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Local maximum at x = −1− √ 15 /3, local minimum at x = −1+ √ 15 /3, global maximum at x = 2 and global minimum at x = −4. For a practical example, [ 6 ] assume a situation where someone has 200 {\displaystyle 200} feet of fencing and is trying to maximize the square footage of a rectangular enclosure, where x {\displaystyle x} is ...
When a minimum point (in ) of a function : [,] is to be found but 's domain is a proper subset of some vector space , then it often technically useful to extend to all of by setting ():= + at every . [1] By definition, no point of belongs to the effective domain of , which is consistent with the desire to find a minimum point of the original ...
(), where (2n − 1)!! is the double factorial of (2n − 1), which is the product of all odd numbers up to (2n − 1). This series diverges for every finite x , and its meaning as asymptotic expansion is that for any integer N ≥ 1 one has erfc x = e − x 2 x π ∑ n = 0 N − 1 ( − 1 ) n ( 2 n − 1 ) ! !
The extreme value theorem was originally proven by Bernard Bolzano in the 1830s in a work Function Theory but the work remained unpublished until 1930. Bolzano's proof consisted of showing that a continuous function on a closed interval was bounded, and then showing that the function attained a maximum and a minimum value.
Stated precisely, suppose that f is a real-valued function defined on some open interval containing the point x and suppose further that f is continuous at x.. If there exists a positive number r > 0 such that f is weakly increasing on (x − r, x] and weakly decreasing on [x, x + r), then f has a local maximum at x.
The maximum of a subset of a preordered set is an element of which is greater than or equal to any other element of , and the minimum of is again defined dually. In the particular case of a partially ordered set , while there can be at most one maximum and at most one minimum there may be multiple maximal or minimal elements.
Is there an efficient way to find the global maximum/minimum? Take for example the sine integral. It has an infinite number of local maxima and minima. So how can one decide which one is the global maximum/minimum? --Abdull 17:04, 17 May 2007 (UTC) Not in the absolutely general case.
If () = = and () () for all x in an open interval that contains c, except possibly c itself, =. This is known as the squeeze theorem . [ 1 ] [ 2 ] This applies even in the cases that f ( x ) and g ( x ) take on different values at c , or are discontinuous at c .