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ans1007 - Homework answers, week of Oct. 7 #1, p. 75...

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Homework answers, week of Oct. 7 #1, p. 75 Because f is continuous on [ a , b ] , the extreme-value theorem implies that there exist real numbers m and M such that m f ( x ) M for all x [ a , b ] , and, moreover, that there exist numbers x m , x M [ a , b ] such that f ( x m ) = m and f ( x M ) = M . Since f is an increasing function, f ( x ) < f ( y ) if and only if x < y . Consequently, if x M < b , then M = f ( x M ) < f ( b ) , a contradiction. Therefore, M = f ( b ) and x M = b . Similarly, if x m > a , then m = f ( x m ) > f ( a ) , which contradicts the definition of m . Hence m = f ( a ) and x m = a . Therefore, f ([ a , b ]) = [ f ( a ) , f ( b )] . #2, p. 75 Let M = sup f ([ a , b ]) . Then there exists a sequence { y n } ⊂ f ([ a , b ]) such that { y n } → M . Consequently, there exists a sequence { x n } ⊂ [ a , b ] such that f ( x n ) = y n for every n . The Bolzano-Weierstrauss theorem implies that there is a subsequence { x n k } that converges to a point p [ a , b ] . Since f is continuous, { f ( x n k ) } → f ( p ) . Furthermore, since f ( x n k ) = y n k for every k , { f ( x n k ) } → M . The uniqueness theorem for limits implies that f ( p ) = M . Hence f
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This note was uploaded on 04/02/2009 for the course MAT 371 taught by Professor Thieme during the Spring '07 term at ASU.

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