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Test Bank For Thomas calculus 11th edition solution manual All Chapters Included
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  • Thomas\' Calculus Early Transcendentals

CHAPTER 1 PRELIMINARIES 1.1 REAL NUMBERS AND THE REAL LINE 1. Executing long division, 0.1, 0.2, 0.3, 0.8, 0.9 " 9 9 9 9 9 œ œ œ œ œ 2 3 8 9 2. Executing long division, 0.09, 0.18, 0.27, 0.81, 0. 11 " œ œ œ œ œ 2 3 9 11 3. NT = necessarily true, NNT = Not necessarily true. Given: 2 &l...

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  • chapter 1 preliminaries 11 real numbers and the r

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CHAPTER 1 PRELIMINARIES

1.1 REAL NUMBERS AND THE REAL LINE
"
1. Executing long division, 9 œ 0.1, 2
9 œ 0.2, 3
9 œ 0.3, 8
9 œ 0.8, 9
9 œ 0.9

"
2. Executing long division, 11 œ 0.09, 2
11 œ 0.18, 3
11 œ 0.27, 9
11 œ 0.81, 11
11 œ 0.99

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. 2x  4 Ê x  2

6. 8  3x 5 Ê 3x 3 Ê x Ÿ 1 ïïïïïïïïïñqqqqqqqqp x
1

7. 5x  $ Ÿ (  3x Ê 8x Ÿ 10 Ê x Ÿ 5
4


8. 3(2  x)  2(3  x) Ê 6  3x  6  2x
Ê 0  5x Ê 0  x ïïïïïïïïïðqqqqqqqqp x
0

"
9. 2x  # 7x  7
6 Ê  "#  7
6 5x
Ê "
5
ˆ 10 ‰
6 x or  "
3 x

6 x 3x4
10. 4  2 Ê 12  2x  12x  16
Ê 28  14x Ê 2  x qqqqqqqqqðïïïïïïïïî x
2

,2 Chapter 1 Preliminaries
"
11. 4
5 (x  2)  3 (x  6) Ê 12(x  2)  5(x  6)
Ê 12x  24  5x  30 Ê 7x  6 or x   67

12.  x2 5 Ÿ 123x
4 Ê (4x  20) Ÿ 24  6x
Ê 44 Ÿ 10x Ê  22
5 Ÿ x qqqqqqqqqñïïïïïïïïî x
22/5

13. y œ 3 or y œ 3

14. y  3 œ 7 or y  3 œ 7 Ê y œ 10 or y œ 4

15. 2t  5 œ 4 or 2t  & œ 4 Ê 2t œ 1 or 2t œ 9 Ê t œ  "# or t œ  9#

16. 1  t œ 1 or 1  t œ 1 Ê t œ ! or t œ 2 Ê t œ 0 or t œ 2

17. 8  3s œ 9
2 or 8  3s œ  #9 Ê 3s œ  7# or 3s œ  25
# Ê sœ
7
6 or s œ 25
6


18. s
#  1 œ 1 or s
#  1 œ 1 Ê s
# œ 2 or s
# œ ! Ê s œ 4 or s œ 0


19. 2  x  2; solution interval (2ß 2)

20. 2 Ÿ x Ÿ 2; solution interval [2ß 2] qqqqñïïïïïïïïñqqqqp x
2 2

21. 3 Ÿ t  1 Ÿ 3 Ê 2 Ÿ t Ÿ 4; solution interval [2ß 4]

22. 1  t  2  1 Ê 3  t  1;
solution interval (3ß 1) qqqqðïïïïïïïïðqqqqp t
3 1

23. %  3y  7  4 Ê 3  3y  11 Ê 1  y  11
3 ;
solution interval ˆ1ß 11 ‰
3


24. 1  2y  5  " Ê 6  2y  4 Ê 3  y  2;
solution interval (3ß 2) qqqqðïïïïïïïïðqqqqp y
3 2

25. 1 Ÿ z
5 1Ÿ1 Ê 0Ÿ z
5 Ÿ 2 Ê 0 Ÿ z Ÿ 10;
solution interval [0ß 10]

26. 2 Ÿ  1 Ÿ 2 Ê 1 Ÿ
3z
#
3z
# Ÿ 3 Ê  32 Ÿ z Ÿ 2;
solution interval  23 ß 2‘ qqqqñïïïïïïïïñqqqqp z
2/3 2

27.  "#  3  "
x  "
# Ê  7#   x"   5# Ê 7
#  "
x  5
#

Ê 2
7 x 2
5 ; solution interval ˆ 27 ß 25 ‰


"
28. 3  2
x 43 Ê 1 2
x ( Ê 1 x
#  7

Ê 2x 2
7 Ê 2
7  x  2; solution interval ˆ 27 ß 2‰ qqqqðïïïïïïïïðqqqqp x
2/7 2

, Section 1.1 Real Numbers and the Real Line 3

29. 2s 4 or 2s 4 Ê s 2 or s Ÿ 2;
solution intervals (_ß 2]  [2ß _)

" "
30. s  3 # or (s  3) # Ê s  5# or s 7
#
Ê s  5# or s Ÿ  7# ;
solution intervals ˆ_ß  7# ‘   5# ß _‰ ïïïïïïñqqqqqqñïïïïïïî s
7/2 5/2

31. 1  x  1 or ("  x)  1 Ê x  0 or x  2
Ê x  0 or x  2; solution intervals (_ß !)  (2ß _)

32. 2  3x  5 or (2  3x)  5 Ê 3x  3 or 3x  7
Ê x  1 or x  73 ;
solution intervals (_ß 1)  ˆ 73 ß _‰ ïïïïïïðqqqqqqðïïïïïïî x
1 7/3

33. r"
# 1 or  ˆ r# 1 ‰ 1 Ê r1 2 or r  1 Ÿ 2
Ê r 1 or r Ÿ 3; solution intervals (_ß 3]  [1ß _)

34. 3r
5 " 2
5 or  ˆ 3r5  "‰  2
5
Ê 3r
5 
or  3r5   53 Ê r  37 or r  1
7
5
solution intervals (_ß ")  ˆ 73 ß _‰ ïïïïïïðqqqqqqðïïïïïïî r
1 7/3

35. x#  # Ê kxk  È2 Ê È2  x  È2 ;
solution interval ŠÈ2ß È2‹ qqqqqqðïïïïïïðqqqqqqp x
È # È#


36. 4 Ÿ x# Ê 2 Ÿ kxk Ê x 2 or x Ÿ 2;
solution interval (_ß 2]  [2ß _) ïïïïïïñqqqqqqñïïïïïïî r
2 2

37. 4  x#  9 Ê 2  kxk  3 Ê 2  x  3 or 2  x  3
Ê 2  x  3 or 3  x  2;
solution intervals (3ß 2)  (2ß 3) qqqqðïïïïðqqqqðïïïïðqqqp x
3 2 2 3

" " " " " " " "
38. 9  x#  4 Ê 3  kxk  # Ê 3 x # or 3  x  #
" "
Ê 3 x or  #"  x   3" ;
#
solution intervals ˆ "# ß  3" ‰  ˆ 3" ß #" ‰ qqqqðïïïïðqqqqðïïïïðqqqp x
1/2 1/3 1/3 1/2

39. (x  1)#  4 Ê kx  1k  2 Ê 2  x  1  2
Ê 1  x  3; solution interval ("ß $) qqqqqqðïïïïïïïïðqqqqp x
1 3

40. (x  3)#  # Ê kx  3k  È2
Ê È2  x  3  È2 or 3  È2  x  3  È2 ;
solution interval Š3  È2ß 3  È2‹ qqqqqqðïïïïïïïïðqqqqp x
3  È # 3  È #

, 4 Chapter 1 Preliminaries

Ê ˆx  12 ‰ <
2
41. x#  x  0 Ê x#  x + 1
4 < 1
4
1
4 ʹx  1
2 ¹< 1
2 Ê  12 < x  1
2 < 1
2 Ê 0 < x < 1.
So the solution is the interval (0ß 1)


42. x#  x  2 0 Ê x#  x + 1
4
9
4 Ê ¹x  1
2 ¹ 3
2 Ê x 1
2
3
2 or ˆx  12 ‰ 3
2 Ê x 2 or x Ÿ 1.
The solution interval is (_ß 1]  [2ß _)

43. True if a 0; False if a  0.

44. kx  1k œ 1  x Í k(x  1)k œ 1  x Í 1  x 0 Í xŸ1

45. (1) ka  bk œ (a  b) or ka  bk œ (a  b);
both squared equal (a  b)#
(2) ab Ÿ kabk œ kak kbk
(3) kak œ a or kak œ a, so kak# œ a# ; likewise, kbk# œ b#
(4) x# Ÿ y# implies Èx# Ÿ Èy# or x Ÿ y for all nonnegative real numbers x and y. Let x œ ka  bk and
y œ kak  kbk so that ka  bk# Ÿ akak  kbkb# Ê ka  bk Ÿ kak  kbk .

46. If a 0 and b 0, then ab 0 and kabk œ ab œ kak kbk .
If a  0 and b  0, then ab  0 and kabk œ ab œ (a)(b) œ kak kbk .
If a 0 and b  0, then ab Ÿ 0 and kabk œ (ab) œ (a)(b) œ kak kbk .
If a  0 and b 0, then ab Ÿ 0 and kabk œ (ab) œ (a)(b) œ kak kbk .

47. 3 Ÿ x Ÿ 3 and x   "# Ê  "
#  x Ÿ 3.

48. Graph of kxk  kyk Ÿ 1 is the interior
of “diamond-shaped" region.




49. Let $ be a real number > 0 and f(x) = 2x + 1. Suppose that | x1 | < $ . Then | x1 | < $ Ê 2| x1 | < 2$ Ê
| 2x  # | < 2$ Ê | (2x + 1)  3 | < 2$ Ê | f(x)  f(1) | < 2$

50. Let % > 0 be any positive number and f(x) = 2x + 3. Suppose that | x  0 | < % /2. Then 2| x  0 | < % and
| 2x + 3 3 | < %. But f(x) = 2x + 3 and f(0) = 3. Thus | f(x)  f(0) | < %.

51. Consider: i) a > 0; ii) a < 0; iii) a = 0.
i) For a > 0, | a | œ a by definition. Now, a > 0 Ê a < 0. Let a = b. By definition, | b | œ b. Since b = a,
| a | œ (a) œ a and | a | œ | a | œ a.
ii) For a < 0, | a | œ a. Now, a < 0 Ê a > 0. Let a œ b. By definition, | b | œ b and thus |a| œ a. So again
| a | œ |a|.
iii) By definition | 0 | œ 0 and since 0 œ 0, | 0 | œ 0. Thus, by i), ii), and iii) | a | œ | a | for any real number.

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