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Unformatted text preview: Problems graded: 29.3 (15 points), 29.13 (10 points), 32.2 (10 points); the other problems were graded based on completion and worth five points each (except that I counted 28.8, which was longer, as two problems). To convert from 65 points to 100, scores were multiplied by 1.5; then we gave everyone 2.5 points, rounding all fractions up at the end. Final homework grade: Students have received nine scores, each out of 100. As per the syllabus, we dropped the two lowest scores for each students and converted the remaining sum to a percentage out of 100, rounding any fractional part up (623/700 = 89 percent but 624/700 = 89.14 percent gets bumped up to 90 percent). 28.8. (Note: I did this question in section on November 18.) (a) Suppose > 0. If | x | < then f ( x ) is equal to either x 2 [0 , ) or 0; in any event, | f ( x ) f (0) | < implying continuity at zero. (b) x rational nonzero: Let = x 2 and suppose > 0. There exists an irrational number q in ( x , x + ) (see Exercise 4.12 from HW 2). However, while | x q | < , | f ( x ) f ( q ) | = ; can be made arbitrarily small so f is NOT continuous at x . x irrational: Let = . 1 x 2 and suppose > 0 (we may further suppose that < . 5 | x | ). There exists a rational number q in ( x , x + ); by the triangle inequality, | q | > . 5 | x | so f ( q ) = . 25 x 2 . Therefore, | f ( x ) f ( q ) | = . 25 x 2 > ; as can be made arbitrarily small, f is not continuous at x . As all nonzero points are either nonzero rational or irrational, f is discon- tinuous for all nonzero x ....
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- Spring '09