Editing Extreme value theorem (section)

{{About|the calculus concept|the statistical concept|Fisher–Tippett–Gnedenko theorem}}
{{short description|Continuous real function on a closed interval has a maximum and a minimum}}
{{More footnotes|date=June 2012}}
[[Image:Extreme Value Theorem.svg|thumb|300px|A continuous function <math>f(x)</math> on the closed interval <math>[a, b]</math> showing the absolute max (red) and the absolute min (blue).]]
In [[calculus]], the '''extreme value theorem''' states that if a real-valued [[Function (mathematics)|function]] <math>f</math> is [[Continuous function|continuous]] on the [[Bounded interval#Classification of intervals|closed]] and [[Bounded set|bounded]] interval <math>[a,b]</math>, then <math>f</math> must attain a [[maximum]] and a [[minimum]], each at least once. 
That is, there exist numbers <math>c</math> and <math>d</math> in <math>[a,b]</math> such that:
<math display="block">f(c) \leq f(x) \leq f(d)\quad \forall x\in [a,b].</math>

The extreme value theorem is more specific than the related  '''boundedness theorem''', which states merely that a continuous function <math>f</math> on the closed interval <math>[a,b]</math> is [[Bounded function|bounded]] on that interval; that is, there exist real numbers <math>m</math> and <math>M</math> such that:
<math display="block">m \le f(x) \le M\quad \forall x \in [a, b].</math>
  
This does not say that <math>M</math> and <math>m</math> are necessarily the maximum and minimum values of <math>f</math> on the interval <math>[a,b],</math> which is what the extreme value theorem stipulates must also be the case.

The extreme value theorem is used to prove [[Rolle's theorem]].  In a formulation due to [[Karl Weierstrass]], this theorem states that a continuous function from a non-empty [[compact space]] to a [[subset]] of the [[real number]]s attains a maximum and a minimum.