solHMK12 - Section 5.6 #1) First suppose f is increasing....

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Section 5.6 #1) First suppose f is increasing. Then for any x [ a, b ], a x b , and hence f ( a ) f ( x ) f ( b ). Hence, f ( a ) is an absolute minimum and f ( b ) is an absolute maximum. If f is strictly increasing, then x ( a, b ) = a < x < b = f ( a ) < f ( x ) < f ( b ), and so f ( a ) and f ( b ) are the unique absolute minimum and maximum. #5) First suppose that f is continuous at a . By Problem 1), f ( a ) is a lower bound for { f ( x ) : x ( a, b ] } . We must show it is the greatest lower bound, or equivalently, show that for all ± > 0, there exists x 0 ( a, b ] such that f ( x 0 ) < f ( a ) + ± . Fix ± > 0. Since f is continuous at a , we can find δ > 0 such that if x [ a, b ] and | x - a | < δ , then f ( a ) - ± < f ( x ) < f ( a ) + ± . Therefore, choosing any a < x 0 < a + δ will work. Conversely, suppose that f ( a ) = inf { f ( x ) : x ( a, b ] } . Fix ± > 0. We can find x 0 ( a, b ] such that f ( a ) f ( x 0 ) < f ( a ) + ± . Set δ = x 0 - a > 0. Suppose x [ a, b ], and | x - a | < δ . Then a x < a + δ = x 0 . Therefore, since f is increasing, f ( a ) f ( x ) f ( x 0 ) < f ( a ) + ± , by the definition of x 0 . Ergo, | f ( x ) - f ( a ) | < ± , and f is continuous at a . #8) Suppose f ( x 0 ) = y = g ( x 1 ) for some x 0 , x 1 I . If x 0 x 1 , then y = f ( x 0 ) f ( x 1 ) since f is increasing. But f ( x 1 ) > g ( x 1 ) = y , so we get y > y , contradiction. Hences, x 0 < x 1 . #9)
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This note was uploaded on 09/30/2009 for the course MATH 444 taught by Professor Junge during the Spring '08 term at University of Illinois at Urbana–Champaign.

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solHMK12 - Section 5.6 #1) First suppose f is increasing....

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