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

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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) If x is rational, then f ( x ) = x is rational. If x is irrational, then f ( x ) = 1 - x is irrational. Suppose therefore, that f ( x 1 ) = f ( x 2 ) = y . If y is rational, then x 1 , x 2 must both be rational and hence y = x 1 = x 2 . If y is irrational, then x 1 , x 2

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