131a7 - Problems graded: 18.4, 18.9, 19.4 (10 points each);...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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,
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 3

131a7 - Problems graded: 18.4, 18.9, 19.4 (10 points each);...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online