# 131a7 - Problems graded 18.4 18.9 19.4(10 points each the...

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Problems graded: 18.4, 18.9, 19.4 (10 points each); the others were worth 5 points for completion (19.1 counted as two problems). To convert from 60 points to 100, we multiplied everyone’s score by 1.5 (rounding up) and then added ten points to everyone’s score. 18.4. We deFne f : S R as follows: f ( x ) = 1 x - x 0 . This is continuous by Theorem 17.4(iii) (the denominator is nonzero in S as x n = x 0 ); however, as x n x 0 , | f ( x ) | → ∞ (by the limit laws from Section 9); since { x n } is a sequence in S , f is unbounded. 18.8. Because the product of zero and any other number is zero while f ( a ) f ( b ) < 0, either f ( a ) > 0 (in which case f ( b ) < 0 so f ( b ) < 0 < f ( a )) or f ( a ) < 0 (in which case f ( b ) > 0 so f ( a ) < 0 < f ( b )); in any case, 0 is between f ( a ) and f ( b ) so the Intermediate Value Theorem states that there exists x ( a, b ) with f ( x ) = 0. 18.9. We write f ( x ) = a 0 + a 1 x + . . . + a n x n , where n is odd and a n n = 0. By dividing by a n (which does not change the location of the roots) we may suppose a n was originally 1. Choose M > | a 0 | + . . . + | a n - 1 | so that M is also greater than 1 and note that the magnitude of the x j term in f ( M ) for j < n is | a j | M j ≤ | a j | M n - 1 so, by the triangle inequality, f ( M ) M j +1 Σ n - 1 j =0 | a j | M j = M j ( M Σ n - 1 j =0 | a j | ) > 0 . Similiarly,

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## This note was uploaded on 08/24/2009 for the course MATH 262447221 taught by Professor Weisbart during the Spring '09 term at UCLA.

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131a7 - Problems graded 18.4 18.9 19.4(10 points each the...

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