[《托马斯微积分》.第11版.(附带习题答案)].Thomas.Calculus.11th.ed.Solution00001.pdf

CHAPTER 1 PRELIMINARIES 1.1 REAL NUMBERS AND THE REAL LINE 1. Executing long division, œ 0.1, 2 œ 0.2, 3 œ 0.3, 8 œ 0.8, 9 œ 0.9 9 9 9 9 9 2. Executing long division, œ 0.09, 2 œ 0.18, 3 œ 0.27, 9 œ 0.81, 11 œ 0.99 11 11 11 11 11 3. NT = necessarily true, NNT = Not necessarily true. Given: 2 x 6. a) NNT. 5 is a counter example. b) NT. 2 x 6 Ê 2 2 x 2 6 2 Ê 0 x 2 2. c) NT. 2 x 6 Ê 2/2 x/2 6/2 Ê 1 x 3. d) NT. 2 x 6 Ê 1/2 1/x 1/6 Ê 1/6 1/x 1/2. e) NT. 2 x 6 Ê 1/2 1/x 1/6 Ê 1/6 1/x 1/2 Ê 6(1/6) 6(1/x) 6(1/2) Ê 1 6/x 3. f) NT. 2 x 6 Ê x 6 Ê (x 4) 2 and 2 x 6 Ê x 2 Ê x 2 Ê x + 4 2 Ê (x 4) 2. The pair of inequalities (x 4) 2 and (x 4) 2 Ê | x 4 | 2. g) NT. 2 x 6 Ê 2 x 6 Ê 6 x 2. But 2 2. So 6 x 2 2 or 6 x 2. h) NT. 2 x 6 Ê 1(2) 1(x) 1(6) Ê 6 x 2 4. NT = necessarily true, NNT = Not necessarily true. Given: 1 y 5 1. a) NT. 1 y 5 1 Ê 1 + 5 y 5 + 5 1 + 5 Ê 4 y 6. b) NNT. y = 5 is a counter example. (Actually, never true given that 4 y 6) c) NT. From a), 1 y 5 1, Ê 4 y 6 Ê y 4. d) NT. From a), 1 y 5 1, Ê 4 y 6 Ê y 6. e) NT. 1 y 5 1 Ê 1 + 1 y 5 + 1 1 + 1 Ê 0 y 4 2. f) NT. 1 y 5 1 Ê (1/2)( 1 + 5) (1/2)(y 5 + 5) (1/2)(1 + 5) Ê 2 y/2 3. g) NT. From a), 4 y 6 Ê 1/4 1/y 1/6 Ê 1/6 1/y 1/4. h) NT. 1 y 5 1 Ê y 5 1 Ê y 4 Ê y 4 Ê y + 5 1 Ê (y 5) 1. Also, 1 y 5 1 Ê y 5 1. The pair of inequalities (y 5) 1 and (y 5) 1 Ê | y 5 | 1. 5