≤ ''a<sub>n</sub>'' < ''b<sub>n</sub>''}} in {{closed-closed|''a'', ''b''}} such that {{math|1=''f&thinsp;''(''a<sub>k</sub>'') = ''f&thinsp;''(''b<sub>k</sub>'')}} for every {{mvar|k}} from 1 to {{mvar|n}}. 
Then there is a number {{mvar|c}} in {{open-open|''a'', ''b''}} such that the {{mvar|n}}th derivative of {{mvar|f}} at {{mvar|c}} is zero.
[[File:Rolle Generale.svg|thumb|290x290px|The red curve is the graph of function with 3 roots in the interval {{closed-closed|−3, 2}}. Thus its second derivative (graphed in green) also has a root in the same interval.]]

The requirements concerning the {{mvar|n}}th derivative of {{mvar|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 {{math|''f&thinsp;''{{isup|(''n'' − 1)}}}} in place of {{mvar|f}}.

Particularly, this version of the theorem asserts that if a function differentiable enough times has {{mvar|n}} roots (so they have the same value, that is 0), then there is an internal point where {{math|''f&thinsp;''{{isup|(''n'' − 1)}}}} vanishes.

===Proof===
The proof uses [[mathematical induction]]. The case {{math|1=''n'' = 1}} is simply the standard version of Rolle's theorem. For {{math|''n'' > 1}}, take as the induction hypothesis that the generalization is true for {{math|''n'' − 1}}. We want to prove it for {{mvar|n}}. Assume the function {{mvar|f}} satisfies the hypotheses of the theorem. By the standard version of Rolle's theorem, for every integer {{mvar|k}} from 1 to {{mvar|n}}, there exists a {{mvar|c<sub>k</sub>}} in the open interval {{open-open|''a<sub>k</sub>'', ''b<sub>k</sub>''}} such that {{math|1=''f&thinsp;''′(''c<sub>k</sub>'') = 0}}. Hence, the first derivative satisfies the assumptions on the {{math|''n'' − 1}} closed intervals {{math|[''c''<sub>1</sub>, ''c''<sub>2</sub>], …, [''c''<sub>''n'' − 1</sub>, ''c<sub>n</sub>'']}}. By the induction hypothesis, there is a {{mvar|c}} such that the {{math|(''n'' − 1)}}st derivative of {{math|''f&thinsp;''′}} at {{mvar|c}} is zero.

== Generalizations to other fields ==
Rolle's theorem is a property of differentiable functions over the real numbers, which are an [[ordered field]]. As such, it does not generalize to other [[field (mathematics)|fields]], but the following corollary does: if a real polynomial factors (has all of its roots) over the real numbers, then its derivative does as well. One may call this property of a field '''Rolle's property'''.{{citation needed|date=September 2018}} More general fields do not always have differentiable functions, but they do always have polynomials, which can be symbolically differentiated. Similarly, more general fields may not have an order, but one has a notion of a root of a polynomial lying in a field.

Thus Rolle's theorem shows that the real numbers have Rolle's property. Any algebraically closed field such as the [[complex numbers]] has Rolle's property. However, the rational numbers do not{{snd}} for example, {{math|1=''x''<sup>3</sup> − ''x'' = ''x''(''x'' − 1)(''x'' + 1)}} factors over the [[rational numbers|rationals]], but its derivative,
<math display="block">3x^2-1 = 3 \left(x - \tfrac{1}{\sqrt 3} \right) \left(x + \tfrac{1}{\sqrt 3} \right),</math>
does not. The question of which fields satisfy Rolle's property was raised in {{Harvnb|Kaplansky|1972}}.<ref>{{ Citation | first = Irving | last = Kaplansky | author-link = Irving Kaplansky | title = Fields and Rings | year = 1972 }}.{{full citation needed|date=July 2024}}</ref> For [[finite field]]s, the answer is that only {{math|'''F'''<sup>2</sup>}} and {{math|'''F'''<sup>4</sup>}} have Rolle's property.<ref>{{ Citation | title = Multiplier sequences for fields | first1 = Thomas | last1 = Craven | first2 = George | last2 = Csordas | journal = Illinois J. Math. | volume = 21 | year = 1977 | pages = 801–817 | url = http://projecteuclid.org/euclid.ijm/1256048929 | issue = 4 | doi = 10.1215/ijm/1256048929 | doi-access = free | url-access = subscription }}.</ref><ref>{{ Citation | title = A Simple Proof of Rolle's Theorem for Finite Fields | first1 = C. | last1 = Ballantine | first2 = J. | last2 = Roberts | journal = [[The American Mathematical Monthly]] | volume = 109 |date=January 2002 | pages = 72–74 | issue = 1 | doi = 10.2307/2695770 | jstor = 2695770 | publisher = Mathematical Association of America }}.</ref>

For a complex version, see [[Voorhoeve index]].

== See also ==
*[[Mean value theorem]]
*[[Intermediate value theorem]]
*[[Linear interpolation]]
*[[Gauss–Lucas theorem]]

== References ==
{{reflist}}

== Further reading ==
* {{cite book |first=Louis |last=Leithold |author-link=Louis Leithold |title=The Calculus, with Analytic Geometry |location=New York |publisher=Harper & Row |edition=2nd |year=1972 |isbn=0-06-043959-9 |pages=201–207 }}
* {{cite book |first=Angus E. |last=Taylor |author-link=Angus Ellis Taylor |title=Advanced Calculus |location=Boston |publisher=Ginn and Company |year=1955 |pages=30–37 }}

== External links ==
* {{springer|title=Rolle theorem|id=p/r082550}}
* [http://www.cut-the-knot.org/Curriculum/Calculus/MVT.shtml Rolle's and Mean Value Theorems] at [[cut-the-knot]].
* [[Mizar system]] proof: http://mizar.org/version/current/html/rolle.html#T2

{{Commons category|Rolle's theorem}}

{{DEFAULTSORT:Rolle's Theorem}}
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