set9 - Lecture 24 Maxima and minima of functions...

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Unformatted text preview: Lecture 24 Maxima and minima of functions (conclusion) The discussion in the previous lecture on Fermats Theorem forms the basis of the following method of finding absolute minimum and maximum values of a continuous function over a closed interval [ a,b ]. You are familiar with this method from your high school calculus course: Closed Interval Method: Assuming that f is continuous on the closed interval [ a,b ]: 1. Find all critical points of f in ( a,b ), i.e., points c ( a,b ) for which f ( c ) = 0 or f ( c ) fails to exist. Let c 1 ,c 2 , ,c n denote all critical points in ( a,b ). Evaluate f at these critical points, i.e., f ( c 1 ) ,f ( c 2 ) , ,f ( c n ). 2. Evaluate f at the endpoints, i.e., f ( a ) and f ( b ). 3. The maximum of the set { f ( c 1 ) , ,f ( c n ) ,f ( a ) ,f ( b ) } is the absolute maximum value of f on [ a,b ] and the minimum (including most negative) of this set is the absolute minimum value of f on [ a,b ]. This method is illustrated in the text by Stewart in Examples 8-10, Section 4.1, pp. 275-276. The Mean Value Theorem We come to a very important result in Calculus that not only leads to a large number of theoretical results but also provides a basis for the approximation of functions. Well state it first and then prove it. Mean Value Theorem: Suppose that a function f is 1. continuous on the closed interval [ a,b ] and 2. differentiable on the open interval ( a,b ). Then there exists a number c ( a,b ) such that f ( c ) = f ( b ) f ( a ) b a (1) 176 B a b f ( a ) f ( b ) y = f ( x ) c A Graphically, the situation is sketched above. The tangent line to f at c is parallel to the secant line AB that joins points ( a,f ( a )) to ( b,f ( b )). Before proceeding with the proof, a couple of comments on the assumptions of the theorem are in order, based on questions that were asked during the lecture. First of all, given a closed interval [ a,b ], they represent a kind of minimal set of assumptions on a function f to get the job done. We assume differentiability on ( a,b ) so that we dont have to worry about the endpoints, since f may not be defined outside the interval [ a,b ]. We could have imposed the condition that one-sided derivatives of f exist at the endpoint, but this is not even necessary: The theorem will work even if one, or both, of the endpoints is a cusp, where the one-sided derivative does not exist. (For example, x at x = 0.) Well need a kind of preliminary result in order to prove the Mean Value Theorem. This result is known as Rolles Theorem. Rolles Theorem: Suppose that a function f is 1. continuous on the closed interval [ a,b ], 2. differentiable on the open interval ( a,b ) and 3. f ( a ) = f ( b )....
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This note was uploaded on 09/22/2011 for the course MATH 137 taught by Professor Speziale during the Fall '08 term at Waterloo.

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set9 - Lecture 24 Maxima and minima of functions...

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