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131a9 - Problems graded 29.3(15 points 29.13(10 points...

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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 ǫ > 0. If | x 0 | < ǫ then f ( x ) is equal to either x 2 [0 , ǫ ) or 0; in any event, | f ( x ) f (0) | < ǫ implying continuity at zero. (b) x rational nonzero: Let ǫ = x 2 and suppose δ > 0. There exists an irrational number q in ( x δ, x + δ ) (see Exercise 4.12 from HW 2). However, while | x q | < δ , | f ( x ) f ( q ) | = ǫ ; δ can be made arbitrarily small so f is NOT continuous at x . x irrational: Let ǫ = . 1 x 2 and suppose δ > 0 (we may further suppose that δ < . 5 | x | ). There exists a rational number q in ( x δ, x + δ ); by the triangle inequality, | q | > . 5 | x | so f ( q ) = . 25 x 2 . Therefore, | f ( x ) f ( q ) | = . 25 x 2 > ǫ ; as δ 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 . (c) If x = 0 and h negationslash = 0, f ( x + h ) f ( x ) h = f ( h ) h , which is equal to h if h is rational and 0 otherwise. In any event, the expression clearly goes to zero as h goes to zero (as it is bounded above in magnitude by | h | ) so f is differentiable at x = 0 with derivative 0.
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