Solutions Manual for Thoma's Calculus Early Transcendentals 15th Edition. by welldoneassistant

CHAPTER 1 FUNCTIONS 1.1

FUNCTIONS AND THEIR GRAPHS

1. domain  (, ); range  [1, )

2. domain  [0, ); range  (, 1]

3. domain  [2, ); y in range and y  5 x  10  0  y can be any positive real number  range  [0, ). 4. domain  (, 0]  [3,  ); y in range and y  x 2  3x  0  y can be any positive real number  range  [0, ). 5. domain  (, 3)  (3, ); y in range and y  3 4 t , now if t  3  3  t  0  3 4 t  0, or if t  3  3  t  0  3 4 t  0  y can be any nonzero real number  range  (, 0)  (0, ).

6. domain  (,  4)  ( 4, 4)  (4, ); y in range and y  2 2 2  2 4  t  4  16  t 2  16  0   16 2

t  16

, now if t  4  t 2  16  0  2 2

t  16 2

t  16

, or if t  4  t  16  0  2 2

t  16

 0, or if

 0  y can be any nonzero

real number  range  (,  18 ]  (0, ). 7. (a) Not the graph of a function of x since it fails the vertical line test. (b) Is the graph of a function of x since any vertical line intersects the graph at most once. 8. (a) Not the graph of a function of x since it fails the vertical line test. (b) Not the graph of a function of x since it fails the vertical line test. 9. base  x; (height)2 

 2x   x2  height  23 x; area is a( x)  12 (base)(height)  12 ( x)  23 x   43 x2 ; 2

perimeter is p ( x)  x  x  x  3 x.

10. s  side length  s 2  s 2  d 2  s  d ; and area is a  s 2  a  12 d 2 . 2

11. Let D  diagonal length of a face of the cube and   the length of an edge. Then  2  D 2  d 2 and 3

  



2 2 D 2  2 2  3 2  d 2    d . The surface area is 6 2  6 d3  2d 2 and the volume is 3  d3



3/2

12. The coordinates of P are x, x so the slope of the line joining P to the origin is m  xx  1 ( x  0).



 

Thus, x, x 



1 , 1 . m2 m

25 13. 2 x  4 y  5  y   12 x  54 ; L  ( x  0)2  ( y  0)2  x 2  ( 12 x  54 )2  x 2  14 x 2  54 x  16



5 x 2  5 x  25  4 4 16

20 x 2  20 x  25  16

20 x 2  20 x  25 4

14. y  x  3  y 2  3  x; L  ( x  4) 2  ( y  0) 2  ( y 2  3  4)2  y 2  ( y 2  1)2  y 2 

y4  2 y2  1  y2 

y4  y2  1

d3 . 3 3

Chapter 1 Functions

15. The domain is (, ).

16. The domain is (, ).

17. The domain is (, ).

18. The domain is (, 0].

19. The domain is (, 0)  (0, ).

20. The domain is (, 0)  (0, ).

21. The domain is (, 5)  (5, 3]  [3, 5)  (5, ).

22. The range is [5, ) .

23. Neither graph passes the vertical line test. (a)

(b)

Section 1.1 Functions and Their Graphs 24. Neither graph passes the vertical line test. (a)

(b)

 x  y 1   y  1 x      x  y 1  or or    x  y  1  y  1  x     

25.

x 0 1 2 y 0 1 0

26.

 4  x 2 , x  1 27. F ( x)   2  x  2 x, x  1

x 0 1 2 y 1 0 0

 1 , x  0 28. G ( x)   x  x, 0  x

29. (a) Line through (0, 0) and (1, 1): y  x; Line through (1, 1) and (2, 0): y   x  2 x, 0  x  1  f ( x)    x  2, 1  x  2  2,  0,  (b) f ( x)    2,  0,

0  x 1 1 x  2 2 x3 3 x 4

30. (a) Line through (0, 2) and (2, 0): y   x  2 0 1 Line through (2, 1) and (5, 0): m  5  2  31   13 , so y   13 ( x  2)  1   13 x  53   x  2, 0  x  2 f ( x)   1 5  3 x  3 , 2  x  5

Chapter 1 Functions 3  0

(b) Line through (1, 0) and (0, 3): m  0  ( 1)  3, so y  3x  3 1  3

Line through (0, 3) and (2, 1) : m  2  0  24  2, so y  2 x  3 3x  3,  1  x  0 f ( x)   2 x  3, 0  x  2

31. (a) Line through (1, 1) and (0, 0): y   x Line through (0, 1) and (1, 1): y  1 0 1 Line through (1, 1) and (3, 0): m  3  1  21   12 , so y   12 ( x  1)  1   12 x  32  x 1  x  0  f ( x)   1 0  x 1  1 3 1 x  3  2 x  2 (b) Line through (2, 1) and (0, 0): y  12 x Line through (0, 2) and (1, 0): y  2 x  2 Line through (1, 1) and (3, 1): y  1





10

 1x 2  x  0  2 f ( x)  2 x  2 0  x  1  1 1 x  3 





32. (a) Line through T2 , 0 and (T, 1): m  T  (T /2)  T2 , so y  T2 x  T2  0  T2 x  1  0, 0  x  T2  f ( x)   T 2  T x  1, 2  x  T

 A, 0  x  T 2    A, T  x  T  2 (b) f ( x )   3T  A, T  x  2    A, 32T  x  2T

33. (a)  x   0 for x  [0, 1)

(b)  x   0 for x  (1, 0]

34.  x    x  only when x is an integer. 35. For any real number x, n  x  n  1, where n is an integer. Now: n  x  n  1   (n  1)   x   n. By definition:   x   n and  x   n    x    n. So   x     x  for all real x. 36. To find f(x) you delete the decimal or fractional portion of x, leaving only the integer part.

Section 1.1 Functions and Their Graphs 37. Symmetric about the origin Dec:   x   Inc: nowhere

38. Symmetric about the y-axis Dec:   x  0 Inc: 0  x  

39. Symmetric about the origin Dec: nowhere Inc:   x  0 0 x

40. Symmetric about the y-axis Dec: 0  x   Inc:   x  0

41. Symmetric about the y-axis Dec:   x  0 Inc: 0  x  

42. No symmetry Dec:   x  0 Inc: nowhere

Chapter 1 Functions

43. Symmetric about the origin Dec: nowhere Inc:   x  

44. No symmetry Dec: 0  x   Inc: nowhere

45. No symmetry Dec: 0  x   Inc: nowhere

46. Symmetric about the y-axis Dec:   x  0 Inc: 0  x  

47. Since a horizontal line not through the origin is symmetric with respect to the y-axis, but not with respect to the origin, the function is even. 48. f ( x)  x 5  15 and f ( x)  ( x) 5  x

    f ( x). Thus the function is odd.

1  1 x5 (  x )5

49. Since f ( x)  x 2  1  ( x) 2  1  f ( x). The function is even. 50. Since [ f ( x)  x 2  x]  [ f ( x)  ( x) 2  x] and [ f ( x)  x 2  x ]  [ f ( x)  ( x) 2  x] the function is neither even nor odd. 51. Since g ( x)  x3  x, g ( x)   x3  x  ( x3  x)   g ( x). So the function is odd. 52. g ( x)  x 4  3 x 2  1  ( x) 4  3( x) 2  1  g (  x), thus the function is even. 53. g ( x) 

1  1  g ( x). Thus the function is even. x 2  1 (  x )2  1

54. g ( x) 

x ; g (  x )   x   g ( x ). So the function is odd. x2  1 x2  1

55. h(t )  t 1 1 ; h(t )   t 1 1 ;  h(t )  1 1 t . Since h(t )   h(t ) and h(t )  h(t ), the function is neither even nor odd.

Section 1.1 Functions and Their Graphs 56. Since |t 3 |  |(t )3 |, h(t )  h( t ) and the function is even. 57. h(t )  2t  1, h(t )  2t  1. So h(t )  h(t ).  h(t )  2t  1, so h(t )   h(t ). The function is neither even nor odd. 58. h(t )  2| t |  1 and h(t )  2|  t |  1  2| t |  1. So h(t )  h(t ) and the function is even. 59. g ( x)  sin 2 x; g ( x)   sin 2 x   g ( x). So the function is odd. 60. g ( x)  sin x 2 ; g ( x)  sin x 2  g ( x ). So the function is even. 61. g ( x)  cos3 x; g ( x)  cos3x  g ( x). So the function is even. 62. g ( x)  1  cos x; g ( x)  1  cos x  g ( x). So the function is even. 63. s  kt  25  k (75)  k  13  s  13 t ; 60  13 t  t  180 64. K  c v 2  12960  c(18)2  c  40  K  40v 2 ; K  40(10) 2  4000 joules 65. r  ks  6  k4  k  24  r  24 ; 10  24  s  12 5 s s k  k  14700  P  14700 ; 23.4  14700  V  24500  628.2 in.3 66. P  Vk  14.7  1000 39 V V

67. V  f ( x)  x(14  2 x)(22  2 x)  4 x3  72 x 2  308 x; 0  x  7

    AB   22  AB  2. So,

68. (a) Let h  height of the triangle. Since the triangle is isosceles, AB

 

h 2  12  2  h  1  B is at (0, 1)  slope of AB  1  The equation of AB is y  f ( x)   x  1; x  [0, 1].

(b) A( x )  2 xy  2 x( x  1)  2 x 2  2 x; x  [0, 1] 69. (a) Graph h because it is an even function and rises less rapidly than does Graph g. (b) Graph f because it is an odd function. (c) Graph g because it is an even function and rises more rapidly than does Graph h. 70. (a) Graph f because it is linear. (b) Graph g because it contains (0, 1). (c) Graph h because it is a nonlinear odd function.

Chapter 1 Functions

71. (a) From the graph, 2x  1  4x  x  (2, 0)  (4, ) (b) 2x  1  4x  2x  1  4x  0 x2  2 x  8 ( x  4)( x  2) x  0: 2x  1  4x  0  0 0 2x 2x  x  4 since x is positive; x2  2 x  8

x  0: 2x  1  4x  0   0 2x  x  2 since x is negative; sign of ( x  4)( x  2)

( x  4)( x  2) 0 2x

Solution interval: (2, 0)  (4, )

72. (a) From the graph, x 3 1  x 2 1  x  (, 5)  (1, 1) 3( x  1)

(b) Case x  1: x 3 1  x 2 1  x  1  2  3 x  3  2 x  2  x  5. Thus, x  (, 5) solves the inequality. 3( x  1) Case 1  x  1: x 3 1  x 2 1  x  1  2  3 x  3  2 x  2  x  5 which is true if x  1. Thus, x  (1, 1) solves the inequality. Case 1  x : x 3 1  x 2 1  3 x  3  2 x  2  x  5

which is never true if 1  x, so no solution here. In conclusion, x  (, 5)  (1, 1).

73. A curve symmetric about the x-axis will not pass the vertical line test because the points (x, y) and ( x,  y ) lie on the same vertical line. The graph of the function y  f ( x)  0 is the x-axis, a horizontal line for which there is a single y-value, 0, for any x. 74. price  40  5 x, quantity  300  25x  R( x)  (40  5 x)(300  25 x) 75. x 2  x 2  h 2  x  h  2

2h ; cost  5(2 x)  10h  C (h)  10 2

   10h  5h  2  2 2h 2

76. (a) Note that 2 mi  10, 560 ft, so there are 8002  x 2 feet of river cable at $180 per foot and (10,560  x) feet of land cable at $100 per foot. The cost is C ( x)  180 8002  x 2  100(10,560 - x). (b) C (0)  $1, 200, 000 C (500)  $1,175,812 C (1000)  $1,186,512 C (1500)  $1, 212, 000 C (2000)  $1, 243, 732 C (2500)  $1, 278, 479 C (3000)  $1,314,870 Values beyond this are all larger. It would appear that the least expensive location is less than 2000 feet from the point P. Copyright  2023 Pearson Education, Inc.

Section 1.2 Combining Functions; Shifting and Scaling Graphs 1.2

COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS

1. D f :   x  , Dg : x  1  D f  g  D fg : x  1

2. D f : x  1  0  x  1, Dg : x  1  0  x  1. Therefore D f  g  D fg : x  1 3. D f :   x  , Dg :   x  , D f /g :   x  , Dg /f :   x   4. D f :   x  , Dg : x  0, D f /g : x  0, Dg /f : x  0 5. (a) 2 (d) ( x  5)2  3  x 2  10 x  22 (g) x  10

(b) 22 (e) 5 (h) ( x 2  3)2  3  x 4  6 x 2  6

6. (a)  13

(b) 2

(c)

(d)

(e) 0

1 x

(g) x  2

(h)

(f ) 1

1 x 1

x 1  x2  x  2 1 1

1 1  x x 1 x 1 3 4

x 1

7. ( f  g h)( x)  f ( g (h( x)))  f ( g (4  x))  f (3(4  x))  f (12  3 x)  (12  3 x)  1  13  3x 8. ( f  g h)( x)  f ( g (h( x)))  f ( g ( x 2 ))  f (2( x 2 )  1)  f (2 x 2  1)  3(2 x 2  1)  4  6 x 2  1

    f  1 4   f  1 x4 x   1 x4 x  1  15x 4 x1 

9. ( f  g h)( x)  f ( g (h( x)))  f g 1x

 



1 x



2x   2  x 2  2 2x 8  3x 3 x     7  2x  f f x 2 2 3 x 3 3x   2  x  1  

 

10. ( f  g h)( x)  f ( g (h( x)))  f g

2 x

11. (a) ( f  g )( x) (d) ( j  j )( x)

(b) ( j  g )( x) (e) ( g h f )( x)

12. (a) ( f  j )( x) (d) ( f  f )( x)

(b) ( g h)( x) (e) ( j  g  f )( x)

g(x)

f (x)

( f  g )( x )

(a) x  7

x7

(b) x  2

13.

3( x  2)  3 x  6

x5

x2  5

(d)

x x 1

x x 1 x 1 x 1

(e)

1 x 1

1  1x

(f )

1 x

 x  (xx  1)  x

Chapter 1 Functions

14. (a) ( f  g )( x)  |g ( x)|  (b) ( f  g )( x) 

1 x 1

g ( x)  1  x x 1  1  g (1x )  x x 1  1  x x 1  g (1x )  x 1 1  g (1x ) , so g ( x )  x  1. g ( x) 2

(c) Since ( f  g )( x)  g ( x)  | x |, g ( x)  x . (d) Since ( f  g )( x)  f x  | x |, f ( x)  x 2 . (Note that the domain of the composition is [0, ).)

 

The completed table is shown. Note that the absolute value sign in part (d) is optional. g(x)

f(x)

( f  g )(x)

(a)

1 x 1

| x|

1 x 1

(b)

x 1

x 1 x

x x 1

(c)

| x|

(d)

15. (a) f ( g (1))  f (1)  1 (d) g ( g (2))  g (0)  0

(b) g ( f (0))  g (2)  2 (e) g ( f (2))  g (1)  1

16. (a) f ( g (0))  f (1)  2  (1)  3, where g (0)  0  1  1 (b) g ( f (3))  g (1)  (1)  1, where f (3)  2  3  1 (c) g ( g (1))  g (1)  1  1  0, where g (1)  (1)  1 (d) f ( f (2))  f (0)  2  0  2, where f (2)  2  2  0 (e) g ( f (0))  g (2)  2  1  1, where f (0)  2  0  2 (f ) f g 12  f  12  2   12  52 , where g 12  12  1   12

    

17. (a) ( f  g )( x)  f ( g ( x))  ( g  f )( x)  g ( f ( x)) 

 

1 1  x 1 x 1

 

1 x x

(b) Domain ( f  g ): (,  1]  (0, ), domain ( g  f ): (1, ) 18. (a) ( f  g )( x)  f ( g ( x))  1  2 x  x ( g  f )( x)  g ( f ( x))  1  | x | (b) Domain ( f  g ): [0, ), domain ( g  f ): (, ) g ( x)

19. ( f  g )( x)  x  f ( g ( x))  x  g ( x )  2  x  g ( x)  ( g ( x)  2) x  x  g ( x)  2 x  g ( x)  x  g ( x)  2 x  g ( x)   1 2xx  x2x 1

20. ( f  g )( x )  x  2  f ( g ( x))  x  2  2( g ( x))3  4  x  2  ( g ( x ))3 

x6 x6  g ( x)  3 2 2

21. V  V ( s )  V ( s (t ))  V (2t  3)  (2t  3) 2  2(2t  3)  3

 4t 2  8t  6

Section 1.2 Combining Functions; Shifting and Scaling Graphs

22. (a) x

4

3

2

1

g ( x)

2

1

0.5

0.2

0.2

0.5

f ( g ( x))

1.3

1.6

1.8

1.5

4

3

2

1

g ( x)

1.5

0.3

0.7

1.5

2.4

2.8

3

2.7

2

f ( g ( x))

0.8

1.9

1.7

1.5

0.7

0.3

0.2

0.5

0.9

(b)

23. (a) y  ( x  7) 2

(b) y  ( x  4)2

24. (a) y  x 2  3

(b) y  x 2  5

25. (a) Position 4

(b) Position 1

(d) Position 3

26. (a) y  ( x  1)2  4

(b) y  ( x  2) 2  3

(d) y  ( x  2)2

27.

28.

Chapter 1 Functions

29.

30.

31.

32.

33.

34.

35.

36.

37.

38.

Section 1.2 Combining Functions; Shifting and Scaling Graphs

39.

40.

41.

42.

43.

44.

45.

46.

47.

48.

49.

50.

Chapter 1 Functions

51.

52.

53.

54.

55.

56.

57. (a) domain: [0, 2]; range: [2, 3]

(b) domain: [0, 2]; range: [–1, 0]

(d) domain: [0, 2]; range: [–1, 0]

Section 1.2 Combining Functions; Shifting and Scaling Graphs

(e) domain: [–2, 0]; range: [0, 1]

(f ) domain: [1, 3]; range: [0,1]

(g) domain: [–2, 0]; range: [0, 1]

(h) domain: [–1, 1]; range: [0, 1]

58. (a) domain: [0, 4]; range: [–3, 0]

(b) domain: [–4, 0]; range: [0, 3]

(d) domain: [–4, 0]; range: [1, 4]

(e) domain: [2, 4]; range: [–3, 0]

(f ) domain: [–2, 2]; range: [–3, 0]

Chapter 1 Functions

(h) domain: [0, 4]; range: [0, 3]

(g) domain: [1, 5]; range: [–3, 0]

59. y  3 x 2  3



60. y  (2 x) 2  1  4 x 2  1



1  1 9 x2 ( x /3) 2

61. y  12 1  12  12  1 2

62. y  1 

63. y  4 x  1

64. y  3 x  1

   12 16  x2

65. y  4  2x

66. y  13 4  x 2

   1  x8

67. y  1  (3x )3  1  27 x3

68. y  1  2x

69. Let y   2 x  1  f ( x) and let g ( x)  x1/2 ,



 , i( x)  2  x  12  , and 1/2   j ( x)    2  x  12    f ( x). The graph of h( x)  

h( x)  x  12

1/2

is the graph of g ( x) shifted left 12 unit; the graph of i ( x) is the graph of h( x) stretched vertically by a factor of 2; and the graph of j ( x)  f ( x) is the graph of i ( x) reflected across the x-axis. 70. Let y  1  2x  f ( x). Let g ( x)  (  x)1/2 , h( x)  (  x  2)1/2 , and i ( x )  1 (  x  2)1/2  2

1  2x  f ( x ). The graph of g ( x) is the graph

of y  x reflected across the x-axis. The graph of h( x) is the graph of g ( x) shifted right two units. And the graph of i ( x) is the graph of h( x) compressed vertically by a factor of 2 .

Section 1.2 Combining Functions; Shifting and Scaling Graphs 71. y  f ( x)  x3 . Shift f ( x) one unit right followed by a shift two units up to get g ( x)  ( x  1)3  2 .

72.

y  (1  x)3  2  [( x  1)3  (2)]  f ( x).

Let g ( x)  x3 , h( x)  ( x  1)3 ,

i( x)  ( x  1)3  (2),

and j ( x)  [( x  1)3  (2)]. The graph of h( x) is the graph of g ( x) shifted right one unit; the graph of i ( x) is the graph of h( x) shifted down two units; and the graph of f ( x) is the graph of i ( x) reflected across the x-axis. 73. Compress the graph of f ( x)  1x horizontally by a factor of 2 to get g ( x)  21x . Then shift g ( x) vertically down 1 unit to get h( x)  21x  1.

74. Let f ( x)  12 and g ( x)  22  1  x



 x/ 2 

1 

1 1  x2   2   

1  1. Since 2  1/ 2 x   





2  1.4, we see

that the graph of f ( x) stretched horizontally by a factor of 1.4 and shifted up 1 unit is the graph of g ( x). 75. Reflect the graph of y  f ( x)  3 x across the x-axis to get g ( x)   3 x .

76.

Chapter 1 Functions

y  f ( x)  (2 x) 2/3  [(1)(2) x]2/3  (1) 2/3 (2 x)2/3  (2 x)2/3 . So the graph of f ( x) is the graph of g ( x)  x 2/3 compressed horizontally by a factor of 2.

77.

78.

79. (a) ( fg )( x)  f ( x) g ( x)  f ( x)( g ( x))  ( fg )( x), odd

  ( x)  (c)   ( x) 

(b)

  ( x), odd    ( x), odd

f g

f ( x) f ( x) f   g ( x)   g g ( x)

g f

g ( x)  g ( x)  f ( x) f ( x)

g f

(d) f 2 ( x)  f ( x) f ( x)  f ( x) f ( x)  f 2 ( x), even (e) g 2 ( x)  ( g ( x)) 2  ( g ( x))2  g 2 ( x), even (f ) ( f  g )( x)  f ( g ( x))  f ( g ( x))  f ( g ( x))  ( f  g )( x), even (g) ( g  f )( x)  g ( f ( x))  g ( f ( x))  ( g  f )( x), even (h) ( f  f )( x)  f ( f ( x))  f ( f ( x))  ( f  f )( x), even (i) ( g  g )( x)  g ( g ( x ))  g ( g ( x))   g ( g ( x ))  ( g  g )( x ), odd 80. Yes, f ( x)  0 is both even and odd since f (  x)  0  f ( x) and f ( x)  0   f ( x). 81. (a)

(b)

Section 1.3 Trigonometric Functions

(d)

(c)

82.

1.3

TRIGONOMETRIC FUNCTIONS

1. (a)

 

s  r  (10) 45  8 m

(b)

 

   55 m s  r  (10)(110) 180  110  18 9

  225 2.   rs  108  54 radians and 54 180 

 

  49  s  (6) 49  8.4 in. (since the diameter  12 in.  radius  6 in.) 3.   80    80 180 

4. d  1 meter  r  50 cm    rs  30  0.6 rad or 0.6 50 

 23

sin 

 23

cos 

1 0

 12

und.

5. 

tan  cot 

und.

sec 

1

csc 

und.

1 3

2  2



3 4 1 2

 1

1

 180    34



 32

 3

 6



5 6

sin 

 23

 12

1 2

cos 

1 2

3 2

1 2

 23

tan 

und.

 3

und.

1

und.

 2

cot 

und.

sec 

und.

csc 

 1

2  2

 1

 3

2 3

2

 1

 3  2

Chapter 1 Functions

7. cos x   54 , tan x   34

8. sin x  2 , cos x  1

9. sin x   38 , tan x   8

, tan x   12 10. sin x  12 13 5

11. sin x   1 , cos x   2

12. cos x   23 , tan x  1

13.

14.

period  4

period  

15.

16.

period  4

period  2

18.

17.

period  1

period  6

20.

19.

period  2

Section 1.3 Trigonometric Functions

21.

22.

period  2

23. period  2 , symmetric about the origin

24. period  1, symmetric about the origin s 3

s =  tan t 1

1 0

26. period  4 , symmetric about the origin

25. period  4, symmetric about the s-axis

27. (a) cos x and sec x are positive for x in the interval  2 , 2 ; and cos x and sec x are negative for x













in the intervals  32 ,  2 and 2 , 32 . sec x is

undefined when cos x is 0. The range of sec x is (, 1]  [1,); the range of cos x is [1, 1].

Chapter 1 Functions

(b) sin x and csc x are positive for x in the intervals  32 ,  and (0,  ); and sin x and csc x are





negative for x in the intervals ( , 0) and  , 32 . csc x is undefined when sin x is 0. The





range of csc x is (, 1]  [1, ); the range of sin x is [1, 1].

28. Since cot x  tan1 x , cot x is undefined when tan x  0 and is zero when tan x is undefined. As tan x approaches zero through positive values, cot x approaches infinity. Also, cot x approaches negative infinity as tan x approaches zero through negative values.

29. D :   x  ; R : y  1, 0, 1

30. D :   x  ; R : y  1, 0, 1





 





 





 





 

31. cos x  2  cos x cos  2  sin x sin  2  (cos x)(0)  (sin x)(1)  sin x 32. cos x  2  cos x cos 2  sin x sin 2  (cos x)(0)  (sin x)(1)   sin x 33. sin x  2  sin x cos 2  cos x sin 2  (sin x)(0)  (cos x)(1)  cos x 34. sin x  2  sin x cos  2  cos x sin  2  (sin x)(0)  (cos x)( 1)   cos x 35. cos( A  B )  cos( A  ( B))  cos A cos( B )  sin A sin( B)  cos A cos B  sin A(  sin B)  cos A cos B  sin A sin B 36. sin( A  B)  sin( A  ( B))  sin A cos( B)  cos A sin( B )  sin A cos B  cos A( sin B )  sin A cos B  cos A sin B 37. If B  A, A  B  0  cos( A  B)  cos 0  1. Also cos( A  B )  cos( A  A)  cos A cos A  sin A sin A  cos 2 A  sin 2 A. Therefore, cos 2 A  sin 2 A  1. 38. If B  2 , then cos( A  2 )  cos A cos 2  sin A sin 2  (cos A)(1)  (sin A)(0)  cos A and sin( A  2 )  sin A cos 2  cos A sin 2  (sin A)(1)  (cos A)(0)  sin A . The result agrees with the fact that the cosine and sine functions have period 2 . 39. cos(  x )  cos  cos x  sin  sin x  ( 1)(cos x )  (0)(sin x )   cos x Copyright  2023 Pearson Education, Inc.

Section 1.3 Trigonometric Functions 40. sin(2  x)  sin 2 cos(  x)  cos(2 ) sin(  x)  (0)(cos( x ))  (1)(sin( x ))   sin x





 





 

41. sin 32  x  sin 32 cos( x)  cos 32 sin(  x)  ( 1)(cos x )  (0)(sin( x))   cos x 42. cos 32  x  cos 32 cos x  sin 32 sin x  (0)(cos x )  (1)(sin x)  sin x





43. sin 712  sin 4  3  sin 4 cos 3  cos 4 sin 3 





        2 2

  cos   2  cos  cos 2  sin  sin 2  44. cos 11 12 4 3 4 3 4 3





2 2

1 2

6 4

3 2

         2 2

1 2

3 2

2 2

2 4

     22    23    22   12 23

 

  cos     cos  cos    sin  sin    1 45. cos 12 3 4 3 4 3 4 2





   

     23   22     12    22   12 23

46. sin 512  sin 23  4  sin 23 cos  4  cos 23 sin  4 

47.

1  cos cos 2 8  2

  49. sin 2 12

 28   1  22  2  2

48.

   1  23  2  3

50. sin 2 38 

1  cos 212 2

1  cos cos 2 512  2

 1012   1    23   2  3 2

   1    22   2  2

1  cos 68 2

51. sin 2   34  sin    23    3 , 23 , 43 , 53 2

cos 

52. sin 2   cos 2   sin 2   cos 2   tan 2   1  tan   1    4 , 34 , 54 , 74 53. sin 2  cos   0  2sin  cos   cos   0  cos  (2sin   1)  0  cos   0 or 2sin   1  0  cos   0 or sin  12    2 , 32 , or   6 , 56    6 , 2 , 56 , 32 54. cos 2  cos   0  2 cos 2   1  cos   0  2 cos 2   cos   1  0  (cos   1)(2 cos   1)  0  cos   1  0 or 2 cos   1  0  cos   1 or cos   12     or   3 , 53    3 ,  , 53 sin( A  B )

sin A cos B  cos A cos B

sin A cos B

A cos B 55. tan( A  B )  cos( A  B )  cos A cos B  sin A sin B  cos cos A cos B

cos A sin B  cos A cos B

sin A sin B  cos cos A cos B A cos B

sin( A  B )

sin A cos B  cos A cos B

sin A cos B

cos A cos B 56. tan( A  B )  cos( A  B )  cos A cos B  sin A sin B  cos A cos B

cos A sin B  cos A cos B

sin A sin B  cos cos A cos B A cos B

tan A  tan B

 1  tan A tan B tan A  tan B

 1  tan A tan B

57. According to the figure in the text, we have the following: By the law of cosines, c 2  a 2  b 2  2ab cos   12  12  2 cos( A  B )  2  2cos( A  B) . By distance formula, c 2  (cos A  cos B)2  (sin A  sin B )2  cos 2 A  2 cos A cos B  cos 2 B  sin 2 A  2sin A sin B  sin 2 B  2  2(cos A cos B  sin A sin B) . Thus

c 2  2  2 cos( A  B )  2  2(cos A cos B  sin A sin B )  cos( A  B )  cos A cos B  sin A sin B .

Chapter 1 Functions

58. (a) cos( A  B )  cos A cos B  sin A sin B sin  cos 2   and cos  sin 2   Let   A  B sin( A  B )  cos  2  ( A  B)   cos  2  A  B   cos 2  A cos B  sin 2  A sin B    sin A cos B  cos A sin B (b) cos( A  B)  cos A cos B  sin A sin B cos( A  ( B ))  cos A cos( B )  sin A sin( B )  cos( A  B)  cos A cos( B )  sin A sin(  B)  cos A cos B  sin A(  sin B)  cos A cos B  sin A sin B Because the cosine function is even and the sine function is odd.























59. c 2  a 2  b 2  2ab cos C  22  32  2(2)(3) cos(60)  4  9  12 cos(60)  13  12 12  7. Thus, c  7  2.65.

60. c 2  a 2  b 2  2ab cos C  22  32  2(2)(3) cos(40)  13  12 cos(40). Thus, c  13  12 cos 40°  1.951. 61. From the figures in the text, we see that sin B  hc . If C is an acute angle, then sin C  bh . On the other hand, if C is obtuse (as in the figure on the right in the text), then sin C  sin(  C )  bh . Thus, in either case, h  b sin C  c sin B  ah  ab sin C  ac sin B. a 2  b2  c2

a 2  c 2  b2

. Moreover, since the sum of the interior By the law of cosines, cos C  and cos B  2 ab 2 ac angles of triangle is  , we have sin A  sin(  ( B  C ))  sin( B  C )  sin B cos C  cos B sin C 

h (2a 2  b 2  c 2  c 2  b 2 )  ah  ah  bc sin A.  hc   a 2bab c    a 2cac b   bh    2abc  bc 2

Combining our results we have ah  ab sin C, ah  ac sin B, and ah  bc sin A. Dividing by abc gives h  sin A  sin C  sin B . bc  a c b  law of sines 62. By the law of sines, sin2 A  sin3 B 

3/2 . By Exercise 59 we know that c  c

7. Thus sin B  3 3  0.982. 2 7

63. From the figure at the right and the law of cosines, b 2  a 2  22  2(2a) cos B



 a 2  4  4a 12  a 2  2a  4.

Applying the law of sines to the figure,

3/2  b  32 a. Thus, combining results, b a 2  2a  4  b 2  32 a 2  0  12 a 2  2a  4  0  a 2  4a  8 . From the quadratic formula and the fact that



2/2  a

sin A sin B  b a

a  0, we have a 

4 

42  4(1)( 8) 4 34  2  1.464. 2

Section 1.3 Trigonometric Functions

64. tan   hc  c  tanh  tan   b h c 

h tan  h  b tan   h  b  tanh 

b tan  tan   h tan   h tan   b tan  tan   h tan   h tan   b tan  tan   h(tan   tan  )  b tan  tan 

h  tan   tan 

65. sin    r r

  sin   r sin   r   sin   r  r sin   r (1  sin  )   r  1sin sin 

66. (a) The graphs of y  sin x and y  x nearly coincide when x is near the origin (when the calculator is in radians mode). (b) In degree mode, when x is near zero degrees the sine of x is much closer to zero than x itself. The curves look like intersecting straight lines near the origin when the calculator is in degree mode. 67. A  2, B  2 , C   , D  1

68. A  12 , B  2, C  1, D  12

69. A   2 , B  4, C  0, D  1

Chapter 1 Functions

70.

A  2L , B  L, C  0, D  0

71–74.

Example CAS commands: Maple: f : x - A*sin((2*Pi/B)*(x-C))D1; A:3; C: 0; D1: 0; f_list : [seq(f(x), B [1,3,2*Pi,5*Pi])]; plot(f_list, x  -4*Pi..4*Pi, scalingconstrained, color [red,blue,green,cyan], linestyle [1,3,4,7], legend ["B1", "B3","B2*Pi","B3*Pi"], title "#71 (Section 1.3)"); Mathematica: Clear[a, b, c, d, f, x] f[x_]: a Sin[2/b (x  c)]  d Plot[f[x]/.{a  3, b  1, c  0, d  0}, {x,  4, 4 }]

71. (a) The graph stretches horizontally.

(b) The period remains the same: period  | B |. The graph has a horizontal shift of 12 period.

72. (a) The graph is shifted right C units.

(b) The graph is shifted left C units. (c) A shift of  one period will produce no apparent shift. | C |  6

Section 1.4 Graphing with Software 73. (a) The graph shifts upwards | D | units for D  0. (b) The graph shifts down | D | units for D  0.

74. (a) The graph stretches | A| units.

(b) For A  0, the graph is inverted.

1.4

GRAPHING WITH SOFTWARE

1–4.

The most appropriate viewing window displays the maxima, minima, intercepts, and end behavior of the graphs and has little unused space.

1. d.

2. c.

3. d.

4. b.

Chapter 1 Functions

5–30. For any display there are many appropriate display widows. The graphs given as answers in Exercises 5–30 are not unique in appearance. 5. [  2, 5] by [  15, 40]

6. [  4, 4] by [  4, 4]

7. [  2, 6] by [  250, 50]

8. [ 1, 5] by [  5, 30]

9. [  4, 4] by [  5, 5]

10. [  2, 2] by [  2, 8]

11. [  2, 6] by [  5, 4]

12. [  4, 4] by [  8, 8]

Section 1.4 Graphing with Software 13. [  1, 6] by [  1, 4]

14. [ 1, 6] by [ 1, 5]

15. [  3, 3] by [0, 10]

16. [ 1, 2] by [0, 1]

17. [ 10, 10] by [ 10, 10]

18. [  5, 1] by [  2, 4]

19. [  4, 4] by [0, 3]

20. [  5, 5] by [  2, 2]

Chapter 1 Functions

21. [ 10, 10] by [  6, 6]

22. [ 5, 5] by [  2, 2]

23. [  6, 10] by [  6, 6]

24. [  3, 5] by [  2, 10]

25. [0.03, 0.03] by [1.25, 1.25]

26. [0.1, 0.1] by [3, 3]

27. [300, 300] by [1.25, 1.25]

28. [50, 50] by [ 0.1, 0.1]

Section 1.4 Graphing with Software 29. [0.25, 0.25] by[0.3, 0.3]

30. [0.15, 0.15] by [0.02, 0.05]

31. x 2  2 x  4  4 y  y 2  y  2   x 2  2 x  8. The lower half is produced by graphing y  2   x 2  2 x  8.

32. y 2  16 x 2  1  y   1  16 x 2 . The upper branch is produced by graphing y  1  16 x 2 .

33.

34.

35.

36.

Chapter 1 Functions

1.5

EXPONENTIAL FUNCTIONS

Section 1.5 Exponential Functions 9.

10.

11. 162 161.75  162 ( 1.75)  160.25  161/4  2 1 1

12. 91/3  91/6  9 3 6  91/2  3 5

44.2  44.23.7  40.5  41/2  2 43.7

13.

14.

35/3  3 3  3  31  3 32/3

15. (251/8 )4  254/8  251/2  5

16.

13 

17. 2 3  7 3  (2  7) 3  14 3

18.

 3   12 

19.

 

2 /2

 132/2  13

1/2

2 2

1/2

1/ 2 4

)



 3  12 

1/2



 36 

 162  4

1/2

 61/2

20.

21. Domain: (, ); y in range  y 

  6 3

(61/ 2 )2 32

 96  32

1 . As x increases, e x becomes infinitely large and y becomes a smaller 2 e x x

and smaller positive real number. As x decreases, e becomes a smaller and smaller positive real number,

 

y  12 , and y gets arbitrarily close to 12  Range: 0, 12 .

22. Domain: (, ); y in range  y  cos(e t ). Since the values of e t are (0, ) and 1  cos x  1  Range: [1, 1]. 23. Domain: (, ); y in range  y  1  3t . Since the values of 3t are (0, )  Range: (1, ). 24. If e 2 x  1, then x = 0  Domain: (, 0)  (0, ); y in range  y 

3 . If x > 0, then 1  e 2 x   1e2 x

  < y < 0. If x < 0, then 0  e 2 x  1  3 < y <   Range: (, 0)  (3, ).

Chapter 1 Functions 26.

25.

x  2.3219

x  1.3863

27.

28.

x  0.6309

x  1.5850

29. Let t be the number of years. Solving 500, 000(1.0375)t  1, 000, 000 graphically, we find that t  18.828. The population will reach 1 million in about 19 years. 30. (a) The population is given by P (t )  6250(1.0275)t , where t is the number of years after 1890. Population in 1915: P(25)  12,315 Population in 1940: P(50)  24,265 (b) Solving P(t) = 50,000 graphically, we find that t  76.651. The population reached 50,000 about 77 years after 1890, in 1967. 31. (a)



A(t )  6.6 12

t /14

(b) Solving A(t) = 1 graphically, we find that t  38. There will be 1 gram remaining after about 38.1145 days. 32. Let t be the number of years. Solving 2300(1.60)t  4150 graphically, we find that t  10.129. It will take about 10.129 years. (If the interest is not credited to the account until the end of each year, it will take 11 years.) 33. Let A be the amount of the initial investment, and let t be the number of years. We wish to solve A(1.0625)t  2 A, which is equivalent to 1.0625t  2. Solving graphically, we find that t  11.433. It will take about 11.433 years. (If the interest is credited at the end of each year, it will take 12 years.)

34. Let A be the amount of the initial investment, and let t be the number of years. We wish to solve Ae0.0575t  3 A, which is equivalent to e0.0575t  3. Solving graphically, we find that t  19.106. It will take about 19.106 years.

Section 1.6 Inverse Functions and Logarithms 35. After t hours, the population is P (t )  2t /0.5 , or equivalently, P (t )  22t . After 24 hours, the population is P (24)  248  2.815  1014 bacteria.

36. (a) Each year, the number of cases is 100%  20% = 80% of the previous year’s number of cases. After t years, the number of cases will be C (t )  10, 000(0.8)t . Solving C(t) = 1000 graphically, we find that t  10.319. It will take 10.319 years. (b) Solving C(t) = 1 graphically, we find that t  41.275. It will take about 41.275 years. 1.6

INVERSE FUNCTIONS AND LOGARITHMS

1. Yes one-to-one, the graph passes the horizontal line test. 2. Not one-to-one, the graph fails the horizontal line test. 3. Not one-to-one since (for example) the horizontal line y = 2 intersects the graph twice. 4. Not one-to-one, the graph fails the horizontal line test. 5. Yes one-to-one, the graph passes the horizontal line test. 6. Yes one-to-one, the graph passes the horizontal line test. 7. Not one-to-one since the horizontal line y = 3 intersects the graph an infinite number of times. 8. Yes one-to-one, the graph passes the horizontal line test. 9. Yes one-to one, graph passes the horizontal line test. 10. Not one-to-one since (for example) the horizontal line y = 1 intersects the graph twice. 11. Domain: 0 < x  1, Range: 0  y

12. Domain: x < 1, Range: y > 0

13. Domain: 1  x  1, Range:  2  y  2

14. Domain:  < x < , Range:  2  y  2

Chapter 1 Functions

15. Domain: 0  x  6, Range: 0  y  3

16. Domain: 2  x  1, Range: 1  y < 3

17. (a) The graph is symmetric about y = x.

(b)

y  1  x 2  y 2  1  x 2  x 2  1  y 2  x  1  y 2  y  1  x 2  f 1 ( x)

18. (a) The graph is symmetric about y = x.

19. Step 1: y  x 2  1  x 2  y  1  x  Step 2: y  x  1  f

1

(b)

y  1x  x  1y  y  1x  f 1 ( x)

y 1

( x)

20. Step 1: y  x 2  x   y , since x  0 Step 2: y   x  f 1 ( x) 21. Step 1: y  x3  1  x3  y  1  x  ( y  1)1/3 Step 2: y  3 x  1  f 1 ( x) 22. Step 1: y  x 2  2 x  1  y  ( x  1)2 

y  x  1, since x  1  x  1  y

Step 2: y  1  x  f 1 ( x) 23. Step 1: y  ( x  1)2  Step 2: y  x  1  f

y  x  1, since x  1  x  1

y 1

( x)

Section 1.6 Inverse Functions and Logarithms 24. Step 1: y  x 2/3  x  y3/2 Step 2: y  x3/2  f 1 ( x) 25. Step 1: y  x5  x  y1/5 Step 2: y  5 x  f 1 ( x) Domain and Range of f 1: all reals f ( f 1 ( x ))  ( x1/5 )5  x and f 1 ( f ( x ))  ( x5 )1/5  x

26. Step 1: y  x 4  x  y1/4 Step 2: y  4 x  f 1 ( x) Domain of f 1: x  0, Range of f 1: y  0 f ( f 1 ( x))  ( x1/4 )4  x and f 1 ( f ( x))  ( x 4 )1/4  x

27. Step 1: y  x3  1  x3  y  1  x  ( y  1)1/3 Step 2: y  3 x  1  f 1 ( x) Domain and Range of f 1: all reals f ( f 1 ( x))  (( x  1)1/3 )3  1  ( x  1)  1  x and f 1 ( f ( x))  (( x3  1)  1)1/3  ( x3 )1/3  x

28. Step 1: y  12 x  72  12 x  y  72  x  2 y  7 Step 2: y  2 x  7  f 1 ( x) Domain and Range of f 1: all reals









f ( f 1 ( x))  12 (2 x  7)  72  x  72  72  x and f 1 ( f ( x))  2 12 x  72  7  ( x  7)  7  x

29. Step 1: y  12  x 2  1y  x  1

1  f 1 ( x ) x

Step 2: y 

Domain of f 1: x > 0, Range of f 1: y > 0 1

f ( f 1 ( x )) 

  1 x



30. Step 1: y  13  x3  x

Step 2: y 

1 1/3

  1 x

 x and f 1 ( f ( x)) 

1 1 x2



 1x 

 x since x > 0

1 1 x 1/3 y y

 3 1x  f 1 ( x)

Domain of f 1: x  0, Range of f 1: y  0 f ( f 1 ( x)) 

1 ( x 1/3 )3



1 x 1

 x and f 1 ( f ( x)) 

  1 x3

1/3



 1x   x 1

Chapter 1 Functions 2 y 3

31. Step 1: y  xx23  y(x  2) = x + 3  xy  2y = x + 3  xy  x = 2y + 3  x  y 1 Step 2: y  2xx13  f 1 ( x) Domain of f 1: x  1, Range of f 1: y  2

 2 x3 3  (2 x 3)3( x 1)  5 x  x and f 1 ( f ( x))  2 xx 23 3  2( x 3)3( x 2)  5 x  x  x1 2 (2 x 3)2( x 1) 5  xx23 1 ( x 3)( x 2) 5

f ( f 1( x ))  2xx13

32. Step 1: y  Step 2: y 



x  y x 3 x 3 1 3x 2

  x  y x  3 y  x  y x  x  3 y  x   y3y1 

 x1   f ( x)

Domain of f 1: (, 0]  (1, ), Range of f 1: [0, 9)  (9, ) 3x  x3x1   x3x1  x 1 3x  0  3x ( x))  ; If x > 1 or x  0     33x  x and x 1 2 2 3 x  3 3 x 3( x 1) 3x 3x 3  3   x 1   x 1  x 1 2

f(f

1

  3 x  x 3 9x   f 1 ( f ( x))    99x  x. 2   xx   3   x x      1    x 3  

 

33. Step 1: y  x 2  2 x, x  1  y  1  ( x  1)2 , x  1   y  1  x  1, x  1  x  1  y  1 Step 2: y  1  x  1  f 1 ( x) Domain of f 1: [1, ), Range of f 1: (, 1]



f ( f 1 ( x))  1  x  1

  2 1  x  1  1  2 x  1  x  1  2  2 x  1  x and 2

f 1 ( f ( x))  1  ( x 2  2 x)  1, x  1  1  ( x  1) 2 , x  1 = 1  |x  1| = 1  (1  x) = x

34. Step 1: y  (2 x3  1)1/5  y 5  2 x3  1  y 5  1  2 x3 

y 5 1 y 5 1  x3  x  3 2 2

Step 2: y  3 x 21  f 1 ( x) Domain of f 1: (, ), Range of f 1: (, ) 1/5

f(f

1

3 (2 x 1) ( f ( x))   2

35.

  5 3  ( x))   2  3 x 21   1      

1

1/5  5



1

     (( x 1)  1)  ( x )  x and 5

 2 x 21  1

3

1/5

5 1/5

(2 x3 1) 1 3 2 x3  2 x 2

xb x22b 2b   1 , x2 x2 x2 2x  b f 1  x   x 1

y

1/5

x

2y  b , y 1

Domain of f 1 :   , 1  1,   , Range of f 1 :   , 2    2,  

Section 1.6 Inverse Functions and Logarithms 36. Since x  b,

x 2  2bx  y  0 ,

x

2b  4b 2  4 y = b  b2  y , 2

f 1  x   b  b 2  x

x  b  b2  y ,



Domain of f 1 :  b 2 ,  , Range of f 1 :   , b   37. (a) y = mx  x  m1 y  f 1 ( x)  m1 x (b) The graph of y  f 1 ( x) is a line through the origin with slope m1 . b ; the graph of f 1 ( x) is a line with slope 1 and 38. y = mx + b  x  m  mb  f 1 ( x)  m1 x  m m y

b. y-intercept  m

39. (a) y = x + 1  x = y  1  f 1 ( x)  x  1 (b) y = x + b  x = y  b  f 1 ( x)  x  b (c) Their graphs will be parallel to one another and lie on opposite sides of the line y = x equidistant from that line. 40. (a) y = x + 1  x = y + 1  f 1 ( x)  1  x; the lines intersect at a right angle. (b) y = x + b  x = y + b  f 1 ( x)  b  x; the lines intersect at a right angle. (c) Such a function is its own inverse.

41. (a) ln 0.75  ln 34  ln 3  ln 4  ln 3  ln 22  ln 3  2 ln 2 (b) ln 94  ln 4  ln 9  ln 22  ln 32  2 ln 2  2 ln 3 (d) ln 3 9  13 ln 9  13 ln 32  23 ln 3

ln 13.5  12 ln13.5  12 ln 27  12 (ln 33  ln 2)  12 (3ln 3  ln 2) 2

1  ln1  3ln 5  3ln 5 42. (a) ln 125

(b) ln 9.8  ln 49  ln 7 2  ln 5  2 ln 7  ln 5 5

(d) ln1225  ln 352  2 ln 35  2 ln 5  2 ln 7

7  ln 7  ln 53  ln 7  3ln 5 (e) ln 0.056  ln 125

(f)

ln 35 ln 17  ln 52lnln75ln 7  12 ln 25

Chapter 1 Functions

   43. (a) ln sin   ln sin5   ln  sin sin     ln 5  5 



(c)





 



(b) ln(3 x 2  9 x)  ln 31x  ln 3 x 3x 9 x  ln( x  3)

  2

1 ln(4t 4 )  ln b  ln 2

4t 4  ln b  ln 2t 2  ln b  ln 2bt

44. (a) ln sec  + ln cos  = ln[(sec )(cos )] = ln 1 = 0





(b) ln(8 x  4)  ln 22  ln(8 x  4)  ln 4  ln 8 x4 4  ln(2 x  1)



46. (a) eln( x  y )  x 2  y 2 47. (a)

(t 1)(t 1) (t 1)

  ln(t 1)

(b)

e ln x 

 12

(c)

eln x ln y  eln( x / y )  xy

(b)

1 e ln 0.3  ln10.3  0.3

(c)

eln  x ln 2  eln( x /2)  2x



2 ln e  2 ln e1/2  (2) 12 ln e  1 2



eln x

(b) ln(ln ee )  ln(e ln e)  ln e  1





48. (a) ln esec   (sec  )(ln e)  sec 

(b) ln e(e )  (e x )(ln e)  e x

   ln x  2 ln x

49. ln y = 2t + 4  eln y  e2t  4  y  e2t  4

50. ln y = t + 5  eln y  et 5  y  et 5

51. ln(y  b) = 5t  eln( y b)  e5t  y  b  e5t  y  e5t  b

  t

52. ln(c  2y) = t  eln(c  2 y )  et  c  2 y  et  2 y  et  c  y   e 2c

53. ln(y  1)  ln 2 = x + ln x  ln(y  1)  ln 2  ln x = x  ln  y  1  2 xe

  y 1 2x

   e x  y 1  e x  xe ln

y 1 2x

 y  2 xe  1

 y 2 1  54. ln( y 2  1)  ln( y  1)  ln(sin x)  ln  y 1   ln(sin x)  ln(y  1) = ln(sin x)  eln( y 1)  eln(sin x )    y  1 = sin x  y = sin x + 1

55. (a) e2k  4  ln e2k  ln 4  2k ln e  ln 22  2k = 2 ln 2  k = ln 2 (b) 100e10 k  200  e10 k  2  ln e10k  ln 2  10k ln e = ln 2  10k = ln 2  k  ln102 k ln e  ln a  k  ln a  k  1000 ln a (c) ek /1000  a  ln ek /1000  ln a  1000 1000

Section 1.6 Inverse Functions and Logarithms 56. (a) e5k  14  ln e5k  ln 41  5k ln e = ln 4  5k = ln 4  k   ln54 (b) 80ek  1  ek  801  ln ek  ln 801  k ln e = ln 80  k = ln 80 (c) e(ln 0.8) k  0.8  (eln 0.8 )k  0.8  (0.8) k  0.8  k = 1 57. (a) e0.3t  27  ln e0.3t  ln 33  (0.3t)ln e = 3 ln 3  0.3t = 3 ln 3  t = 10 ln 3 (b) ekt  12  ln ekt  ln 21  kt ln e   ln 2  t   lnk2 0.4 (c) e(ln 0.2)t  0.4  (eln 0.2 )t  0.4  0.2t  0.4  ln 0.2t  ln 0.4  t ln 0.2 = ln 0.4  t  ln ln 0.2

58. (a) e0.01t  1000  ln e0.01t  ln1000  (0.01t)ln e = ln 1000  0.01t = ln 1000  t = 100 ln 1000 1  ln ekt  ln101  kt ln e   ln10  kt = ln 10  t   ln10 (b) ekt  10 k

60. e x e2 x 1  et  e x  2 x 1  et  ln e x  2 x 1  ln et  t  x 2  2 x  1 61. e2t  3et  (et ) 2  3et  et (et  3)  0  et  3  0  et  3  ln et  ln 3  t  ln 3 62. e2t  6  5et  (et )2  5et  6  0  (et  3)(et  2)  0  et  3  0 or et  2  0  et  3 or et  2  ln et  ln 3 or ln et  ln 2  t  ln 3 or t  ln 2  t   ln 3 or t   ln 2

 

63. ln t t 1  2  e

   e2  t  e2  t  e2t  e2  e2  e2t  t  (e2  1)t  t  e2

ln t t 1

t 1

e2  1

64. ln(t  2)  ln 8  ln t  ln(t  2)  ln t  ln 8  ln((t  2)t )  ln 8  eln(t  2t )  eln8  t 2  2t  8  t 2  2t  8  (t  4)(t  2)  0  t  4  0  t  4

65. (a) 5log5 7  7

(b)

8log8

 2

(c)

1.3log3 75  75

10log10 (1/2)  12

(c)

 log 7  7

(d) log 4 16  log 4 42  2 log 4 4  2 1  2 (e) log3 3  log3 31/2  12 log3 3  12 1  12  0.5 (f) 66. (a)

 

log 4 14  log 4 41  1log 4 4  1 1  1

2log 2 3  3

(b)

(d) log11 121  log11 112  2 log11 11  2 1  2

 

(e) log121 11  log121 1211/2  12 log121 121  (f)



 12  1  12

log3 19  log3 32  2 log3 3  2 1  2

Chapter 1 Functions

67. (a) Let z  log 4 x  4 z  x  22 z  x  (2 z )2  x  2 z  x (b) Let z  log3 x  3z  x  (3z )2  x 2  32 z  x 2  9 z  x 2 (c) log 2 (e(ln 2) sin x )  log 2 2sin x  sin x 68. (a) Let z  log5 (3x 2 )  5 z  3 x 2  25 z  9 x 4 (b) log e (e x )  x x

x (c) log 4 (2e sin x )  log 4 4(e sin x )/2  e sin 2

69. (a) (c)

log 2 x x ÷ ln x  ln x  ln 3  ln 3  ln log 3 x ln 2 ln 3 ln 2 ln x ln 2

(b)

log 2 x x ÷ ln x  ln x  ln 8  3ln 2  3  ln log8 x ln 2 ln 8 ln 2 ln x ln 2

ln x a a ÷ ln a  ln a  ln x 2  2 ln x  2  ln ln 2 a ln x ln x 2 ln x ln a ln x x

70. (a) (b) (c)

log 9 x x ÷ ln x  ln x  ln 3  1  ln log3 x ln 9 ln 3 2 ln 3 ln x 2

log 10 x log 2

 x

ln x ÷ ln x  ln 10 ln 2

 

1 ln 2 ln x  2 ln 2  ln10 1 ln10 ln x 2



 

log a b ln b ÷ ln a  ln b  ln b  ln b 2  ln a ln b log b a ln a ln a ln a

71. (a)

 6

(b)



(c)

 3

72. (a)



(b)

3 4

(c)



73. (a) arccos(1) =  since cos() = 1 and 0    .

 

(b) arccos(0)  2 since cos 2  0 and 0  2   .

  (b) arcsin   1    4 since sin   4    1 and  2   4  2 . 2 2

74. (a) arcsin(1)   2 since sin  2  1 and  2   2  2 .

75. The function g(x) is also one-to-one. The reasoning: f(x) is one-to-one means that if x1  x2 then f ( x1 )  f ( x2 ), so  f ( x1 )   f ( x2 ) and therefore g ( x1 )  g ( x2 ). Therefore g(x) is one-to-one as well. 76. The function h(x) is also one-to-one. The reasoning: f(x) is one-to-one means that if x1  x2 then f ( x1 )  f ( x2 ), so f (1x )  f (1x ) , and therefore h( x1 )  h( x2 ). 1

77. The composition is one-to-one also. The reasoning: If x1  x2 then g ( x1 )  g ( x2 ) because g is one-to-one. Since g ( x1 )  g ( x2 ), we also have f ( g ( x1 ))  f ( g ( x2 )) because f is one-to-one; thus, f  g is one-to-one because x1  x2  f ( g ( x1 ))  f ( g ( x2 )).

Section 1.6 Inverse Functions and Logarithms

78. Yes, g must be one-to-one. If g were not one-to-one, there would exist numbers x1  x2 in the domain of g with g ( x1 )  g ( x2 ). For these numbers we would also have f ( g ( x1 ))  f ( g ( x2 )), contradicting the assumption that f  g is one-to-one. 79. (a)

1 2





x   log 2 100  1   log 2 y



100  y y



  log 

y 2 100  y





Interchange x and y: y  log 2 100x x  f 1 ( x)  log 2 100x x Verify.



 



( f  f 1 )( x)  f log 2 100x x

( f 1  f )( x)  f 1

(b)

y



100 1 2 x







100 1 2

log 2

100x x 



100



log 2 100  x x 1 2



 x x 100  100  100 x 100 x 100 1 100x x



 100 x    100  log 2  1 2 100   log 2    log 2  100  x  100(1 2 x ) 100   1 2  



   log (2 )  x 1 2 x







50  1  1.1 x  50  1.1 x  50  1  log (1.1 x )  log 50 50 1.1 1.1 y  1   x  log1.1 y  1 y y 11.1 x





x   log1.1 50  1   log1.1 y

   log   50  y y



y 1.1 50  y





Interchange x and y: y  log1.1 50x x  f 1 ( x)  log1.1 50x x Verify.



 

50 11.1 x

  log 

( f  f 1 )( x)  f log1.1 50x x ( f 1  f )( x)  f 1

(c)





y  100 x  1  2 x  100  2 x  100  1  log 2 (2 x )  log 2 100  1   x  log 2 100 1 y y y y





x  50 x  x  x 50 50  x 50



50 11.1 x 50  50 x 11.1

  x 50 1   log1.1    log1.1 1.1 x  log1.1 (1.1 )  x  50(11.1 x ) 50   



log1.1 50  x x

11.1





log1.1 x 50  x

11.1

1.1



 1  50x x

 

y 1

y  e x  1  (e x  1) y  e x  1  e x y  y  e x  1  y  1  e x  e x y  (1  y )e x  e x  1  y  e 1

 

y 1

ln e x  ln 1  y  x  ln 1  y

 

Interchange x and y: y  ln 1x x1  f 1 ( x)  ln 1x x1 Verify.

  

( f  f 1 )( x)  f ln 1x x1 (f

(d)

1

 

x  f )( x)  f 1 e x  1 e 1

x 1  1 x x 11 x   1 x 1 x  1 x  12xx  1 2 x  x  x 1  1 x x 11 x ln  x  1  e 1  x 1 1 x 1 x 1 x

e

ln x  1 1  x 1

 ex  1  1  x x  e x  1  (e x  1)   ln  e  1x   ln  x  ln 22e  ln e x  x  x  1 e  1 x   e  1  (e  1)   e 1 

 

y  2 lnlnx x  (2  ln x ) y  ln x  2 y  y ln x  ln x  2 y  y ln x  ln x  2 y  ( y  1) ln x  ln x  y 1  2y

eln x  e y  1  x  e y  1 2x

Interchange x and y: y  e x  1  f 1 ( x)  e x  1

Chapter 1 Functions Verify.  e x2x1  ln 2x   2x    x 1  x 1  2x  2x  x ( f  f 1 )( x)  f  e x  1     2  ln  e x2x1  2  x2x1 x  1 2( x  1)  2 x 2    



( f 1  f )( x)  f 1 2 lnlnx x







 2 ln x   2 ln x   2ln x  2ln x  ln x  1   e  2ln x   e  ln x (2ln x )   e 2  eln x  x





80. sin 1 (1)  cos 1 (1)  2  0  2 ; sin 1 (0)  cos 1 (0)  0  2  2 ; and sin 1 (1)  cos 1 (1)   2    2 . If x  (1, 0) and x = a, then sin 1 ( x)  cos1 ( x)  sin 1 ( a)  cos 1 (a )   sin 1 a  (  cos1 a)    (sin 1 a  cos 1 a )    2  2 from Equations (3) and (4) in the text.

81. (a) (b) (c) (d) (e)

Begin with y = ln x and reduce the y-value by 3  y = ln x  3. Begin with y = ln x and replace x with x  1  y = ln(x  1). Begin with y = ln x, replace x with x + 1, and increase the y-value by 3  y = ln(x + 1) + 3. Begin with y = ln x, reduce the y-value by 4, and replace x with x  2  y = ln(x  2)  4. Begin with y = ln x and replace x with x  y = ln(x).

(f) Begin with y = ln x and switch x and y  x = ln y or y  e x . 82. (a) Begin with y = ln x and multiply the y-value by 2  y = 2 ln x.

 

(b) Begin with y = ln x and replace x with 3x  y  ln 3x . (c) Begin with y = ln x and multiply the y-value by 14  y  14 ln x. (d) Begin with y = ln x and replace x with 2x  y = ln 2x. 83. From zooming in on the graph at the right, we estimate the third root to be x  0.76666.

84. The functions f ( x)  xln 2 and g ( x)  2ln x appear to have identical graphs for x > 0. This is no accident, because xln 2  eln 2ln x  (eln 2 )ln x  2ln x.

 t /12 t /12 t /12 3 t  3  t  36  1   12   81   12    12   12 (b) 8  12 

85. (a)

Amount  8 12

t /12

There will be 1 gram remaining after 36 hours. ln(2)

86. 500(1.0475)t  1000  1.0475t  2  ln(1.0475t )  ln(2)  t ln(1.0475)  ln(2)  t  ln(1.0475)  14.936 It will take about 14.936 years. (If the interest is paid at the end of each year, it will take 15 years.) Copyright  2023 Pearson Education, Inc.

Chapter 1 Practice Exercises



87. 375, 000(1.0225)t  1, 000, 000  1.0225t  83  ln(1.0225t )  ln 83  t ln(1.0225)  ln 83



ln 8

3  t  ln(1.0225)  44.081

It will take about 44.081 years. 88.

y  y0 e0.18t represents the decay equation; solving (0.9) y0  y0 e0.18t  t  0.18  0.585 days. ln(0.9)

CHAPTER 1 PRACTICE EXERCISES

1. The area is A   r 2 and the circumference is C  2 r. Thus, r  2C  A  

 2C   C4 . 2

 4S  . The volume is V  43  r 3  r  3 34V . Substitution into the formula 2/3 for surface area gives S  4 r 2  4  34V  .

2. The surface area is S  4 r 2  r 

1/2

3. The coordinates of a point on the parabola are (x, x2). The angle of inclination  joining this point to the origin 2 satisfies the equation tan   xx  x. Thus the point has coordinates ( x, x 2 )  (tan  , tan 2  ). h  h  500 tan  ft  500 4. tan   rise run

Symmetric about the origin

Symmetric about the y-axis

Neither 9.

Symmetric about the y-axis

y ( x)  ( x) 2  1  x 2  1  y ( x); even

10. y ( x)  ( x)5  ( x)3  ( x)   x5  x3  x   y ( x); odd 11.

y ( x)  1  cos( x)  1  cos x  y ( x); even

Chapter 1 Functions

12.

y ( x)  sec( x) tan( x) 

13.

y ( x) 

14.

y ( x)  ( x)  sin( x)  ( x)  sin x  ( x  sin x)   y ( x); odd

15.

y ( x)   x  cos( x)   x  cos x; neither even nor odd

16.

y ( x)  ( x) cos( x)   x cos x   y ( x); odd

sin   x 

cos 2   x 

  sin2 x   sec x tan x   y ( x); odd cos x

  x 4  1 x4  1 x4  1   y ( x); odd  3   3 3 x  2x   x   2  x   x  2 x

17. Since f and g are odd  f ( x)   f ( x) and g ( x)   g ( x). (a) ( f  g )( x)  f ( x) g ( x)  [ f ( x)] [ g ( x)]  f ( x) g ( x)  ( f  g )( x)  f  g is even. (b)

f 3 ( x)  f ( x) f ( x) f ( x)  [ f ( x)] [ f  x ] [ f ( x)]   f ( x)  f ( x)  f ( x)   f 3 ( x)  f 3 is odd.

(c) f (sin( x))  f (sin( x))   f (sin( x))  f (sin( x)) is odd. (d) g (sec( x))  g (sec( x))  g (sec( x)) is even. (e) | g ( x)|  |  g ( x)|  | g ( x) |  | g | is even. 18. Let f (a  x)  f (a  x) and define g ( x)  f ( x  a ). Then g ( x)  f (( x)  a )  f (a  x)  f (a  x)  f ( x  a)  g ( x)  g ( x)  f ( x  a ) is even. 19. (a) The function is defined for all values of x, so the domain is (, ) . (b) Since | x | attains all nonnegative values, the range is [2, ) . 20. (a) Since the square root requires 1  x  0 , the domain is (,1] . (b) Since 1  x attains all nonnegative values, the range is [2, ) . 21. (a) Since the square root requires 16  x 2  0, the domain is [4, 4] . (b) For values of x in the domain, 0  16  x 2  16, so 0  16  x 2  4. The range is [0, 4] . 22. (a) The function is defined for all values of x, so the domain is (, ) . (b) Since 32 x attains all positive values, the range is (1, ) . 23. (a) The function is defined for all values of x, so the domain is (, ) . (b) Since 2e x attains all positive values, the range is (3, ) . 24. (a) The function is equivalent to y  tan 2 x, so we require 2 x  k2 for odd integers k. The domain is given by x  k4 for odd integers k.

(b) Since the tangent function attains all values, the range is (, ) . 25. (a) The function is defined for all values of x, so the domain is (, ). (b) The sine function attains values from –1 to 1, so 2  2sin (3x   )  2 and hence 3  2sin (3 x   )  1  1. The range is [3, 1]. 26. (a) The function is defined for all values of x, so the domain is (, ). 5

(b) The function is equivalent to y  x 2 , which attains all nonnegative values. The range is [0, ) . 27. (a) The logarithm requires x  3  0, so the domain is (3, ). (b) The logarithm attains all real values, so the range is (, ). Copyright  2023 Pearson Education, Inc.

Chapter 1 Practice Exercises

28. (a) The function is defined for all values of x, so the domain is (, ). (b) The cube root attains all real values, so the range is (, ). 29.

y  5  ( x  3)( x  1) so the domain  (, 1]  [3, ); so the range  (, 5].

30.

y  2  32x

31.

y  4sin 1x so the domain  (, 0)  (0, ); if 32  x  2 , then 1  sin 1x  1, so the range  [4, 4].

32.

y  3cos x  4sin x so the domain  (, );

x 4

( x  3)( x  1)  0 and can be any positive number,

so the domain  (, ); 0  32x  3 so the range  [2, 5). x 4



and

 



32  42  5 so 3cos x  4sin x  5 53 cos x  54 sin x



 5(cos  cos x  sin  sin x)  5cos(  x), where and 1  cos(  x)  1 so the range  [5, 5].

33. (a) Increasing because volume increases as radius increases. (b) Neither, since the greatest integer function is composed of horizontal (constant) line segments. (c) Decreasing because as the height increases, the atmospheric pressure decreases. (d) Increasing because the kinetic (motion) energy increases as the particles velocity increases. 34. (a) Increasing on [2, ) (c) Increasing on (, )

(b) Increasing on [1, ) (d) Increasing on  12 , 



35. (a) The function is defined for 4  x  4, so the domain is [4, 4]. (b) The function is equivalent to y  | x |,  4  x  4, which attains values from 0 to 2 for x in the domain. The range is [0, 2]. 36. (a) The function is defined for 2  x  2, so the domain is [2, 2]. (b) The range is [1, 1]. 0 1

37. First piece: Line through (0, 1) and (1, 0); m  1  0  11  1  y   x  1  1  x 0 1

Second piece: Line through (1, 1) and (2, 0); m  2  1  11  1  y   ( x 1)  1   x  2  2  x 1  x, 0  x  1 f ( x)    2  x, 1  x  2 50

38. First piece: Line through (0, 0) and (2, 5); m  2  0  52  y  52 x 05

Second piece: Line through (2, 5) and (4, 0); m  4  2  25   52  y   52 ( x  2)  5   52 x  10  10  52x 5 x, 0  x  2   2 f ( x)   (Note: x  2 can be included on either piece.) 5x 10  2 , 2  x  4

 39. (a) ( f  g )(1)  f ( g (1))  f  

 1  f (1)  11  1 1  2 

Chapter 1 Functions



(b) ( g  f )(2)  g ( f (2))  g 12 

1 1 2 2

1 2.5



2 5

 1x   1/1x  x, x  0

40. (a) ( f  g )(1)  f ( g (1))  f

1  x  2 

1 1 2 x2



4 x2

1 2 x  2

 3 1  1  f (0)  2  0  2

(b) ( g  f )(2)  f ( g (2))  g (2  2)  g (0)  3 0  1  1 (c) ( f  f )( x)  f ( f ( x))  f (2  x)  2  (2  x)  x





(d) ( g  g )( x)  g ( g ( x))  g 3 x  1  3 3 x  1  1 41. (a) ( f  g )( x)  f ( g ( x))  f

 x  2   2   x  2    x, x  2 2

( g  f )( x)  g ( f ( x))  g (2  x 2 ) 

(b) Domain of f  g : [2, ). Domain of g  f : [2, 2].

 2  x2   2  4  x2

 1 x   1 x  4 1 x ( g  f )( x)  g ( f ( x))  g  x   1  x

42. (a) ( f  g )( x)  f ( g ( x))  f

(b) Domain of f  g : (, 1] Domain of g  f : [0, 1] 43.

y  f ( x)

44.

Chapter 1 Practice Exercises

45.

46.

The graph of f 2 ( x)  f1 (| x |) is the same as the graph of f1 ( x) to the right of the y-axis. The graph of f 2 ( x) to the left of the y-axis is the reflection of y  f1 ( x), x  0 across the y-axis. 47.

It does not change the graph.

48.

Whenever g1 ( x) is positive, the graph of y  g 2 ( x)  | g1 ( x)| is the same as the graph of y  g1 ( x). When g1 ( x) is negative, the graph of y  g 2 ( x) is the reflection of the graph of y  g1 ( x) across the x-axis.

Whenever g1 ( x) is positive, the graph of y  g 2 ( x)  g1 ( x) is the same as the graph of y  g1 ( x). When g1 ( x) is negative, the graph of y  g 2 ( x) is the reflection of the graph of y  g1 ( x) across the x-axis.

50.

49.

Whenever g1 ( x) is positive, the graph of y  g 2 ( x)  | g1 ( x)| is the same as graph of y  g1 ( x). When g1 ( x) is negative, the graph of y  g 2 ( x) is the reflection of the graph of y  g1 ( x) across the x-axis.

Chapter 1 Functions

51.

52.





y  g ( x  3)  12

(b)

y  g (  x) y  5  g ( x)

y  g x  23  2

(d) (f )

y   g ( x) y  g (5 x)

(b) Horizontally compress the graph of f by a factor of 4. 54. (a) Shift the graph of f right 5 units. (c) Horizontally compress the graph of f by a factor of 3 and then reflect the graph about the y-axis. (d) Horizontally compress the graph of f by a factor of 2 and then shift the graph left 12 unit. (e) Horizontally stretch the graph of f by a factor of 3 and then shift the graph down 4 units. (f ) Vertically stretch the graph of f by a factor of 3, then reflect the graph about the x-axis, and finally shift the graph up 14 unit. 55. Reflect the graph of y  x about the x-axis followed by a horizontal compression by a factor of 1 , then shift left 2 units. 2

56. Reflect the graph of y  x about the x-axis, followed by a vertical compression of the graph by a factor of 3, then shift the graph up 1 unit.

57. Vertical compression of the graph of y  12 by a x

factor of 2, then shift the graph up 1 unit.

Chapter 1 Practice Exercises

58. Reflect the graph of y  x1/3 about the y-axis, then compress the graph horizontally by a factor of 5.

60.

59.

period  4

period   62.

61.

period  4

period  2 64.

63.

period  2

period  2 65. (a) sin B  sin 3  bc  b2  b  2sin 3  2

   3 . By the Pythagorean theorem, 3 2

a 2  b2  c2  a  c2  b2  4  3  1 .

(b) sin B  sin 3  bc  2c  c 

66. (a) sin A  ac  a  c sin A

2  sin 3

  3 2

 4 . Thus, a  c 2  b 2  3

   (2)  4 3

(b) tan A  ba  a  b tan A

4  2 3. 3 3

Chapter 1 Functions

67. (a)

tan B  ba  a  tanb B

(b) sin A  ac  c  sina A (b) sin A  ac 

68. (a) sin A  ac

c 2  b2 c

69. Let h  height of vertical pole, and let b and c denote the distances of points B and C from the base of the pole, measured along the flat ground, respectively. Then, tan 50  hc , tan 35  bh , and b  c  10. Thus, h  c tan 50 and h  b tan 35  (c  10) tan 35  c tan 50  (c  10) tan 35  c(tan 50  tan 35)  10 tan 35 c

10 tan 35  h  c tan 50 tan 50  tan 35

tan 35 tan 50  16.98 m.  10 tan 50  tan 35

70. Let h  height of balloon above ground. From the figure at the right, tan 40  ah , tan 70  bh , and a  b  2 . Thus, h  b tan 70  h  (2  a ) tan 70 and h  a tan 40  (2  a ) tan 70  a tan 40  a(tan 40  tan 70)  2 tan 70 2 tan 70  a  tan 40  h  a tan 40   tan 70 2 tan 70 tan 40  1.3 km.  tan 40  tan 70

71. (a)

(b) The period appears to be 4 . (c)

f ( x  4 )  sin( x  4 )  cos



x  4 2

  sin( x  2 )  cos   2   sin x  cos x 2

since the period of sine and cosine is 2 . Thus, f(x) has period 4 . 72. (a)

(b) Domain: (, 0)  (0, ); Range: [1, 1]

x 2

Chapter 1 Practice Exercises

 21  kp   f  21   sin 2  0 for all integers k. 1   . But then f  21  kp   sin  (1/(21))  kp   0 Choose k so large that 21  kp  1  0  1/(2 )   kp

which is a contradiction. Thus f has no period, as claimed. 73. (a) Domain:   x  

(b) Domain: x  0

74. (a) Domain: (, 0)  (0, )

(b) Domain: (,  2)  (2, 2)  (2, )

75. (a) Domain: 3  x  3

(b) Domain: 0  x  4

76. (a) Domain: [1, 1)

(b) Domain: [1, 1]

77. ( f  g )( x)  ln(4  x 2 ) and domain:  2  x  2 ( g  f )( x)  4  (ln x) 2 and domain: x  0 ( f  f )( x)  ln(ln x) and domain: x  1 ( g  g )( x)   x 4  8 x 2  12 and domain:   x  

78. (a) Even (c) Neither even nor odd

(b) Neither even nor odd (d) Even

79. This requires Grapher Technology. 80. For c  0, Domain: (, ) For c  0, Domain: (,  c )  ( c , ) y 4

y  ln( x 2  c )

81. This requires Grapher Technology. 82. For large values of x, y  a x has the largest values; y  log a x has the smallest. 83. (a) Domain: (, ); Range:  2 , 2 

(b) Domain: [1, 1]; Range: [1, 1]

Chapter 1 Functions

84. (a) Domain:   x  ; Range: 0  y  

(b) Domain: 1  x  1; Range: 1  y  1

y 

y  cos 1 (cos x )

y  cos(cos1 x )

1 0.5

2



2



0.5



85. (a) No

(b) Yes

86. Answers depend on the view screen used. For [15, 17]  [5 106 , 107 ] it appears that e x  107 for x  16.128. 87. (a)

 3  x, g ( f ( x))  3 x3  x

f ( g ( x))  3 x

(b) y

y  x3

y  x1/ 3

yx



88. (a) h(k ( x))  14 (4 x)1/3

3  x, k (h( x))   4  x4   x 3

1/3

(b) y 4

y

x3 4

y  (4 x )1/ 3

yx

Chapter 1 Additional and Advanced Exercises

CHAPTER 1 ADDITIONAL AND ADVANCED EXERCISES

1. There are (infinitely) many such function pairs. For example, f ( x)  3 x and g ( x)  4 x satisfy f ( g ( x))  f (4 x)  3(4 x)  12 x  4(3x)  g (3x)  g ( f ( x)). 2. Yes, there are many such function pairs. For example, if g ( x)  (2 x  3)3 and f ( x)  x1/3 , then ( f  g )( x)  f ( g ( x))  f ((2 x  3)3 )  ((2 x  3)3 )1/3  2 x  3.

3. If f is odd and defined at x, then f ( x)   f ( x). Thus g (  x)  f (  x)  2   f ( x)  2 whereas  g ( x)  ( f ( x)  2)   f ( x)  2. Then g cannot be odd because g (  x)   g ( x)   f ( x) 2   f ( x)  2  4  0, which is a contradiction. Also, g ( x) is not even unless f ( x)  0 for all x. On the other hand, if f is even, then g ( x)  f ( x)  2 is also even: g (  x)  f (  x)  2  f ( x)  2  g ( x). 4. If g is odd and g(0) is defined, then g (0)  g ( 0)   g (0). Therefore, 2 g (0)  0  g (0)  0 . 5. For (x, y) in the 1st quadrant, | x |  | y |  1  x  x  y  1  x  y  1 . For (x, y) in the 2nd quadrant, | x |  | y |  x  1   x  y  x  1  y  2 x  1. In the 3rd quadrant, | x |  | y |  x  1   x  y  x  1  y  2 x  1. In the 4th quadrant, | x |  | y |  x  1  x  ( y )  x  1  y  1. The graph is given at the right. 6. We use reasoning similar to Exercise 5. (1) 1st quadrant: y  | y |  x  | x |  2 y  2 x  y  x. (2) 2nd quadrant: y  | y |  x  | x |  2 y  x  ( x)  0  y  0. (3) 3rd quadrant: y  | y |  x  | x |  y  ( y )  x  ( x)  0  0  all points in the 3rd quadrant satisfy the equation. (4) 4th quadrant: y  | y |  x  | x |  y  ( y )  2 x  0  x. Combining these results we have the graph given at the right. sin 2 x

7. (a) sin 2 x  cos 2 x  1  sin 2 x  1  cos 2 x  (1  cos x) (1  cos x)  1  cos x   1  cos x  sin x 1  cos x

(b) Using the definition of the tangent function and the double angle formulas, we have 1  cos  2  x   2   x .    cos2  2x   1  cos2 2 2x   11  cos cos x

tan 2 2x

sin 2 2x

8. The angles labeled  in the accompanying figure are equal since both angles subtend arc CD. Similarly, the two angles labeled α are equal since they both subtend arc AB. Thus, triangles AED and BEC are similar which implies

ac 2 a cos   b  ac  (a  c)(a  c)  b(2a cos   b) b

 a 2  c 2  2ab cos   b 2  c 2  a 2  b 2  2ab cos  .

1  cos x  sin x

Chapter 1 Functions

9. As in the proof of the law of sines of Section 1.3, Exercise 61, ah  bc sin A  ab sin C  ac sin B  the area of ABC  12 (base)(height)  12 ah  12 bc sin A  12 ab sin C  12 ac sin B .

10. As in Section 1.3, Exercise 61, (Area of ABC ) 2  14 (base)2 (height) 2  14 a 2 h 2  14 a 2b 2 sin 2 C  14 a 2 b 2 (1  cos 2 C ) . By the law of cosines, c 2  a 2  b 2  2ab cos C  cos C 

a 2  b2  c 2 . Thus, 2 ab

  a 2  b2  c 2 2  2 2 2 2  a 2b 2 1  ( a  b  c )  (area of ABC ) 2  14 a 2b 2 (1  cos 2 C )  14 a 2b 2 1       2 2 2 ab 4    4a b     





1 4a 2 b 2  ( a 2  b 2  c 2 ) 2  1 [(2ab  ( a 2  b 2  c 2 )) (2ab  ( a 2  b 2  c 2 ))]  16 16 1 [(( a  b)  c )(c  (a  b) 2 )]  1 [(( a  b)  c)((a  b)  c)(c  ( a  b))(c  ( a  b))]  16 16

 



abc 2



a  b  c 2



abc 2





abc  abc  s ( s  a)( s  b)( s  c), where s  . 2 2 

s ( s  a )( s  b)( s  c) .

Therefore, the area of ABC equals

11. If f is even and odd, then f (  x)   f ( x) and f (  x)  f ( x)  f ( x)   f ( x) for all x in the domain of f. Thus 2 f ( x)  0  f ( x)  0. f ( x)  f ( x) f (  x )  f (  (  x )) f ( x)  f ( x)  E ( x)    E ( x)  E is an even 2 2 2 f (  x)  f  ( x)  f ( x)  f ( x) f ( x)  f ( x)  . Then O( x)  function. Define O( x)  f ( x)  E ( x)  f ( x)  2 2 2 f ( x)  f ( x) f ( x)  f ( x)    O( x)  O is an odd function  f ( x)  E ( x)  O( x) is the sum of an 2 2

12. (a) As suggested, let E ( x) 





even and an odd function. (b) Part (a) shows that f ( x)  E ( x)  O( x) is the sum of an even and an odd function. If also f ( x)  E1 ( x)  O1 ( x), where E1 is even and O1 is odd, then f ( x)  f ( x)  0  ( E1 ( x)  O1 ( x))  ( E ( x)  O( x)) . Thus, E ( x)  E1 ( x)  O1 ( x)  O( x) for all x in the domain of f (which is the same as the domain of E  E1 and O  O1 ). Now ( E  E1 )( x)  E (  x)  E1 ( x)  E ( x)  E1 ( x) (since E and E1 are even)  ( E  E1 )( x)  E  E1 is even. Likewise, (O1  O)( x)  O1 ( x )  O(  x )  O1 ( x)  (O( x)) (since O and O1 are odd)   (O1 ( x)  O( x))  (O1  O) ( x)  O1  O is odd. Therefore, E  E1 and O1  O are both even and odd so they must be zero at each x in the domain of f by Exercise 11. That is, E1  E and O1  O, so the decomposition of f found in part (a) is unique. 13.









y  ax 2  bx  c  a x 2  ba x  b 2  4ba  c  a x  2ba  4ba  c 4a (a) If a  0 the graph is a parabola that opens upward. Increasing a causes a vertical stretching and a shift of the vertex toward the y-axis and upward. If a  0 the graph is a parabola that opens downward. Decreasing a causes a vertical stretching and a shift of the vertex toward the y-axis and downward. (b) If a  0 the graph is a parabola that opens upward. If also b  0, then increasing b causes a shift of the graph downward to the left; if b  0, then decreasing b causes a shift of the graph downward and to the right. If a  0 the graph is a parabola that opens downward. If b  0, increasing b shifts the graph upward to the right. If b  0, decreasing b shifts the graph upward to the left. (c) Changing c (for fixed a and b) by c shifts the graph upward c units if c  0, and downward c units if c  0.

Chapter 1 Additional and Advanced Exercises

14. (a) If a  0, the graph rises to the right of the vertical line x  b and falls to the left. If a < 0, the graph falls to the right of the line x  b and rises to the left. If a  0 , the graph reduces to the horizontal line y  c. As | a | increases, the slope at any given point x  x0 increases in magnitude and the graph becomes steeper. As | a | decreases, the slope at x0 decreases in magnitude and the graph rises or falls more gradually. (b) Increasing b shifts the graph to the left; decreasing b shifts it to the right. (c) Increasing c shifts the graph upward; decreasing c shifts it downward. 15. Each of the triangles pictured has the same base b  vt  v (1 sec) . Moreover, the height of each triangle is the same value h. Thus 12 (base)(height)  12 bh  A1  A2  A3  … . In conclusion, the object sweeps out equal areas in each one second interval.

16. (a) Using the midpoint formula, the coordinates of P are



a0 b0 , 2 2

   , . Thus the slope a b 2 2

y

 ba . of OP  x  ba /2 /2 b0

(b) The slope of AB  0  a   ba . The line segments AB and OP are perpendicular when the product of their slopes is 1 

 ba   ba    ba . Thus, b2  a2  a  b (since both are positive). Therefore, AB is 2

perpendicular to OP when a  b. 17. From the figure we see that 0    2 and AB  AD  1. From trigonometry we have the following: sin  . We can see that: sin   EB  EB, cos   AE  AE , tan   CD  CD, and tan   EB  cos AD  AB AB AE

  area ADC  1 ( AE )( EB)  1 ( AD) 2   1 ( AD) (CD) area AEB  area sector DB 2 2 2 sin  .  12 sin  cos   12 (1)2   12 (1)(tan  )  12 sin  cos   12   12 cos 

18. ( f  g )( x)  f ( g ( x))  a(cx  d )  b  acx  ad  b and ( g  f )( x)  g ( f ( x))  c (ax  b)  d  acx  cb  d Thus ( f  g )( x)  ( g  f )( x)  acx  ad  b  acx  bc  d  ad  b  bc  d . Note that f (d )  ad  b and g (b)  cb  d , thus ( f  g )( x)  ( g  f )( x) if f (d )  g (b). 19. (a) The expression a(bc  x )  d is defined for all values of x, so the domain is (, ). Since bc  x attains all positive values, the range is (d, ) if a > 0 and the range is (d, ) if a < 0. (b) The expression a logb ( x  c)  d is defined when x  c > 0, so the domain is (c, ). Since a logb ( x  c)  d attains every real value for some value of x, the range is (, ).

Chapter 1 Functions

20. (a) Suppose f ( x1 )  f ( x2 ). Then: ax1 b ax b  cx 2 d cx1  d 2

(ax1  b)(cx2  d )  (ax2  b)(cx1  d ) acx1 x2  adx1  bcx2  bd  acx1 x2  adx2  bcx1  bd adx1  bcx2  adx2  bcx1 (ad  bc) x1  (ad  bc) x2 .

Since ad  bc  0, this means that x1  x2 . ax b y  cx d

(b)

cxy  dy  ax  b (cy  a ) x   dy  b  dy b

x  cy  a

Interchange x and y. dx  b y cx  a f 1 ( x)  cxdxab

21. (a) y = 100,000  10,000x, 0  x  10 (b) y  55, 000 100, 000  10, 000 x  55, 000 10, 000 x  55, 000 x  4.5 The value is $55,000 after 4.5 years. 22. (a) f (0) = 90 units (b) f (2) = 90  52 ln 3  32.8722 units (c)

ln(10/3)

5000  10  ln(1.08)t  ln 10  t ln1.08  ln 10  t  23. 1500(1.08)t  5000  1.08t  1500  15.6439 3 3 3 ln1.08

It will take about 15.6439 years. (If the bank only pays interest at the end of the year, it will take 16 years.) 24.

70  70 A(t )  A0 ert ; A(t )  2 A0  2 A0  A0 ert  ert  2  rt = ln 2  t  lnr 2  t  0.7  100 r r ( r %) x

25. ln x( x )  x x ln x and ln( x x ) x  x ln( x x )  x 2 ln x; then, x x ln x  x 2 ln x  x x  x 2  x ln x  2 ln x  x  2. x

Therefore, x( x )  ( x x ) x when x = 2 and when x = 1.

Chapter 1 Additional and Advanced Exercises 26. (a) No, there are two intersections: one at x = 2 and the other at x = 4.

(b) Yes, because there is only one intersection.

27.

log 4 x  log 2 x

28. (a)

 lnln 4x   ln x  ln 2  ln 2  ln 2  1  lnln 2x  ln 4 ln x ln 4 2 ln 2 2

ln 2 , g ( x)  ln x f ( x)  ln ln 2 x

(b) f is negative when g is negative, positive when g is positive, and undefined when g = 0; the values of f decrease as those of g increase.

CHAPTER 2 2.1

LIMITS AND CONTINUITY

RATES OF CHANGE AND TANGENT LINES TO CURVES

1. (a)

f f (3)  f (2)  3 2  2819  19 x

(b)

f f (1)  f ( 1)  1( 1)  22 0  1 x

2. (a)

g (3)  g (1) ( ) g  3  1  3  21  2 x

(b)

g (4)  g ( 2) g  4  ( 2)  8 6 8  0 x

3. (a)

h  t

(b)

h  t

4. (a)

     11   4

h 34  h 4 3   4 4



g g ( )  g (0) (2 1) (2 1)   0    2  0 t

(b)

R  R (2)  R (0)   20

P  P (2)  P (1)  (816 10)  (1 4  5)  2  2  0  21 1

7. (a) (b) 8. (a)

     0 3  3 3

h 2  h 6   2



g g ( )  g (  ) (2 1) (2 1)   (  )  0 t 2

81  1 31  2 1 2

2 ((2 h )2 5) (22 5) 4 4 h  h 2 51 y    4h h h  4  h. As h  0, 4  h  4  at P (2, 1) the slope is 4. h h x

y  ( 1)  4( x  2)  y  1  4 x  8  y  4 x  9

2 y (7(2 h )2 ) (7 22 ) 744 h h 2 3    4 hhh  4  h. As h  0,  4  h  h h x

4  at P (2, 3) the slope

is 4. (b) y  3  ( 4)( x  2)  y  3  4 x  8  y  4 x  11 y

((2  h )2  2(2  h ) 3)  (22  2(2) 3)

y

((1 h ) 2  4(1 h )) (12  4(1))

y

(2  h )3  23

y

2 (1 h)3 (2 13 )

4  4 h  h 2  4  2 h 3( 3)

 9. (a) x  h h P(2,  3) the slope is 2. (b) y  (3)  2( x  2)  y  3  2 x  4  y  2 x  7 1 2 h  h 2  4  4 h  ( 3)

 2h h h  2  h. As h  0, 2  h  2  at

  h h 2h  h  2. As h  0, h  2  2  at P(1,  3) the 10. (a) x  h h slope is 2. (b) y  ( 3)  ( 2)( x  1)  y  3  2 x  2  y  2 x  1 2

 812h  4hh  h 8  12 h  4hh  h  12  4h  h 2 . As h  0, 12  4h  h 2  12,  at P (2, 8) 11. (a) x  h the slope is 12. (b) y  8  12( x  2)  y  8  12 x  24  y  12 x  16 2

12. (a) x   213h h3h  h 1  3h 3hh  h  3  3h  h 2 . As h  0, 3 3h  h 2  3,  at h P(1, 1) the slope is 3. (b) y  1  ( 3)( x  1)  y  1  3 x  3  y  3x  4

Chapter 2 Limits and Continuity (1 h )3 12(1 h ) (13 12(1))

y

13h 3h 2  h3 1212 h ( 11)

13. (a) x    9h 3hh  h  9  3h  h 2 h h As h  0, 9  3h  h 2  9  at P(1,  11) the slope is 9. (b) y  (11)  (9)( x  1)  y  11  9 x  9  y  9 x  2

14. (a)

(2  h )3 3(2  h ) 2  4 (23 3(2)2  4) 812 h  6 h 2  h3 12 12 h 3h 2  4  0 3h 2  h3 y    h  3h  h 2 h x h 2

As h  0, 3h  h  0  at P (2, 0) the slope is 0. (b) y  0  0( x  2)  y  0 15. (a)

1  1 y 2  ( 2  h ) 2  h 2   2( 2 h)  1h  2( 21 h) x h





As h  0, 2( 21 h)  41 ,  at P 2, 21 the slope is 41 . (b)

 

y  21  41 ( x  (2))  y  12  41 x  12  y  41 x  1 (4  h )

16. (a)

(b) 17. (a)

y  x

4 h  4  h

As h  0, (b) 18. (a)

19. (a)





(4  h )  4 4 h  2  4 h  2   h 4 h  2 h ( 4 h  2)

1  4 h  2

1  1, 4 2 4

1 4 h  2

 at P (4, 2) the slope is 14 .

y  2  14 ( x  4)  y  2  14 x  1  y  14 x  1 y  x

7 ( 2 h )  7 ( 2)  h

As h  0, (b)

 y 4  h  2( 2 h ) 1  2 (4  hh) 2  4  42hh  12  1h   h  21h  21 h x 2  h As h  0, 21 h  12 ,  at P (4,  2) the slope is 12 . y  (2)  12 ( x  4)  y  2  12 x  2  y  12 x  4

1  9 h 3

9  h 3  h

1  1 , 6 9 3

(9  h) 9 9  h 3 9  h  3    h 9  h 3 h( 9  h 3)

1 9 h 3

 at P (2, 3) the slope is 61 .

y  3  61 ( x  (2))  y  3  61 x  13  y  61 x  83

Q Q1 (10, 225) Q2 (14,375)

Q3 (16.5, 475) Q4 (18,550)

p

Slope of PQ  t

650  225  42.5 m/sec 2010 650 375  45.83 m/sec 20 14 650  475  50.00 m/sec 20 16.5 650 550  50.00 m/sec 2018

(b) At t  20, the sportscar was traveling approximately 50 m/sec or 180 km/h. 20. (a)

Q Q1 (5, 20) Q2 (7, 39)

Q3 (8.5,58) Q4 (9.5, 72)

p

Slope of PQ  t

80  20  12 m/sec 10 5 80 39  13.7 m/sec 10 7 80 58  14.7 m/sec 10 8.5 8072  16 m/sec 10 9.5

(b) Approximately 16 m/sec

Section 2.1 Rates of Change and Tangent Lines to Curves 21. (a)

Profit (1000s)

200 160 120 80 40 0

(b) (c)

2010 2011 2012 2013 2014 Ye ar

p 174 62  112  56 thousand dollars per year  2014 2 t 2012 p 62 27  35 thousand dollars per year. The average rate of change from 2018 to 2019 is t  2012 2011 p 111 62  49 thousand dollars per year. The average rate of change from 2019 to 2020 is t  20132012 So, the rate at which profits were changing in 2019 is approximately 12 (35  49)  42 thousand dollars

per year.

22. (a) F ( x)  ( x  2)/( x  2) x 1.2 1.1 F ( x) 4.0 3.4

1.01 3.04

1.001 3.004

1.0001 3.0004

1 3

F  3.4 ( 3)  4.4 1.11 x F  3.004 ( 3)  4.004 1.0011 x

F  4.0 ( 3)  5.0 1.2 1 x F  3.04 ( 3)  4.04 1.011 x 3.0004  ( 3) F   4.0004 1.00011 x

(b) The rate of change of F ( x) at x  1 is 4. 23. (a)

g g (2)  g (1)  21  2211  0.414213 x g g (1 h )  g (1)  (1 h ) 1  1hh 1 x

g g (1.5)  g (1) 1  1.51  1.5  0.449489 x 0.5

(b) g ( x)  x 1 h

1.1

1.01

1.001

1.0001

1.00001

1.000001

1 h

1.04880

1.004987

1.0004998

1.0000499

1.000005

1.0000005

0.4880

0.4987

0.4998

0.499

0.5

 1  h  1 /h

11 1 f (3)  f (2) 3 2 6     16 3 2 1 1 1 1 2 T f (T )  f (2) T  TT  22  2TT  22T  2T2(TT 2)  2T2(2   21T , T  2 T ) T 2

(b) T f (T ) ( f (T )  f (2))/(T  2)

2.1 2.01 2.001 0.476190 0.497512 0.499750 0.2381 0.2488 0.2500 (c) The table indicates the rate of change is 0.25 at t  2.

2.0001 0.4999750 0.2500

2.00001 0.499997 0.2500

NOTE: Answers will vary in Exercises 25 and 26. 0  15 mph; [1, 2.5]: s  20 15  10 mph; [2.5, 3.5]: s  30 20  10 mph 25. (a) [0, 1]: st  15 t t 10 2.51 3 3.5 2.5

2.000001 0.499999 0.2500

Chapter 2 Limits and Continuity





(b) At P 12 , 7.5 : Since the portion of the graph from t  0 to t  1 is nearly linear, the instantaneous rate of

change will be almost the same as the average rate of change, thus the instantaneous speed at t  12 is 157.5  15 mi/hr. At P (2, 20): Since the portion of the graph from t  2 to t  2.5 is nearly linear, the 10.5  20  0 mi/hr. instantaneous rate of change will be nearly the same as the average rate of change, thus v  20 2.5 2 For values of t less than 2, we have the following.

Q Q1 (1, 15)

Q2 (1.5, 19) Q3 (1.9, 19.9)

Slope of PQ  st 15 20  5 mi/hr 1 2 19  20  2 mi/hr 1.5 2 19.9  20  1 mi/hr 1.9  2

Thus, it appears that the instantaneous speed at t  2 is 0 mi/hr. At P(3, 22):

Q Q1 (4, 35) Q2 (3.5, 30) Q3 (3.1, 23)

Slope of PQ  st 35 22  13 mi/hr 4 3 30  22  16 mi/hr 3.53 23 22  10 mi/hr 3.13

Q Q1 (2, 20) Q2 (2.5, 20)

Q3 (2.9, 21.6)

Slope of PQ  st 20 22  2 mi/hr 2 3 20 22  4 mi/hr 2.53 21.6 22  4 mi/hr 2.9 3

Thus, it appears that the instantaneous speed at t  3 is about 7 mi/hr. (c) It appears that the curve is increasing the fastest at t  3.5. Thus for P(3.5, 30) Slope of PQ  st Slope of PQ  st Q Q Q1 (4, 35) Q2 (3.75, 34) Q3 (3.6, 32)

3530  10 mi/hr 43.5 34 30  16 mi/hr 3.753.5 32 30  20 mi/hr 3.6 3.5

Q1 (3, 22)

Q2 (3.25, 25) Q3 (3.4, 28)

2230  16 mi/hr 33.5 2530  20 mi/hr 3.253.5 2830  20 mi/hr 3.4 3.5

Thus, it appears that the instantaneous speed at t  3.5 is about 20 mi/hr. gal

gal

15  2.2 26. (a) [0, 3]: At  10315  1.67 day ; [0, 5]: At  3.9 ; [7, 10]: At  0101.4  0.5 day 7 0 5 0 day

(b) At P(1, 14):

Q Q1 (2, 12.2) Q2 (1.5, 13.2) Q3 (1.1, 13.85)

Slope of PQ  At 12.2 14  1.8 gal/day 21 13.2 14  1.6 gal/day 1.51 13.8514  1.5 gal/day 1.11

Q Q1 (0, 15) Q2 (0.5, 14.6) Q3 (0.9, 14.86)

Slope of PQ  At 1514  1 gal/day 0 1 14.6 14  1.2 gal/day 0.51 14.86 14  1.4 gal/day 0.9 1

Thus, it appears that the instantaneous rate of consumption at t  1 is about 1.45 gal/day. At P (4, 6): Slope of PQ  At Q Slope of PQ  At Q Q1 (5, 3.9) Q2 (4.5, 4.8)

Q3 (4.1, 5.7)

3.9 6  2.1 gal/day 5 4 4.86  2.4 gal/day 4.5 4 5.7 6  3 gal/day 4.1 4

Q1 (3, 10)

Q2 (3.5, 7.8)

Q3 (3.9, 6.3)

10 6  4 gal/day 3 4 7.86  3.6 gal/day 3.5 4 6.36  3 gal/day 3.9  4

Thus, it appears that the instantaneous rate of consumption at t  1 is 3 gal/day. (solution continues on next page)

Section 2.2 Limit of a Function and Limit Laws At P (8, 1):

Q Q1 (9, 0.5) Q2 (8.5, 0.7)

Q3 (8.1, 0.95)

Slope of PQ  At

Q1 (7, 1.4)

0.51  0.5 gal/day 9 8 0.7 1  0.6 gal/day 8.58 0.951  0.5 gal/day 8.18

Q2 (7.5, 1.3) Q3 (7.9, 1.04)

Slope of PQ  At 1.4 1  0.6 gal/day 7 8 1.31  0.6 gal/day 7.58 1.04 1  0.6 gal/day 7.98

Thus, it appears that the instantaneous rate of consumption at t  1 is 0.55 gal/day. (c) It appears that the curve (the consumption) is decreasing the fastest at t  3.5. At P(3.5, 7.8): Slope of PQ  st A Q Slope of PQ  Q t 11.2 7.8   3.4 gal/day Q1 (2.5, 11.2) 4.8  7.8 2.53.5 Q1 (4.5, 4.8)   3 gal/day 4.53.5 10 7.8  4.4 gal/day Q (3, 10) 2 6 7.8  3.6 gal/day 33.5 Q2 (4, 6) 4 3.5 8.2 7.8  4 gal/day Q (3.4, 8.2) 3 7.4  7.8 3.4 3.5 Q3 (3.6, 7.4)  4 gal/day 3.6 3.5 Thus, it appears that the rate of consumption at t  3.5 is about 4 gal/day. 2.2

LIMIT OF A FUNCTION AND LIMIT LAWS

1. (a) Does not exist. As x approaches 1 from the right, g ( x) approaches 0. As x approaches 1 from the left, g ( x) approaches 1. There is no single number L that all the values g ( x) get arbitrarily close to as x  1. (b) 1 (c) 0 (d) 0.5 2. (a) 0 (b) 1 (c) Does not exist. As t approaches 0 from the left, f (t ) approaches 1. As t approaches 0 from the right, f (t ) approaches 1. There is no single number L that f (t ) gets arbitrarily close to as t  0. (d) 1 3. (a) (d) (g) (j)

True False True True

(b) (e) (h) (k)

4. (a) False (d) True (g) False

True False False False

(b) False (e) True (h) True

5. lim | xx | does not exist because | xx |  xx  1 if x  0 and | xx |  xx   1 if x  0. As x approaches 0 from the left, | xx | x0

approaches 1. As x approaches 0 from the right, | xx | approaches 1. There is no single number L that all the function values get arbitrarily close to as x  0.

1 become increasingly large and negative. As x approaches 1 6. As x approaches 1 from the left, the values of x 1 from the right, the values become increasingly large and positive. There is no number L that all the function values get arbitrarily close to as x  1, so lim x11 does not exist. x 1

7. Nothing can be said about f ( x) because the existence of a limit as x  x0 does not depend on how the function is defined at x0 . In order for a limit to exist, f ( x) must be arbitrarily close to a single real number L when x is close enough to x0 . That is, the existence of a limit depends on the values of f ( x) for x near x0 , not on the definition of f ( x) at x0 itself. Copyright  2023 Pearson Education, Inc.

Chapter 2 Limits and Continuity

8. Nothing can be said. In order for lim f ( x) to exist, f ( x) must close to a single value for x near 0 regardless of x 0

the value f (0) itself.

9. No, the definition does not require that f be defined at x  1 in order for a limiting value to exist there. If f (1) is defined, it can be any real number, so we can conclude nothing about f (1) from lim f ( x)  5. x 1

10. No, because the existence of a limit depends on the values of f ( x) when x is near 1, not on f (1) itself. If lim f ( x) exists, its value may be some number other than f (1)  5. We can conclude nothing about lim f ( x), x 1

whether it exists or what its value is if it does exist, from knowing the value of f (1) alone. 11.

x 1

lim ( x 2  13)  ( 3)2  13  9  13  4

x3

12. lim ( x 2  5 x  2)  (2)2  5(2)  2  4  10  2  4 x 2

13. lim 8(t  5)(t  7)  8(6  5)(6  7)  8 t 6

14.

15.

16.

17.

lim ( x3  2 x 2  4 x  8)  (2)3  2(2)2  4( 2)  8  8  8  8  8  16

x 2

2(2) 5 lim 2 x 53  11(2)3 x2 11 x

    2  23   1  (8  2)  43   1  (6)  13   2

lim (8  3s )(2s  1)  8  3 23

t 2/3

x 1/2

y2 2

y 2 y 5 y  6

20.

       (2)   23  4   (2)  25    252

lim 4 x (3x  4)2  4  12 3  12  4

18. lim

19.

 93  3



2 2 4 4 1  410  20 6 5 (2) 2 5(2)  6

y 3



lim

z 2  10  42  10  16  10  6

lim (5  y ) 4/3  [5  (3)]4/3  (8)4/3  (8)1/3

z 4

21. lim

3  3h 1 1

22. lim

5h  4  2  lim h h 0

h 0

  24  16 4

3  3  32 3(0) 1 1 1 1 (5h  4)  4 5 h  4  2 5h  4  2   lim h 5 h  4  2 h 0 h 5 h  4  2





 lim

h 0 h

 5h  4  2 

x 5  lim x 5  lim 1  1  1 2 x 5 x  25 x 5 ( x 5)( x 5) x 5 x 5 55 10

23. lim 24.

x 3 lim 2 x 3  lim  lim 1  1   12 x 3 x  4 x 3 x 3 ( x 3)( x 1) x 3 x 1 31

 lim

h0

5  5h 4  2

5  5 4 4 2

Section 2.2 Limit of a Function and Limit Laws 25.

2 ( x 5)( x  2) lim x x3x510  lim  lim ( x  2)  5  2  7 x 5 x 5 x 5

x 5

( x 5)( x  2)  lim ( x  5)  2  5  3 x2 x 2

26. lim x x7x210  lim x 2

x2

(t  2)(t 1)

27. lim t 2 t  2  lim (t 1)(t 1)  lim tt12  1112  32 t 1 t 1

28. 29.

t 1

(t  2)(t 1)

lim t 23t  2  lim (t  2)(t 1)  lim tt  22  11 22   13

t 1 t t  2

t 1

5 y 3 8 y 2

4 2 y 0 3 y 16 y

32.

y 2 (5 y 8)

 lim

2 2 y 0 y (3 y 16) 1 x

1

5 y 8

2 y 0 3 y 16



 816   12

 x1

lim xx 11  lim x x 1  lim 1xx  x11  lim  1x  1

x1

x 1

x1

( x 1)  ( x 1) ( x 1)( x 1)

1  1

lim x 1 x x 1  lim

x 0

33. lim u 3 1  lim

(u 2 1)(u 1)(u 1)

v  2 v 16

 lim

(u 2 1)(u 1)

u 1

(v  2)( v 2  2v  4)

34. lim v4 8  lim



2x 1 x 0 ( x 1)( x 1) x

 lim

2 u 1 (u  u 1)(u 1)

u 1 u 1

35.

t 1

2( x  2) lim 32 x  42  lim 2  lim 22  42   12 x 2 x  2 x x 2 x ( x  2) x 2 x

30. lim

31.

t 1

u 2 u 1

  lim

2  21  2 x 0 ( x 1)( x 1)

(11)(11)

 111  43

v 2  2v  4  4 4  4  12  3 2 (4)(8) 32 8 v 2 (v  2)( v  4)

 lim

v 2 (v  2)(v  2)( v  4)

x 3 lim xx93  lim  lim 1  1  16 ( 3)( 9 3 x x 3) x 9 x 3  9 9 x x 2



x (4  x ) x (2 x )(2 x )  lim  lim x 2  2 x x 4 2 x x 4 x 4

36. lim 4 x  x  lim x 4 2 x

 x 3  2  lim ( x1) x3  2  lim x  3  2  4  2  4   x 1  x 3 2  x 3  2  x1 ( x 3)4 ( x 1)

37. lim

x 1  lim x 3  2 x 1

38.

x 2 8 3  lim x 1 x 1

x 1

lim

x 1

 lim

x 2

x 2 12  4  lim x 2 x 2

 lim

x2

 x 8 3 x 8 3  lim ( x 8)9  lim ( x1)( x1) x 1 ( x 1) x 83  x 1 ( x 1) x 8 3 ( x 1) x 8 3 2

x 1

x 1

39. lim



x  4(2  2)  16

x 2 8  3

 323   13

 x 12 4 x 12 4  lim ( x 12)16  lim ( x2)( x2) x 2 ( x  2) x 12  4  x 2 ( x  2) x 12  4  ( x  2) x 12  4  2

x2

x 2 12  4



4  12 16  4

Chapter 2 Limits and Continuity

 x 5 3  lim ( x2) x 5 3  lim ( x2) x 5 3 40. lim  lim ( x 5) 9 x 2 ( x  2)( x  2) x 2 x 5 3 x 2  x 5 3 x 5 3 x 2 2

x 2

x 2 5 3  x 2

 lim

41.

( x  2)

x2

9 3   32 4

 2 x 5  2 x 5   lim 4( x 5)  lim 9 x x 3 ( x  3) 2  x 5  x 3 ( x 3) 2  x 5  ( x 3) 2  x 5  (3 x )(3 x ) 3 x  6  3  lim  lim x 3 ( x 3) 2 x 5  x 3 2  x 5 2  4 2 2

2 lim 2 x x3 5  lim x 3 x 3

  lim (4 x)5 x 9   lim (4 x)5 x 9  25( x 9) 16  x x4 x 4 5 x 9 x 4  5 x 9  5 x 9  x  4 (4  x ) 5 x 9   lim  lim 5 x 9  5 25  5 4 x

42. lim

 lim



(4  x ) 5 x 2 9 2

x 4

(4  x )(4 x )

x 4

4 x

43. lim (2sin x  1)  2 sin 0  1  0  1  1

44.

45. lim sec x  lim cos1 x  cos1 0  11  1

46.

x 0

lim sin 2 x   lim sin x   (sin 0) 2  02  0 x 0  x 0  sin x  sin 0  0  0 lim tan x  lim cos x cos 0 1

x 0

x sin x  1 0 sin 0  1 0  0  1 47. lim 13cos x 3cos 0 3 3 x 0

48. lim ( x 2  1)(2  cos x)  (02  1)(2  cos 0)  (1)(2  1)  (1)(1)  1 x 0

49.

lim

x 

x  4 cos( x   )  lim

50. lim 7  sec2 x  x 0

x  4  lim cos( x   )    4  cos 0  4    1  4  

x 

lim (7  sec2 x)  7  lim sec2 x  7  sec2 0  7  (1) 2  2 2

x 0

51. (a) Quotient Rule (c) Sum and Constant Multiple Rules

(b) Difference and Power Rules

52. (a) Quotient Rule (c) Difference and Constant Multiple Rules

(b)

Power and Product Rules

53. (a) lim f ( x) g ( x)   lim f ( x)   lim g ( x)   (5)( 2)  10 x c  x c   x c     (b) lim 2 f ( x) g ( x)  2  lim f ( x )   lim g ( x)   2(5)(2)  20 x c  x c   x c  (c) lim [ f ( x )  3 g ( x)]  lim f ( x)  3 lim g ( x)  5  3( 2)  1 x c

(d)

x c x c lim f ( x ) f ( x) xc lim  lim f ( x )  lim g ( x )  5(52)  75 x c f ( x )  g ( x ) x c

x c

Section 2.2 Limit of a Function and Limit Laws

54. (a) lim [ g ( x)  3]  lim g ( x)  lim 3  3  3  0 x 4

x4

x 4

(b) lim xf ( x )  lim x  lim f ( x)  (4)(0)  0 x 4

x4

g ( x)

(d) lim f ( x ) 1  limx f (4x )  lim 1  031  3 x 4

x4

x 4

55. (a) lim [ f ( x)  g ( x )]  lim f ( x)  lim g ( x )  7  (3)  4 x b

x b

(b) lim f ( x)  g ( x)   lim f ( x )   lim g ( x)   (7)(3)  21 x b  x b   xb  (c) lim 4 g ( x)   lim 4   lim g ( x)   (4)(3)  12 x b  x b   x b  (d) lim f ( x)/g ( x)  lim f ( x)/ lim g ( x)  73   73 x b

56. (a)

x b

lim [ p ( x)  r ( x)  s ( x)]  lim p ( x)  lim r ( x)  lim s ( x )  4  0  ( 3)  1

x 2

(b) lim p( x)  r ( x)  s ( x )   lim p ( x )   lim r ( x)   lim s ( x)   (4)(0)(3)  0 x 2  x 2   x2   x 2    (c) lim [4 p ( x)  5r ( x)]/s ( x)   4 lim p ( x)  5 lim r ( x )  lim s ( x)  [4(4)  5(0)]/  3  16 3 x 2 x 2  x2  x2 2 h (2  h ) (1 h ) 2 12  lim 1 2 h h h 1  lim h  lim (2  h)  2 h h 0 h 0 h 0 h 0

57. lim

2 h ( h  4) ( 2  h ) 2 ( 2)2  lim 4 4h h h  4  lim h  lim (h  4)  4 h h 0 h 0 h 0 h 0

58. lim

[3(2  h )  4][3(2)  4]  lim 3hh  3 h h 0 h 0

59. lim

60. lim

h 0

61. lim

h 0

 21 h  12   lim 22 h 1  lim 2(2 h)  lim h 0 2 h

7h  7  lim h h 0

h0 2 h ( 2  h )

 7 h  7  7 h  7   lim (7 h)7  lim h  lim h 7  h  7  h 0 h 7  h  7  h 0 h 7  h  7  h0

3(0  h ) 1  3(0) 1  lim h h 0 h 0

62. lim

63. lim

x 0

h   14 h 0 h (4  2 h ) 1  1 7h  7 2 7

 3h 1 1 3h 1 1  lim (3h1)1  lim 3h  lim 3  3 h 3h 1 1 h 0 h 3h 1 1 h 0 h 3h 1 1 h0 3h11 2

5  2 x 2  5  2(0)2  5 and lim

x0

5  x 2  5  (0) 2  5; by the Sandwich Theorem, lim f ( x)  5 x 0

64. lim (2  x 2 )  2  0  2 and lim 2 cos x  2(1)  2; by the Sandwich Theorem, lim g ( x)  2 x 0

x 0





x 0

x 1 65. (a) lim 1  x6  1  06  1 and lim 1  1; by the Sandwich Theorem, lim 2x2sin cos x x0

x0

x 0

Chapter 2 Limits and Continuity (b) For x  0, y  ( x sin x)/(2  2 cos x) lies between the other two graphs in the figure, and the graphs converge as x  0.

    lim  lim x2 24

1 x 0 2

66. (a) lim

1 x 0 2

x 2  1  0  1 and lim 1  1 ; by the Sandwich Theorem, lim 1cos x  1 . 2 24 2 2 2 2 x 0 x0 2 x 0 x 2

(b) For all x  0, the graph of f ( x)  (1  cos x)/x lies between the line y  12 and the parabola y  12  x 2 /24, and the graphs converge as

x  0.

67. (a)

f ( x)  ( x 2  9)/( x  3)

3.1

3.01

3.001

3.0001

3.00001

3.000001

f ( x)

6.1

6.01

6.001

6.0001

6.00001

6.000001

2.9

2.99

2.999

2.9999

2.99999

2.999999

f ( x)

5.9

5.99

5.999

5.9999

5.99999

5.999999

The estimate is lim f ( x)  6. x 3

(b)

( x 3)( x 3)  x  3 if x  3, and lim ( x  3)  3  3  6 x 3 x 3



68. (a) g ( x)  ( x 2  2)/ x  2

x g ( x)

1.4 2.81421



1.41 2.82421

1.414 2.82821

1.4142 2.828413

1.41421 2.828423

1.414213 2.828426

Section 2.2 Limit of a Function and Limit Laws (b)

 x  2  x 2   x  2 if x  2, and lim x  2  2  2  2 2     x 2  x 2

69. (a) G ( x)  ( x  6)/( x 2  4 x  12) x 5.9 5.99 G ( x ) .126582 .1251564

x G ( x)

6.1 .123456

6.01 .124843

5.999 .1250156

5.9999 .1250015

5.99999 5.999999 .1250001 .1250000

6.001 .124984

6.0001 .124998

6.00001 .124999

6.000001 .124999

(b)

x 6 x6  ( x  6)(  x 1 2 if x  6, and lim x 1 2  61 2   18  0.125 x  2) ( x 2  4 x 12) x 6

70. (a) h( x)  ( x 2  2 x  3)/( x 2  4 x  3)

x h( x )

2.9 2.052631

2.99 2.005025

2.999 2.000500

2.9999 2.000050

2.99999 2.000005

2.999999 2.0000005

x 3.1 h( x) 1.952380

3.01 1.995024

3.001 1.999500

3.0001 1.999950

3.00001 1.999995

3.000001 1.999999

(b)

( x 3)( x 1)

x 3

Chapter 2 Limits and Continuity

71. (a) f ( x )  ( x 2  1)/(| x |  1)

x f ( x)

1.1 2.1

1.01 2.01

1.001 2.001

1.0001 2.0001

1.00001 2.00001

1.000001 2.000001

x f ( x)

.9 1.9

.99 1.99

.999 1.999

.9999 1.9999

.99999 1.99999

.999999 1.999999

(b)

(c)

 ( x 1)( x 1)  x  1, x  0 and x  1 2  x 1 x  1 f ( x)   , and lim (1  x)  1  (1)  2 x 1  ( x 1)( x 1) x 1  ( x 1)  1  x, x  0 and x  1 

72. (a) F ( x)  ( x 2  3 x  2)/(2 | x |)

x F ( x)

2.1 1.1

2.01 1.01

2.001 1.001

2.0001 1.0001

2.00001 1.00001

2.000001 1.000001

x F ( x)

1.9 .9

1.99 .99

1.999 .999

1.9999 .9999

1.99999 .99999

1.999999 .999999

(b)

(c)

 ( x  2)( x 1) , x0 2  2 x x  3 x  2 F ( x)   , and lim ( x  1)  2  1  1 2 x ( x  2)( x 1) x 2       x 1, x 0 and x 2   2 x

73. (a) g ( )  (sin  )/



.1 .998334

.01 .999983

.001 .999999

.0001 .999999

.00001 .999999

.000001 .999999



.1 .998334

.01 .001 .999983 .999999

.0001 .999999

.00001 .999999

.000001 .999999

g ( )

lim g( )  1

 0

Section 2.2 Limit of a Function and Limit Laws (b)

74. (a) G (t )  (1  cos t )/t 2

t G (t )

.1 .499583

.01 .499995

.001 .499999

.0001 .5

t G (t )

.1 .499583

.01 .499995

.001 .499999

.0001 .00001 .000001 .5 .5 .5

f ( x)  x1/(1 x ) x .9 .99 f(x) .348678 .366032

.999 .367695

.9999 .367861

.99999 .367877

.999999 .367879

x f(x)

1.001 .368063

1.0001 .367897

1.00001 .367881

1.000001 .367878

.00001 .5

.000001 .5

lim G (t )  0.5

t 0

(b)

75. (a)

1.1 .385543

1.01 .369711

lim f ( x)  0.36788

x 1

(b)

Graph is NOT TO SCALE. Also, the intersection of the axes is not the origin: the axes intersect at the point (1, 2.71820).

Chapter 2 Limits and Continuity

76. (a)

f ( x)  (3x  1)/x x .1 .01 f(x) 1.161231 1.104669

.1 1.040415

x f(x)

.01 1.092599

.001 1.099215

.0001 1.098672

.00001 1.098618

.000001 1.098612

.001 1.098009

.0001 1.098551

.00001 1.098606

.000001 1.098611

lim f ( x)  1.0986

x 1

(b)

77. lim f ( x) exists at those points c where lim x 4  lim x 2 . Thus, c 4  c 2  c 2 (1  c 2 )  0  c  0, 1, or 1. x c

x c

Moreover, lim f ( x)  lim x 2  0 and lim f ( x )  lim f ( x)  1. x 0

x 0

x 1

x 1

78. Nothing can be concluded about the values of f , g , and h at x  2. Yes, f (2) could be 0. Since the conditions of the sandwich theorem are satisfied, lim f ( x )  5  0. x 2

lim f ( x )  lim 5 lim f ( x ) 5 f ( x ) 5 x4 x4 x 4    lim f ( x)  5  2(1)  lim f ( x)  2  5  7 lim x  lim 2 4 2 x4 x 4 x 4 x  2

79. 1  lim

x 4

80. (a) 1  lim

f ( x)

2 x 2 x

(b) 1  lim

f ( x)

2 x 2 x

x4

lim f ( x )

 x 2

lim x 2

x 2

  lim  x 2

lim f ( x )

 x 24

f ( x)  x 

 lim f ( x)  4 x 2

 lim 1    lim f ( x )   x 2 x   x 2 x 



f ( x)  2  12   xlim 2 x



f ( x ) 5 f ( x ) 5 81. (a) 0  3  0   lim x  2   lim ( x  2)   lim  x  2 ( x  2)   lim [ f ( x )  5]   x2  x 2   x2  x 2   lim f ( x )  5  lim f ( x)  5 x2

x2

f ( x ) 5 (b) 0  4  0   lim x  2   lim ( x  2)   lim f ( x)  5 as in part (a) x 2  x 2   x 2  2

f ( x) f ( x) f ( x) 82. (a) 0  1  0   lim 2   lim x    lim 2   lim x 2   lim  2  x 2   lim f ( x). That is,  x 0  x0 x   x 0  x 0  x  x 0 x   x 0  lim f ( x)  0. x 0

f ( x) f ( x) f ( x) f ( x) (b) 0  1  0   lim 2   lim x   lim  2  x   lim x . That is, lim x  0.  x0 x0  x 0 x   x 0  x0  x

Section 2.3 The Precise Definition of a Limit 83. (a) lim x sin 1x  0 x 0

(b) 1  sin 1x  1 for x  0: x  0   x  x sin 1x  x  lim x sin 1x  0 by the Sandwich Theorem; x 0

x  0   x  x sin 1x  x  lim x sin 1x  0 by the Sandwich Theorem. x 0

84. (a) lim x 2 cos x 0

(b) 1  cos

 0 1 x3

   1 for x  0   x  x cos    x  lim x cos    0 by the Sandwich Theorem 2

1 x3

since lim x 2  0.

1 x3

x 0

1 x3

x 0

85–90. Example CAS commands: Maple: f : x - (x^4  16)/(x  2); x0 : 2; plot( f (x), x  x0-1..x0 1, color  black, title  "Section 2.2, #85(a)" );

limit( f (x), x  x 0 );

In Exercise 87, note that the standard cube root, x^(1/3), is not defined for x<0 in many CASs. This can be overcome in Maple by entering the function as f : x - (surd(x 1, 3)  1)/x. Mathematica: (assigned function and values for x0 and h may vary) Clear[f , x] f[x _]: (x 3  x 2  5x  3)/(x  1) 2 x0  1; h  0.1; Plot[f[x],{x, x0  h, x0  h}] Limit[f[x], x  x0]

2.3

THE PRECISE DEFINITION OF A LIMIT

1. Step 1: x  5      x  5      5  x    5 Step 2:   5  7    2,or   5  1    4. A value of δ which assures x  5    1  x  7 is the smaller value,   2, or any smaller positive value.

Chapter 2 Limits and Continuity

Step 1: Step 2:

x  2      x  2      2  x    2   2  1    1, or   2  7    5. A value of  which assures x  2    1  x  7 is the smaller value,   1, or any smaller positive value.

Step 1: Step 2:

x  (3)      x  3      3  x    3   3   72    12 , or   3   12    52 . A value of  which assures x  (3)   

 72  x   12 is the smaller value,   12 , or any smaller positive value.

 

Step 1:

x   32      x  32      32  x    32

Step 2:

  32   72    2, or   32   12    1. A value of  which assures x   32   

 72  x   12 is the smaller value,   1, or any smaller positive value.

 

5. Step 1: Step 2:

x  12      x  12      12  x    12

1 , or   1  4    1 . A value of  which assures x  1    4  x  4 is   12  94    18 2 7 14 9 7 2 1 , or any smaller positive value. the smaller value,   18

Step 1: Step 2:

7. Step 1: Step 2:

8. Step 1: Step 2:

x  3      x  3      3  x    3   3  2.7591    0.2409, or   3  3.2391    0.2391. A value of  which assures x  3    2.7591  x  3.2391 is the smaller value,   0.2391, or any smaller positive value. x  5      x  5      5  x    5 From the graph,   5  4.9    0.1, or   5  5.1    0.1; thus   0.1, or any smaller positive value in either case. x  (3)      x  3      3  x    3 From the graph,   3  3.1    0.1, or   3  2.9    0.1; thus   0.1, or any smaller positive value.

Section 2.3 The Precise Definition of a Limit

9. Step 1: Step 2:

x  1      x  1      1  x    1 9    7 , or   1  25    9 ; thus   7 , or any smaller positive From the graph,   1  16 16 16 16 16

value. 10. Step 1: Step 2:

x  3      x  3      3  x    3 From the graph,   3  2.61    0.39, or   3  3.41    0.41; thus   0.39, or any smaller positive value.

11. Step 1: Step 2:

x  2      x  2      2  x    2 From the graph,  2  3    2  3  0.2679, or   2  5    5  2  0.2361; thus   5  2, or any smaller positive value.

12. Step 1:

x  (1)      x  1      1  x    1

Step 2:

From the graph,   1   25   



5 2  0.118 or   1   23    22 3  0.1340; thus 2

5 2 , or any smaller positive value. 2

13. Step 1: Step 2:

x  (1)      x  1      1  x    1 9  0.36; thus   9  0.36, or any From the graph,   1   16    79  0.77, or   1   16  25 9 25 25 smaller positive value.

14. Step 1:

x  12      x  12      12  x    12

Step 2:

1    1  1  0.00248, or   1  1    1  1  0.00251; From the graph,   12  2.01 2 2.01 2 1.99 1.99 2 thus   0.00248, or any smaller positive value.

15. Step 1: Step 2:

( x  1)  5  0.01  x  4  0.01  0.01  x  4  0.01  3.99  x  4.01 x  4      x  4      4  x    4    0.01, or any smaller positive value.

16. Step 1:

(2 x  2)  (6)  0.02  2 x  4  0.02  0.02  2 x  4  0.02  4.02  2 x  3.98  2.01  x  1.99 x  (2)      x  2      2  x    2    0.01, or any smaller positive value.

Step 2: 17. Step 1: Step 2:

18. Step 1: Step 2:

x  1  1  0.1  0.1  x  1  1  0.1  0.9  x  1  1.1  0.81  x  1  1.21

 0.19  x  0.21 x  0      x   . Then,    0.19    0.19 or   0.21; thus,   0.19, or any smaller positive value. x  12  0.1  0.1  x  12  0.1  0.4  x  0.6  0.16  x  0.36

x  14      x  14   . Then   14  0.16    0.09 or   14  0.36    0.11; thus   0.09, or any smaller positive value.

Chapter 2 Limits and Continuity

19. Step 1: Step 2:

20. Step 1:

19  x  3  1  1  19  x  3  1  2  19  x  4  4  19  x  16

 4  x  19  16  15  x  3 or 3  x  15 x  10      x  10      10  x    10. Then   10  3    7, or   10  15    5; thus   5, or any smaller positive value. x  7  4  1  1  x  7  4  1  3  x  7  5  9  x  7  25  16  x  32

Step 2:

x  23      x  23      23  x    23. Then   23  16    7, or   23  32    9; thus   7, or any smaller positive value.

21. Step 1:

1  1  0.05  0.05  1  1  0.05  0.2  1  0.3  10  x  10 or 10  x  5 x 4 x 4 2 3 3 x or   23 , or   4  5 or   1; x  4      x  4      4  x    4. Then   4  10 3 thus   23 , or any smaller positive value.

Step 2:

22. Step 1: Step 2:

x 2  3  0.1  0.1  x 2  3  0.1  2.9  x 2  3.1  2.9  x  3.1 x  3      x  3      3  x    3. Then   3  2.9

   3  2.9  0.0291, or   3  3.1    3.1  3  0.0286; thus   0.0286, or any smaller positive value.

23. Step 1: Step 2:

x 2  4  0.5  0.5  x 2  4  0.5  3.5  x 2  4.5  3.5  x  4.5   4.5  x   3.5,

for x near 2 x  (2)      x  2      2  x    2. Then   2   4.5    4.5  2  0.1213, or   2   3.5    2  3.5  0.1292; thus   4.5  2  0.12, or any smaller positive value.

24. Step 1: Step 2:

25. Step 1: Step 2:

1  ( 1)  0.1  0.1  1  1  0.1   11  1   9   10  x   10 or  10  x   10 10 10 11 9 x x x 9 11 10 1 x  (1)      x  1      1  x    1. Then   1   9    9 , or   1   10 11 1 ; thus   1 , or any smaller positive value.    11 11

( x 2  5)  11  1  x 2  16  1  1  x 2  16  1  15  x 2  17  15  x  17 x  4      x  4      4  x    4. Then   4  15    4  15  0.1270, or

  4  17    17  4  0.1231; thus   17  4  0.12, or any smaller positive value. 26. Step 1: Step 2:

27. Step 1: Step 2:

120  5  1  1  120  5  1  4  120  6  1  x  1  30  x  20 or 20  x  30 x x x 4 120 6

x  24      x  24      24  x    24. Then   24  20    4, or   24  30    6; thus    4, or any smaller positive value. mx  2m  0.03  0.03  mx  2m  0.03  0.03  2m  mx  0.03  2m  2  0.03  x  2  0.03 m m x  2      x  2      2  x    2. Then   2  2  0.03    0.03 , or m m

  2  2  0.03    0.03 . In either case,   0.03 , or any smaller positive value. m m m 28. Step 1:

mx  3m  c  c  mx  3m  c  c  3m  mx  c  3m  3  mc  x  3  mc

Section 2.3 The Precise Definition of a Limit Step 2:

x  3      x  3      3  x    3. Then   3  3  mc    mc , or

  3  3  mc    mc . In either case,   mc , or any smaller positive value.





29. Step 1:

(mx  b)  m2  b  c  c  mx  m2  c  c  m2  mx  c  m2  12  mc  x  12  mc

Step 2:

x  12      x  12      12  x    12 . Then   12  12  mc    mc , or   12  12  mc    mc . In either case,   mc , or any smaller positive value.

30. Step 1:

(mx  b)  (m  b)  0.05  0.05  mx  m  0.05  0.05  m  mx  0.05  m

 1  0.05  x  1  0.05 m m

Step 2:

x  1      x  1      1  x    1. Then   1  1  0.05    0.05 , or m m

  1  1  0.05    0.05 . In either case,   0.05 , or any smaller positive value. m m m 31. lim (3  2 x)  3  2(3)  3 x 3

(3  2 x)  (3)  0.02  0.02  6  2 x  0.02  6.02  2 x  5.98  3.01  x  2.99 or 2.99  x  3.01 Step 2: 0  x  3      x  3      3  x    3. Then   3  2.99    0.01, or   3  3.01    0.01; thus   0.01, or any smaller positive value.

Step 1:

32.

lim (3 x  2)  (3)(1)  2  1

x  1

Step 1: Step 2:

(3 x  2)  1  0.03   0.03  3 x  3  0.03  0.01  x  1  0.01  1.01  x  0.99 x  (1)      x  1      1  x    1. Then   1  1.01    0.01, or   1  0.99    0.01; thus   0.01, or any smaller positive value. ( x  2)( x  2)  lim ( x  2)  2  2  4, x  2 ( x  2) x 2 x 2 x 2  4  4  0.05  0.05  ( x  2)( x  2) 4  0.05  3.95  x  2  4.05, x  2  x 2 ( x 2)

33. lim xx 24  lim x 2

Step 1: Step 2:

34.

35.

 

 1.95  x  2.05, x  2 x  2      x  2      2  x    2. Then   2  1.95    0.05, or   2  2.05    0.05; thus   0.05, or any smaller positive value.

x 2 6x 5  lim (x 5)(x 1)  lim ( x  1)  4, x  5  (x 5) x 5 x 5 x 5 x5

lim

  (4)  0.05  0.05 

Step 1:



Step 2:

 5.05  x   4.95, x  5 x  (5)      x  5      5  x    5. Then   5  5.05    0.05, or   5  4.95    0.05; thus   0.05, or any smaller positive value.

lim

x  3

x 2  6 x 5 x 5

( x  5)( x 1)  4  0.05  4.05  x  1  3.95, ( x  5)

x  5

1  5 x  1  5(3)  16  4

Step 1:

1  5 x  4  0.5  0.5  1  5 x 4  0.5  3.5  1  5 x  4.5  12.25  1  5 x  20.25

Step 2:

 11.25  5 x  19.25  3.85  x  2.25 x  (3)      x  3      3  x    3. Then   3  3.85    0.85, or   3  2.25  0.75; thus   0.75, or any smaller positive value.

Chapter 2 Limits and Continuity

36. lim 4x  24  2 x2

Step 1: Step 2:

37. Step 1: Step 2:

4  2  0.4  0.4  4  2  0.4  1.6  4  2.4  10  x  10  10  x  10 or 5  x  5 16 4 24 4 6 x x x 3 2 5 5 1 x  2      x  2      2  x    2. Then   2  3    3 , or   2  2    12 ; thus   13 , or any smaller positive value.

(9  x)  5      4  x      4   x    4    4  x  4    4    x  4   x  4      x  4      4  x    4. Then   4    4    , or

  4    4    . Thus choose   , or any smaller positive value. 38. Step 1: Step 2:

(3 x  7)  2      3 x  9    9    3x  9    3  3  x  3  3

x  3      x  3      3  x    3. Then   3  3  3    3 , or

  3  3  3    3 . Thus choose   3 , or any smaller positive value. 39. Step 1:

x  5  2      x  5 2    2    x  5  2    (2   ) 2  x  5  (2   )2  (2  ) 2  5  x  (2   )2  5

Step 2:

x  9      x  9      9  x    9. Then   9   2  4  9    4   2 , or

  9   2  4  9    4   2 . Thus choose the smaller distance,   4   2 , or any smaller positive value. 40. Step 1:

4  x  2      4  x  2    2    4  x  2    (2   ) 2  4  x  (2   ) 2

  (2   )2  x  4  (2   ) 2  (2   )2  4  x  (2   ) 2  4

Step 2:

x  0      x   . Then   (2   )2  4   2  4    4   2 , or

   (2  ) 2  4  4   2 . Thus choose the smaller distance,   4   2 , or any smaller positive value. 41. Step 1:

For x  1, x 2  1      x 2  1    1    x 2  1    1    x  1    1    x  1   near x  1

Step 2:

x  1      x  1      1  x    1. Then   1  1      1  1   , or





  1  1      1    1. Choose   min 1  1   , 1    1 , that is, the smaller of the two distances, or any smaller positive value. 42. Step 1:

For x  2, x 2  4      x 2  4    4    x 2  4    4    x  4     4    x   4   near x  2

Step 2:

x  (2)      x  2      2  x    2. Then   2   4      4    2, or

  2   4      2  4   . Choose   min

 4    2, 2  4    , or any smaller positive

value. 43. Step 1: Step 2:

1  1      1  1    1    1  1    1  x  1 x x x 1 1

x  1      x  1    1    x  1   . Then 1    11    1  11  1  , or 1    11    11  1  1  . Choose   1  , the smaller of the two distances, or any smaller positive value.

Section 2.3 The Precise Definition of a Limit 44. Step 1:

Step 2:

1  1      1  1    1    1  1    13  1  1 3 3 3 3 3 x2 3 x2 3 x2 x2 2  133  x  133  133  x  133 , or 133  x  133 for x near

x  3      x  3    3    x  3   . Then   3

  min 45. Step 1: Step 2:

46. Step 1:

3 , or 13

3  

3 , 13

3  13

 3

3   13



3  13

3  

3 1 3

3. Choose

3 , or any smaller positive value.

   (6)      ( x  3)  6  , x  3    x  3      3  x    3 x 2 9 x 3

x  (3)      x  3      3  x    3. Then   3    3    , or   3    3    . Choose   , or any smaller positive value.

   2      (x  1)  2  , x  1  1    x  1   x 2 1 x 1

Step 2:

x  1      x  1    1    x  1   . Then 1    1      , or 1    1      . Choose   , or any smaller positive value.

47. Step 1:

x  1: (4  2 x)  2    0  2  2 x   since x  1. Thus, 1  2  x  0; x  1: (6 x  4)  2    0  6 x 6   since x  1. Thus, 1  x  1  6 . x  1      x  1    1    x  1   . Then 1    1  2    2 , or 1    1  6    6 .

Step 2:

Choose   6 , or any smaller positive value. 48. Step 1:

x  0: 2 x  0      2 x  0   2  x  0; x  0: 2x  0    0  x  2.

Step 2:

x  0      x   . Then    2    2 , or   2   2. Choose   2 , or any smaller

positive value. 49. By the figure,  x  x sin 1x  x for all x  0 and  x  x sin 1x  x for x  0. Since lim ( x)  lim x  0, then by x 0

the Sandwich Theorem, in either case, lim x sin 1x  0.

x 0

50. By the figure,  x 2  x 2 sin 1x  x 2 for all x except possibly at x  0. Since lim ( x 2 )  lim x 2  0, then by the Sandwich Theorem, lim x 2 sin 1x  0.

x 0

51. As x approaches the value 0, the values of g ( x) approach k. Thus for every number   0, there exists a   0 such that 0  x  0    g ( x)  k  . 52. Write x  h  c. Then 0  x  c      x  c   , x  c    (h  c) c   , h  c  c    h   , h  0  0  h  0   . Thus, lim f ( x)  L  for any   0, there exists   0 such that f ( x)  L   x c

whenever 0  x  c    f (h  c)  L   whenever 0  h  0    lim f (h  c)  L. h0

53. Let f ( x)  x 2 . The function values do get closer to 1 as x approaches 0, but lim f ( x)  0, not 1. The function f ( x)  x 2 never gets arbitrarily close to 1 for x near 0.

x 0

Chapter 2 Limits and Continuity

54. Let f ( x)  sin x, L  12 , and x0  0. There exists a value of x (namely x  6 ) for which sin x  12   for any

given   0. However, lim sin x  0, not 12 . The wrong statement does not require x to be arbitrarily close to x0 . x 0

As another example, let g ( x)  sin 1x , L  12 , and x0  0. We can choose infinitely many values of x near 0 such that sin 1x  12 as you can see from the accompanying figure. However, lim sin 1x fails to exist. The wrong x 0

statement does not require all values of x arbitrarily close to x0  0 to lie within   0 of L  12 . Again you can see from the figure that there are also infinitely many values of x near 0 such that sin 1x  0. If we choose   14

we cannot satisfy the inequality sin 1x  12   for all values of x sufficiently near x0  0.

55.

A  9  0.01  0.01  

 2x   9  0.01  8.99  π 4x  9.01  π4 (8.99)  x2  π4 (9.01) 2

 2 8.99  x  2 9.01 or 3.384  x  3.387. To be safe, the left endpoint was rounded up and  

the right endpoint was rounded down.

56. V  RI  VR  I  VR  5  0.1  0.1  120  5  0.1  4.9  120  5.1  10  R  10 49 120 51 R R (120)(10)

(120)(10)

 51  R  49  23.53  R  24.48. To be safe, the left endpoint was rounded up and the right endpoint was rounded down.

57. (a)   x  1  0  1    x  1  f ( x)  x. Then f ( x)  2  x  2  2  x  2  1  1. That is, f ( x)  2  1  12 no matter how small  is taken when 1    x  1  lim f ( x)  2. x 1

(b) 0  x  1    1  x  1    f ( x)  x  1. Then f ( x)  1  ( x  1)  1  x  x  1. That is, f ( x)  1  1 no matter how small  is taken when 1  x  1    lim f ( x)  1. x 1

(c)   x  1  0  1    x  1  f ( x)  x. Then f ( x)  1.5  x  1.5  1.5  x  1.5  1  0.5. Also, 0  x  1    1  x  1    f ( x)  x  1. Then f ( x)  1.5  ( x  1)  1.5  x  0.5  x  0.5  1  0.5  0.5. Thus, no matter how small  is taken, there exists a value of x such that   x  1   but f ( x)  1.5  12  lim f ( x)  1.5. x 1

58. (a) For 2  x  2    h( x)  2  h( x)  4  2. Thus for   2, h( x)  4   whenever 2  x  2   no matter how small we choose   0  lim h( x)  4. x2

(b) For 2  x  2    h( x)  2  h( x)  3  1. Thus for   1, h( x)  3   whenever 2  x  2   no matter how small we choose   0  lim h( x)  3. 2

x2

(c) For 2    x  2  h( x)  x so h( x)  2  x 2  2 . No matter how small   0 is chosen, x 2 is close to 4 when x is near 2 and to the left on the real line  x 2  2 will be close to 2. Thus if   1, h( x)  2   whenever 2    x  2 no matter how small we choose   0  lim h( x)  2. x2

Section 2.3 The Precise Definition of a Limit

59. (a) For 3    x  3  f ( x)  4.8  f ( x)  4  0.8. Thus for   0.8, f ( x)  4   whenever 3    x  3 no matter how small we choose   0  lim f ( x)  4. x 3

(b) For 3  x  3    f ( x)  3  f ( x)  4.8  1.8. Thus for   1.8, f ( x)  4.8   whenever 3  x  3   no matter how small we choose   0  lim f ( x)  4.8. x 3

(c) For 3    x  3  f ( x)  4.8  f ( x)  3  1.8. Again, for   1.8, f ( x)  3   whenever 3    x  3 no matter how small we choose   0  lim f ( x)  3. x 3

60. (a) No matter how small we choose   0, for x near 1 satisfying 1    x  1   , the values of g ( x) are near 1  g ( x)  2 is near 1. Then, for   12 we have g ( x)  2  12 for some x satisfying 1    x  1   , or 0  x  1    lim g ( x)  2. x 1

(b) Yes, lim g ( x )  1 because from the graph we can find a   0 such that g ( x)  1   if 0  x  (1)   . x 1

Example CAS commands (values of del may vary for a specified eps): 61–66. Maple: f : x - (x^4-81)/(x-3); x0 : 3;

plot( f (x), x  x0-1..x0 1, color  black, title "Section 2.3, #61(a)" );

# (a)

L : limit( f (x), x  x0 );

# (b)

epsilon : 0.2; # (c) plot( [f (x), L-epsilon,L  epsilon], x  x0-0.01..x0  0.01, color  black, linestyle [1,3,3], title "Section 2.3, #61(c)" ); q : fsolve( abs( f (x)-L )  epsilon, x  x0-1..x0 1 ); # (d) delta : abs(x0-q); plot( [f (x), L-epsilon, L  epsilon], x  x0-delta..x0  delta, color  black, title "Section 2.3, #61(d)" ); for eps in [0.1, 0.005, 0.001 ] do # (e) q : fsolve( abs( f (x)-L )  eps, x  x0-1..x0 1 ); delta : abs(x0-q); head : sprintf ("Section 2.3, #61(e)\n epsilon  %5f , delta  %5f \n", eps, delta ); print(plot( [f (x), L-eps, L  eps], x  x0-delta..x0  delta, q : fsolve( abs( f (x)-L )  epsilon, x  x0-1..x0 1 ); # (d) delta : abs(x0-q); plot( [f (x), L-epsilon, L  epsilon], x  x0-delta..x0  delta, color  black, title "Section 2.3, #61(d)" ); for eps in [0.1, 0.005, 0.001 ] do # (e) q : fsolve( abs( f (x)-L )  eps, x  x0-1..x0 1 ); delta : abs(x0-q); head : sprintf ("Section 2.3, #61(e)\n epsilon  %5f , delta  %5f \n", eps, delta ); print(plot( [f (x), L-eps, L  eps], x  x0-delta..x0  delta, color  black, linestyle [1,3,3], title  head ));

end do:

Chapter 2 Limits and Continuity Mathematica (assigned function and values for x0, eps and del may vary): Clear[f , x] y1:  L  eps; y2:  L  eps; x0  1; f[x _]:  (3x 2  (7x  1)Sqrt[x]  5)/(x  1) Plot[f [x], {x, x0  0.2, x0  0.2}] L:  Limit[f [x], x  x0] eps  0.1; del  0.2; Plot[{f [x], y1, y2}, {x, x0  del, x0  del}, PlotRange  {L  2eps, L  2eps}]

2.4

ONE-SIDED LIMITS

1. (a) True (e) True (i) False

(b) True (f) True (j) False

(d) True (h) False (l) False

2. (a) True (e) True (i) True

(b) False (f) True (j) False

(d) True (h) True

3. (a)

lim f ( x)  22  1  2, lim f ( x)  3  2  1 x  2

x 2

(b) No, lim f ( x) does not exist because lim f ( x)  lim f ( x) x  2 4 4 lim f ( x )  2  1  3, lim f ( x)  2  1  3 x 4 x 4

x 2

x 2

(c)

(d) Yes, lim f ( x)  3 because 3  lim f ( x)  lim f ( x) x  4

x 4

4. (a)

x  4

lim f ( x)  22  1, lim f ( x)  3  2  1, f (2)  2 x 2

x  2

(b) Yes, lim f ( x)  1 because 1  lim f ( x)  lim f ( x ) x 2

x 2

(c)

lim

x 1

f ( x)  3  (1)  4, lim

x 1

f ( x)  3  (1)  4

(d) Yes, lim f ( x )  4 because 4  lim

x 1

x 1

x  2

f ( x)  lim

x 1

f ( x)



5. (a) No, lim f ( x) does not exist since sin 1x does not approach any single value as x approaches 0 (b)

x 0 

lim f ( x)  lim 0  0

x 0 

x 0

x 0

6. (a) Yes, lim g ( x)  0 by the Sandwich Theorem since  x  g ( x)  x when x  0 x 0 

(b) No, lim g ( x) does not exist since x is not defined for x  0 x 0 

x 0

Section 2.4 One-Sided Limits

7. (a)

(b)

lim f ( x)  1  lim f ( x )

x 1

x 1

limits exist and equal 1

(b)

8. (a)

lim f ( x)  0  lim f ( x )

x 1

x 1

limits exist and equal 0

9. (a) Domain: 0  x  2 Range: 0  y  1 and y  2 (b) lim f ( x) exists for c belonging to (0, 1)  (1, 2) x c

10. (a) Domain:   x   Range: 1  y  1 (b) lim f ( x) exists for c belonging to x c

(,  1)  (1, 1)  (1, ) (c) none (d) none

11.

13. 14.

lim

x 0.5

x2  x 1

0.5 2  0.51

3/2  1/2

12.

lim

x 1

x 1  x2

11  1 2

 2( 2) 5  2 x  5  2 x  (2)  12   1  x 1   x  x   2 1   ( 2)  ( 2)  x 2 lim

lim

x 1



 x11   xx 6   37 x    111   116   371    12   71   72   1

0 0

86 15.

Chapter 2 Limits and Continuity lim

h 0 

 lim   h 0 

h2  4h 5  5  h

h 0  h

lim

h 0



6  5h 2 11h  6  h

h ( h  4)

 lim

16.

( h  4 h 5) 5 h 2  4 h 5  5   h 2  4h 5  5    2   lim h   h  4 h  5  5  h 0 h h 2  4 h  5  5

 h  4h 5  5  2

 lim  h 0  

 lim

h 0  h

17. (a)

(b)

0 4  2 5 5 5



6  5h 2 11h  6   6  5h 2 11h  6     h   6  5h2 11h  6  2 6 (5h 11h  6)  h (5h 11)

 lim

 6  5h 11h6 

h 0 h

x2

( x  2)

lim ( x  3) x  2  lim ( x  3) ( x  2)

x 2



 6  5h 11h6  2





(0 11)   11 6 6 2 6

(|x  2|  ( x  2) for x  2)

x 2

 lim ( x  3)  ((2)  3)  1

x 2 x2 ( x  2) lim ( x  3) x  2  lim ( x  3)  ( x  2)      x 2 x 2

(|x  2|  ( x  2) for x  2)

 lim ( x  3)(1)  (2  3)  1 x 2

18. (a)

lim



x 1

2 x ( x 1)  lim x 1 x 1

 lim

x 1

(b)

lim



x 1

2 x ( x 1) ( x 1)

2 x ( x 1)  lim x 1 x 1

(|x  1|  x  1 for x  1)

2x  2 2 x ( x 1) ( x 1)

(|x  1|  ( x  1) for x  1)

 lim  2 x   2 x 1



sin x  x

(b) If 2  x  0, then sin x  0, so that lim

sin x  x

19. (a) If 0  x  2 , then sin x  0, so that lim x 0

x 0



lim sinx x  lim 1  1

x 0 

x 0

x  lim 1  1 lim  sin x

x 0 

20. (a) If 0  x  2 , then cos x  1, so that lim (b)

21. (a) 22. (a)

23.

1cos x  lim 1cos x  lim 1cos x  lim 1  1  (cos x 1) 1 cos x x 0 cos x 1 x 0 x 0  x 0    cos x  1 cos x  1 If 2  x  0, then cos x  1, so that lim  lim (cos x 1)  lim 1  1 x 0 cos x 1 x 0 x 0

  lim    33  1  3 

(b)

lim (t  t  )  4  4  0

(b)

t  4

lim sin 2  lim sinx x  1

 0

2

x 0

  lim    23  3 

lim (t  t  )  4  3  1

t  4

(where x  2 )

kt  lim k sin   k lim sin   k  1  k 24. lim sint kt  lim k sin   kt

(where   kt )

sin 3 y 3sin 3 y sin 3 y  14 lim 3 y  43 lim 3 y  43 lim sin   43 y 4 y 0 y 0 y 0  0

(where   3y )

t 0

25. lim

t 0

 0

Section 2.4 One-Sided Limits 26.



   1 1  1 sin  3  3     0 

1 1  1  13  sin3h3h   13 hlim  3  lim 0  h 0

lim sinh3h  lim

h 0 



x 0

2t  2 lim 28. lim tan t



csc 2 x  lim 29. lim xcos 5x



t 0

x  1 x 0 sin 2 x cos 5 x

x 0

1   lim 2sin 2 x   1  2  2  cos 2 x   x 0 2 x  0 x  

x 0 x cos 2 x

  cos t  2  lim cos t   1  2 lim tsin    lim sin t   2 1 1  2 t  t 0   t 0 t   t 0

t sin t cos t

t 0

(where   3h)

sin 2 x  cos 2x   lim sin 2 x   lim

27. lim tanx2 x  lim x 0

sin 3h 3h



2 x   lim 1   1 1 (1)  1    12 xlim    2  2 0 sin 2 x   x 0 cos 5 x  2





6 x cos x  lim 3cos x  x  2 x  3 1 1  3 30. lim 6 x 2 (cot x)(csc 2 x)  lim sin x sin 2 x sin x sin 2 x x 0

x 0

x  x cos x  lim 31. lim sin x cos x

x 0



x  x cos x x 0 sin x cos x sin x cos x

x 0

x   xlim  x  1   xlim 0 sin x cos x 0 sin x

     lim  sin1 x   lim cos1 x  lim  sin1 x   (1)(1)  1  2 x 0  x  x 0 x 0  x 



32. lim x  x2xsin x  lim





   0  12  12 (1)  0

x  1  1 sin x x x 0 2 2 2

x 0

(1cos  )(1 cos  )

cos   lim cos  sin   lim (2sin 1cos  lim (2sin  cos 33. lim 1sin 2 (2sin  cos  )(1 cos  )  )(1 cos  )  )(1 cos  )

 0

 0 0  lim (2 cos sin  (2)(2) )(1 cos  )  0

34. lim x  x 2cos x  lim x 0 sin 3 x

x (1cos x )

x 0

sin 3 x

 lim

x 0

x (1 cos x ) 9 x2 sin 2 3 x 9 x2

 lim

x 0

1 cos x 9x

 sin3 x3 x 

 2



1 lim 1 cos x 9 x0 x

lim

x 0

 sin3 x3 x 



1 (0)

 9 2 0

sin(1cos t )  lim sin   1 since   1  cos t  0 as t  0  0 t 0 1cos t

35. lim

36. lim

h 0

sin(sin h )  lim sin   1 since   sin h  0 as h  0 sin h  0

sin   lim 37. lim sin 2



sin 5 x  lim 38. lim sin 4x



 0

x 0







sin   2  1 lim sin   2  12 1 1  12 2  0  sin 2  0 sin 2 2

sin 5 x  4 x  5 x 0 sin 4 x 5 x 4

  54 xlim  sin5 x5x  sin4 x4 x   54 11  54 0

39. lim  cos   0 1  0  0

2  lim sin  cos 2 2  1 40. lim sin  cot 2  lim sin  cos  lim 2cos sin 2 2sin  cos  cos  2

 0

Chapter 2 Limits and Continuity

  xlim  sin 3x  1  8 x  3  0 cos 3 x sin 8 x 3 x 8  83 lim  cos13 x   sin3 x3 x   sin8 x8 x   83 1 1 1  83 x0

3 x  lim 41. lim tan sin 8 x

42. lim

y 0



sin 3 x  1 x 0 cos 3 x sin 8 x

x 0

             1 1 1 1  

sin 3 y cot 5 y sin 3 y sin 4 y cos 5 y sin 3 y  lim y cos 4 y sin 5 y  lim y cot 4 y y y 0 y 0



sin 3 y 3y y 0

 lim



sin 4 y 4y

5y sin 5 y

sin 4 y cos 4 y

cos 5 y cos 4 y

sin 

34 5



tan   lim cos   lim sin  sin 3  lim sin  2 2 3  0  cot 3  0  cos  0  cos  cos 3  0  sin 3

43. lim

cos 5 y sin 5 y

345 y 345 y

12 5

  sin33   cos 3cos 3   (1)(1)  113   3

4 2  cos  cos 4 (4sin 2 cos 2 )  cos 4 (2sin  cos  )2  cot 4  lim sin 4  lim  2cos 42sin 2  lim  lim 2 2 2 2 2 2 2  0 sin  cos 2 sin 4  0 sin  cot 2  0 sin 2 cos 2  0 sin  cos 2 sin 4  0 sin  cos 2 sin 4

44. lim

sin 2 2

 lim

4 cos 4 cos 2 2

 0 cos 2 sin 4

45.

  1 11  4   cos 4 cos    lim  1   coscos4 cos   1  1   1 2   0 sin 4  cos 2   0  2

 lim

sin 4 4

x  lim 1cos3 x  1 cos3 x  lim 1cos 3 x  lim sin 3 x 3x lim 1cos3  lim 23  sin3 x3 x  1sin 2x 2x 1 cos3 x 2 x (1 cos3 x ) 2 x (1 cos3 x ) cos3 x

x 0

   0 (where   3x )

x 0

  3 (1) 0  lim 32  sin   1sin cos 2 11  0

46.

cos x (cos x 1) cos x (cos x 1) cos x 1 cos x (cos 2 x 1) cos x(  sin 2 x )  lim  cos x 1  lim 2  lim 2 2 2 x x x 0 x 0 x 0 x (cos x 1) x 0 x (cos x 1)

lim cos x 2 cos x  lim

x 0





cos x  (1)(1)  1  1  lim  sinx x  sinx x  cos x 1 11 2 x 0

47. Yes. If lim f ( x)  L  lim f ( x), then lim f ( x)  L. If lim f ( x)  lim f ( x), then lim f ( x) does not exist.

x a 

xa 

x a 

x a

x a 

x a

48. Since lim f ( x)  L if and only if lim f ( x)  L and lim f ( x)  L, then lim f ( x) can be found by x c

calculating lim f ( x).

x c 

x c 

x c

x c 

49. If f is an odd function of x, then f ( x)   f ( x). Given lim f ( x)  3, then lim f ( x)  3. x 0 

x 0 

50. If f is an even function of x, then f ( x)  f ( x). Given lim f ( x)  7 then lim can be said about lim

x 2

x 2

f ( x ) because we don’t know lim f ( x).

x 2

f ( x)  7. However, nothing

x  2

51. I  (5, 5   )  5  x  5   . Also, x  5    x  5   2  x  5   2 . Choose    2 , or any smaller positive value  lim x  5  0. x 5

52. I  (4   , 4)  4    x  4. Also, 4  x    4  x   2  x  4   2 . Choose    2 , or any smaller positive value  lim 4  x  0. x  4

Section 2.5 Continuity

53. As x  0 the number x is always negative. Thus, x  (1)    xx  1    0   which is always true x

x  1. x 0  x

independent of the value of x. Hence we can choose any   0 with   x  0  lim

54. Since x  2 we have x  2 and x  2  x  2. Then, x  2  1  xx  22  1    0   which is always true so x2

x  2  1. x 2 x  2

long as x  2. Hence we can choose any   0, and thus 2  x  2    x  2  1  . Thus, lim x2

55. (a)

lim

x 400

 x   400. Just observe that if 400  x  401, then  x   400. Thus if we choose   1, we have for

any number   0 that 400  x  400     x   400  400  400  0  . (b) lim  x   399. Just observe that if 399  x  400 then  x   399. Thus if we choose   1, we have for x 400

any number   0 that 400    x  400   x   399  399  399  0  . (c) Since lim  x   lim  x  we conclude that lim  x  does not exist. x 400

56. (a)

x 400

lim f ( x)  lim

x 0 

x 400

x  0  0;

x  0      x    0  x   2 for x positive. Choose    2

 lim f ( x)  0. x 0 

(b)



lim f ( x)  lim x 2 sin 1x  0 by the Sandwich Theorem since  x 2  x 2 sin 1x  x 2 for all x  0.

x 0 

x 0



Since x 2  0   x 2  0  x 2   whenever x   , we choose    and obtain x 2 sin 1x  0  

if   x  0. (c) The function f has limit 0 at x0  0 since both the right-hand and left-hand limits exist and equal 0. 2.5

CONTINUITY

1. No, discontinuous at x  2, not defined at x  2 2. No, discontinuous at x  3, 1  lim g ( x)  g (3)  1.5 x 3

3. Continuous on [1, 3] 4. No, discontinuous at x  1, 1.5  lim k ( x)  lim k ( x)  0 x 1

x 1

5. (a) Yes

(b) Yes, lim

(d) Yes

6. (a) Yes, f (1)  1

x 1

f ( x)  0

(b) Yes, lim f ( x)  2

(d) No

7. (a) No

(b) No

x 1

8. [1, 0)  (0, 1)  (1, 2)  (2, 3) 9. f (2)  0, since lim f ( x)  2(2)  4  0  lim f ( x) x 2

x  2

Chapter 2 Limits and Continuity

10. f (1) should be changed to 2  lim f ( x ) x 1

11. Nonremovable discontinuity at x  1 because lim f ( x) fails to exist ( lim f ( x)  1 and lim f ( x)  0). x 1

x 1

x 1

Removable discontinuity at x  0 by assigning the number lim f ( x)  0 to be the value of f (0) rather x 0

than f (0)  1.

12. Nonremovable discontinuity at x  1 because lim f ( x) fails to exist ( lim f ( x)  2 and lim f ( x)  1). x 1

x 1

x 1

Removable discontinuity at x  2 by assigning the number lim f ( x)  1 to be the value of f (2) rather than x 2

f (2)  2.

14. Discontinuous only when ( x  2) 2  0  x  2

13. Discontinuous only when x  2  0  x  2

15. Discontinuous only when x 2  4 x  3  0  ( x  3)( x  1)  0  x  3 or x  1 16. Discontinuous only when x 2  3 x  10  0  ( x  5)( x  2)  0  x  5 or x  2 17. Continuous everywhere. (|x  1|  sin x defined for all x; limits exist and are equal to function values.) 18. Continuous everywhere. (|x|  1  0 for all x; limits exist and are equal to function values.) 19. Discontinuous only at x  0 20. Discontinuous at odd integer multiples of 2 , i.e., x  (2n  1) 2 , n an integer, but continuous at all other x. 21. Discontinuous when 2x is an integer multiple of  , i.e., 2 x  n , n an integer  x  n2 , n an integer, but continuous at all other x. 22. Discontinuous when 2x is an odd integer multiple of 2 , i.e., 2x  (2n  1) 2 , n an integer  x  2n  1, n an integer (i.e., x is an odd integer). Continuous everywhere else. 23. Discontinuous at odd integer multiples of 2 , i.e., x  (2n  1) 2 , n an integer, but continuous at all other x. 24. Continuous everywhere since 1  sin x  1  0  sin 2 x  1  1  sin 2 x  1; limits exist and are equal to the function values.



25. Discontinuous when 2 x  3  0 or x   32  continuous on the interval   32 ,  .



26. Discontinuous when 3 x  1  0 or x  13  continuous on the interval  13 ,  . 27. Continuous everywhere: (2 x  1)1/3 is defined for all x; limits exist and are equal to function values. 28. Continuous everywhere: (2  x)1/5 is defined for all x; limits exist and are equal to function values. 2

( x 3)( x  2)  lim ( x  2)  5  g (3) x 3 x 3 x 3

29. Continuous everywhere since lim x xx36  lim x 3

Section 2.5 Continuity 30. Discontinuous at x  2 since lim f ( x) does not exist while f ( 2)  4. x 2

31. Discontinuous at x  1; lim ( x 2  2)  3, but lim e x  e, so that lim f ( x) does not exist while f (1)  e; x 1

x 1

x 1

and lim (1  x)  1  lim e , so that lim f ( x)  1  f (0) x 0 

x 0 

x 0

32. Discontinuous at x  ln 2, since 2  e x  0  e x  2  ln e x  ln 2  x  ln 2 33. lim sin( x  sin x)  sin(  sin  )  sin(  0)  sin   0, and function continuous at x   . x 





 

34. lim sin( 2 cos(tan t ))  sin( 2 cos(tan(0)))  sin 2 cos(0)  sin 2  1, and function continuous at t  0. t 0

35. lim sec ( y sec2 y  tan 2 y  1)  lim sec ( y sec2 y  sec2 y )  lim sec (( y  1) sec2 y )  sec ((1  1)sec2 1)  y 1

y 1

sec 0  1, and function continuous at y  1.





 

36. lim tan  4 cos(sin x1/3 )   tan  4 cos(sin(0))   tan 4 cos(0)  tan 4  1, and function continuous at x  0.   x 0       37. lim cos   cos   cos   cos 4  22 , and function continuous at t  0. 16 t 0  193 sec 2t   193 sec 0 

38. lim

x  6

 

   9  3, and function continuous

 

csc2 x  5 3 tan x  csc2 6  5 3 tan 6  4  5 3 1

at x  6 . 39.

40.

41. 42.

43.

44.



  

 

lim sin 2 e x  sin 2 e0  sin 2  1, and the function is continuous at x = 0.

x 0 









lim cos 1 ln x  cos 1 ln 1  cos 1 (0)  2 , and the function is continuous at x = 1.

x 1









 

   1  sec 1   tan 4 sec0  1   sec  tan 4   sec ( )  1 lim sec e x   tan 4 sec x      

x 0









 

0 lim sin tan x 2tansecx x  sin tan0tan  sin  2  1 2sec0

x 0



lim tan 1 

t 0

sin t t

  tan 1  lim

sin t    tan (1  1)  0 t 0 t 









( x 3)( x 3)  x  3, x  3  g (3)  lim ( x  3)  6 ( x 3) x 3

 

1   cos   1  1   cos   lim    cos    lim cos sin     sin 1   0   0  lim sin      0 

45. g ( x)  xx 39 

Chapter 2 Limits and Continuity 2

46. h(t )  t t3t210  3

47. f ( s )  s3 1  s 1

48. g ( x) 

(t 5)(t  2)  t  5, t  2  h(2)  lim (t  5)  7 t 2 t 2



2 2 ( s 2  s 1)( s 1)  s ss11 , s  1  f (1)  lim s ss11 ( s 1)( s 1) s 1



3 2

   85

x 2 16  ( x  4)( x  4)  x  4 , x  4  g (4)  lim x  4  x 1 ( x  4)( x 1) x 2 3 x  4 x 4 x 1

49. As defined, lim f ( x)  (3)2  1  8 and lim (2a )(3)  6a. For f ( x) to be continuous we must have x 3 6a  8  a  43 .

x 3

50. As defined, lim g ( x)  2 and lim g ( x)  b(2)2  4b. For g ( x) to be continuous we must have x 2 4b  2  b   12 .

x 2

51. As defined, lim f ( x)  12 and lim f ( x)  a 2 (2)  2a  2a 2  2a. For f ( x) to be continuous we must have 2

x  2

x 2

12  2a  2a  a  3 or a  2.

52. As defined, lim g ( x)  0bb1  bb1 and lim g ( x)  (0) 2  b  b. For g ( x) to be continuous we must have x 0 

x 0  b  b  b  0 or b  2. b 1

53. As defined, lim

x 1

f ( x)  2 and lim

x 1

f ( x)  a (1)  b  a  b, and lim f ( x)  a (1)  b  a  b and x 1

lim f ( x)  3. For f ( x) to be continuous we must have 2  a  b and a  b  3  a  52 and b  12 .

x 1

54. As defined, lim g ( x)  a (0)  2b  2b and lim g ( x)  (0)2  3a  b  3a  b, and lim g ( x)  (2)2  3a  b  x 0 

x 0 

x  2

4  3a  b and lim g ( x)  3(2)  5  1. For g ( x) to be continuous we must have 2b  3a  b and 4  3a  b  1  x 0  3 a   2 and b   32 .

55. The function can be extended: f (0)  2.3.

56. The function cannot be extended to be continuous at x  0. If f (0)  2.3, it will be continuous from the right. Or if f (0)  2.3, it will be continuous from the left.

Section 2.5 Continuity 57. The function cannot be extended to be continuous at x  0. If f (0)  1, it will be continuous from the right. Or if f (0)  1, it will be continuous from the left.

58. The function can be extended: f (0)  7.39.

59. f ( x) is continuous on [0, 1] and f (0)  0, f (1)  0  by the Intermediate Value Theorem f ( x) takes on every value between f (0) and f (1)  the equation f ( x)  0 has at least one solution between x  0 and x  1.

   

 

60. cos x  x  (cos x)  x  0. If x   2 , cos  2   2  0. If x  2 , cos 2  2  0. Thus cos x  x  0 for some x between  2 and 2 according to the Intermediate Value Theorem, since the function cos x  x is continuous.

61. Let f ( x)  x3  15 x  1, which is continuous on [4, 4]. Then f (4)  3, f (1)  15, f (1)  13, and f (4)  5. By the Intermediate Value Theorem, f ( x)  0 for some x in each of the intervals 4  x  1, 1  x  1, and 1  x  4. That is, x3  15 x  1  0 has three solutions in [4, 4]. Since a polynomial of degree 3 can have at most 3 solutions, these are the only solutions. 62. Without loss of generality, assume that a  b. Then F ( x)  ( x  a )2 ( x  b)2  x is continuous for all values of x, so it is continuous on the interval [ a, b]. Moreover F ( a )  a and F (b)  b. By the Intermediate Value Theorem, since a  a 2 b  b, there is a number c between a and b such that F ( x)  a 2 b . 63. Answers may vary. Note that f is continuous for every value of x. (a) f (0)  10, f (1)  13  8(1)  10  3. Since 3    10, by the Intermediate Value Theorem, there exists a c so that 0  c  1 and f (c)   . (b) f (0)  10, f ( 4)  (4)3  8( 4)  10  22. Since 22   3  10, by the Intermediate Value Theorem, there exists a c so that 4  c  0 and f (c)   3. (c) f (0)  10, f (1000)  (1000)3  8(1000)  10  999,992, 010. Since 10  5, 000, 000  999,992, 010, by the Intermediate Value Theorem, there exists a c so that 0  c  1000 and f (c)  5, 000, 000. 64. All five statements ask for the same information because of the intermediate value property of continuous functions. (a) A root of f ( x)  x3  3 x  1 is a point c where f (c)  0. (b) The point where y  x3 crosses y  3 x  1 have the same y-coordinate, or y  x3  3 x  1  f ( x)  x3  3 x  1  0. (c) x3  3x  1  x3  3 x  1  0. The solutions to the equation are the roots of f ( x )  x3  3 x  1. Copyright  2023 Pearson Education, Inc.

Chapter 2 Limits and Continuity (d) The points where y  x3  3 x crosses y  1 have common y-coordinates, or y  x3  3 x  1  f ( x)  x3  3 x  1  0. (e) The solutions of x3  3x  1  0 are those points where f ( x)  x3  3 x  1 has value 0. sin( x  2)

65. Answers may vary. For example, f ( x)  x  2 is discontinuous at x  2 because it is not defined there. However, the discontinuity can be removed because f has a limit (namely 1) as x  2. 66. Answers may vary. For example, g ( x)  x11 has a discontinuity at x  1 because lim g ( x) does not exist. x 1

   lim  g ( x)   and lim  g ( x)  .  x 1  x 1 

67. (a) Suppose x0 is rational  f ( x0 )  1. Choose   12 . For any   0 there is an irrational number x (actually infinitely many) in the interval ( x0   , x0   )  f ( x)  0. Then 0  |x  x0 |   but | f ( x)  f ( x0 )|  1  12  , so lim f ( x) fails to exist  f is discontinuous at x0 rational. x  x0

On the other hand, x0 irrational  f ( x0 )  0 and there is a rational number x in ( x0   , x0   )  f ( x)  1. Again lim f ( x) fails to exist  f is discontinuous at x0 irrational. That is, f is discontinuous at every point. x  x0

(b) f is neither right-continuous nor left-continuous at any point x0 because in every interval ( x0   , x0 ) or ( x0 , x0   ) there exist both rational and irrational real numbers. Thus neither limits lim f ( x) and x  x0

lim f ( x) exist by the same arguments used in part (a).

x  x0

f ( x)

68. Yes. Both f ( x)  x and g ( x)  x  12 are continuous on [0, 1]. However g ( x ) is undefined at x  12 since



f ( x)

g 12  0  g ( x ) is discontinuous at x  12 .

69. No. For instance, if f ( x)  0, g ( x)   x  , then h( x)  0   x    0 is continuous at x  0 and g ( x) is not. 70. Let f ( x)  x11 and g ( x)  x  1. Both functions are continuous at x  0. The composition f  g  f ( g ( x)) 

1  1 is discontinuous at x  0, since it is not defined there. Theorem 10 requires that f ( x ) be continuous ( x 1) 1 x

at g (0), which is not the case here since g (0)  1 and f is undefined at 1.

71. Yes, because of the Intermediate Value Theorem. If f ( a ) and f (b) did have different signs then f would have to equal zero at some point between a and b since f is continuous on [ a, b]. 72. Let f ( x ) be the new position of point x and let d ( x)  f ( x)  x. The displacement function d is negative if x is the left-hand point of the rubber band and positive if x is the right-hand point of the rubber band. By the Intermediate Value Theorem, d ( x)  0 for some point in between. That is, f ( x)  x for some point x, which is then in its original position. 73. If f (0)  0 or f (1)  1, we are done (i.e., c  0 or c  1 in those cases). Then let f (0)  a  0 and f (1)  b  1 because 0  f ( x)  1. Define g ( x)  f ( x)  x  g is continuous on [0, 1]. Moreover, g (0)  f (0)  0  a  0 and g (1)  f (1)  1  b  1  0  by the Intermediate Value Theorem there is a number c in (0, 1) such that g (c)  0  f (c)  c  0 or f (c)  c. f (c )

74. Let   2  0. Since f is continuous at x  c there is a   0 such that x  c    f ( x)  f (c)    f (c)    f ( x)  f (c)  . If f (c)  0, then   12 f (c)  12 f (c)  f ( x)  23 f (c)  f ( x)  0 on the interval (c   , c   ). Copyright  2023 Pearson Education, Inc.

Section 2.6 Limits Involving Infinity; Asymptotes of Graphs

If f (c)  0, then    12 f (c)  23 f (c)  f ( x)  12 f (c)  f ( x)  0 on the interval (c   , c   ).

75. By Exercise 52 in Section 2.3, we have lim f ( x )  L  lim f (c  h )  L. x c

h 0

Thus, f ( x) is continuous at x  c  lim f ( x)  f (c)  lim f (c  h)  f (c). x c

h 0

76. By Exercise 75, it suffices to show that lim sin(c  h)  sin c and lim cos(c  h)  cos c. h 0

h 0

Now lim sin(c  h)  lim  (sin c)(cos h)  (cos c)(sin h)  (sin c)  lim cos h   (cos c)  lim sin h . h 0 h 0  h 0   h 0  By Example 12 Section 2.2, lim cos h  1 and lim sin h  0. So lim sin(c  h)  sin c and thus f ( x)  sin x is h 0

h 0

continuous at x  c. Similarly, lim cos(c  h)  lim  (cos c)(cos h)  (sin c)(sin h)  h 0

h 0

 (cos c)  lim cos h   (sin c)  lim sin h   cos c. Thus, g ( x)  cos x is continuous at x  c.  h 0   h 0 

77. x  1.8794,  1.5321,  0.3473

78. x  1.4516,  0.8547, 0.4030

79. x  1.7549

80. x  1.5596

81. x  3.5156

82. x  3.9058, 3.8392, 0.0667

83. x  0.7391

84. x  1.8955, 0, 1.8955

2.6

LIMITS INVOLVING INFINITY; ASYMPTOTES OF GRAPHS

1. (a) lim f ( x)  0 (c) (e)

x 2

lim

x 3

f ( x)  2

lim f ( x)  1

x 0 

(g) lim f ( x)  does not exist (i)

x 0

lim f ( x)  0

(b)

lim

x 3

f ( x)  2

(d) lim f ( x)  does not exist (f) (h)

x 3

lim f ( x)  

x 0 

lim f ( x)  1

x 

x 

Chapter 2 Limits and Continuity

2. (a) lim f ( x)  2 (c) (e) (g) (i)

(b)

x 4

lim f ( x)  1

x  2

(d) lim f ( x)  does not exist

f ( x)  

(f)

lim f ( x)  

(h)

lim f ( x)  

(j)

lim

x 3

lim f ( x )  3

x  2

x 3

x 0 

(k) lim f ( x)  0

(l)

x 

Note: In these exercises we use the result lim

x   x

m/ n

Theorem 8 and the Power Rule in Theorem 1: lim

x 2

lim

x  3

lim f ( x)   

x  0

lim f ( x)  does not exist

x 0

lim f ( x)  1

x 

 0 whenever mn  0. This result follows immediately from

  1

m/n x   x



 lim 1x x  

3. (a) 3

(b) 3

4. (a) 

(b) 

5. (a)

1 2

(b)

1 2

6. (a)

1 8

(b)

1 8

7. (a)  53 8. (a)

f ( x)   

m/ n

    lim 1x   x  

m/ n

 0m / n  0.

(b)  53

3 4

(b)

3 4

9.  1x  sinx2 x  1x  lim sinx2 x  0 by the Sandwich Theorem x 

  1  lim cos   0 by the Sandwich Theorem 10.  31  cos 3 3   3

t 

r  lim 12. lim 2 r r 7 sin5sin r

14. (a)

   lim 10  1   r  200 2

1 sinr r

7 sin r r  2  r 5 r

r 

13. (a)

   010  1   10

2 1 sin t t cos t t  1 t

 sin t  lim t 11. lim 2tt cos t

lim 52xx73  lim

x 

2  3x 7

x  5 x

 52

(b)

2  73  x  2 x3  7  lim 2 3 2 7 1 1 x  x  x  x  7 x  1 x  2  3

lim

15. (a)

1 1

x x2 lim x21  lim 0 3

x  x 3

x  1 2

(b)

2 (same process as part (a)) 5

2 (same process as part (a))

(b) 0 (same process as part (a))

Section 2.6 Limits Involving Infinity; Asymptotes of Graphs 16. (a)

17. (a)

lim 3 x2  7  lim

3 7 x x2

0

x  1 2

x  x  2

(b) 0 (same process as part (a))

7 x3  lim 37 9  7 2 x  x 3 x  6 x x  1 x  2

(b) 7 (same process as part (a))

9  13 9 x4  x x  lim  92 4 2 5 6 1 x  2 x 5 x  x  6 x  2  2  3  4

(b) 92 (same process as part (a))

lim

18. (a)

lim

19. (a) 20. (a) (b)

21. (a) (b)

22. (a) (b)

23.

x 

x 1 x  x 1

3

lim 3 x 35 x 1  lim 3 x 5x2  x3  , since x  n  0 and 3x 4  . 7

x 6 x 7 x 3

x 67 x 3 x

7 2 4 1 3 lim 3 x 35 x 1  lim 3 x 5x2  x3  , since x  n  0 and 3x 4  . x 6 x 7 x 3 x 67 x 3 x

2

lim 5 x 2 x 5 9  lim 5 x 25 x 49 x

x 3 x 4 x

5

 , since x  n  0, 5x 3  , and the denominator  4.

3 x  x 4

x

8 3 3 2 5 lim 5 x 2 x 5 9  lim 5 x 25 x 49 x  , since x  n  0, 5x 3  , and the denominator  4.

x 3 x 4 x

8 x 2 3  lim 2 x2  x x 





25.

3 lim 12 x 7 x  x x 

29.

x  1 x  x

x  x 1

24.

27.

1

x  x 1

x

x 



lim

(b) 0 (same process as part (a))

3 2 1 lim x 2 7 x  2  lim x 71 2 x 2  , since x  n  0 and x  7  .

lim

x 

0

x 

lim x 2 7 x  2  lim x 71 2 x 2  , since x  n  0 and x  7  .

x 

2 lim x 2 x 1 x  8 x 3

26.

10  1  31 x x 2 x6

lim 10 x  6x 31  lim

1/3



x 

1

8 32 x 2  1x



3 x  x 4 8 32 x 1

lim

x  2 x 1/3

 1 1x  12  x   lim  3 x   8 2  x  

8 0  2 0



42 1/3

1 1x  12   x    lim  x  8 32  x  



 18000



1/3



 18

1 5 x x2

1 12  23 x



lim

1 5 x x2 1

x  1 2  3



 12



 10000  0  0

 2   1   1/ 2   2   x  0 3 7x x 

 lim  x

28.

 2  1  1/ 2    1 x   2  1 1/ 2 x 

lim 2 x  lim  x

x  2  x

1 21  x 5 x  x /15  1 x (1/5)  (1/3)  lim  lim 1 (1/5)  (1/3) 3 5 x  1 x x  1 1  x  x  x  x 2 /15 

lim

1/3

1 x  12  x    5 x x2   01   lim  7    lim 7 0        1 1 x   x  x   x 

x 2 5 x  lim x3  x  2 x 

lim 2 3xx7x

x

98 30.

32.

Chapter 2 Limits and Continuity 1

x  12

4

lim x2  x 3  lim

x  x  x

x 1 1x

x 



31.

5/3

1/3

lim 2 x8/5  x

 7  lim 3 x  x x 

x  x

1  7 2 x1/15  19/15 8/5 x

3 1 1 3/5  11/10 x



1  5 3 x x 5 x  3 x 2/3  lim   52 2/3 1 4 x  2 x  x  4 x  2 1/3  x 3

lim

33.

34.

35.

36.

37.

39.

41.

2 2 x 2 1  lim x 1/ x2  lim x 1 x  ( x 1)/ x x 

lim

x 

lim

x 

( x 2 1)/ x 2 x 2 1 x 2 1/ x 2 11/ x 2  lim  lim  lim  ( 1100)  1 x 1 x  ( x 1)/(  x ) x  ( 11/ x ) x  ( x 1)/ x 2

lim

x 3 2

x 

4 x  25

 lim

x 

( x 3)/ x 2 2

4 x  25 / x

x 

x 9

x 

x 

( x 3)/ x 2

(4 x  25)/ x

 lim

x 

(13/ x ) 4  25/ x

(1 0)  12 4 0





40.



negative positive



42.



positive positive



44. lim



lim x2x8   x 8 4  2 x 7 ( x 7) 2

lim



(b)



(b)

1/3

x 0  3 x

lim

1/5

x 0  x

4  lim 47. lim 2/5 x 0 x

1/5 2

x 0 ( x

)

 

5   2x



positive negative

lim

1  x 3



positive positive

lim 2 x3x10   x 5



negative negative

1   2 x 0 x ( x1)



negative positivepositive

x 3

lim

1/3

x 0  3 x

lim

1/5

x 0  x

x 0 x

50.



 

lim

 

x  2



sec x  

lim (1  csc  )   lim (2  cot  )   and lim (2  cot  )  , so the limit does not exist

53. (a)

 0 

lim

x  2 x  4

 lim ( x  2)(1 x  2)   x 2



1 positivepositive



1/3 2

x 0 ( x

 0 

 0 



lim

x 0 

1  lim 48. lim 2/3



lim tan x  

x  2

1 0

19/ x

positive negative

3   x 2

lim

x 

38.

positive positive

x  2

( x 9)/ x





lim 31x  

46. (a)

52.

 lim

x 9 / x

x 0 

45. (a)

51.

3 (4 3 x3 )/ x 6 (43 x3 )/(  x3 ) ( 4/ x3 3) (0 3)  lim  lim  3 lim 463 x  lim 6 6 6 6 6

43. lim

49.

( x 2 1)/ x 2 1/ x 2  (110)0  1  lim (111/ x) ( x 1)/ x x 

)





Section 2.6 Limits Involving Infinity; Asymptotes of Graphs (b) (c) (d)

54. (a) (b) (c) (d)

lim

x  2 x  4 1

lim

  

 lim ( x  2)(1 x  2)   x 2

x 2 x  4 1

x 2 x  4

 lim ( x  2)(1 x  2)   x 2  lim ( x  2)(1 x  2)   x 2

      

x  lim x  2 ( x 1)( x 1) x 1 x 1 x 1 lim 2x  lim ( x 1)(x x 1)   x 1 x 1 x 1

lim



lim 2x  lim ( x 1)(x x 1) x 1 x 1 x 1 lim 2x  lim ( x 1)(x x 1) x 1 x 1 x 1 x2 2

1 x

x2 2

1 x

x2 2

1 x

22/3 2

x2 x 1 2

1 x

1 2

lim

x 0



x 0  3

x 2

56. (a) (b) (c) (d)

57. (a) (b) (c) (d)

x 0 

x 0 

   

negative positivenegative

negative negativenegative

 

1 negative

1 x

1 positive

1/3





0

3 2

lim 2xx 14   x 2



positive positive



positive lim 2xx 14   negative x 2 2 ( x 1)( x 1) lim 2xx 14  lim  2204  0 2 x4   x 1 x 1 2 lim 2xx 14  41  x 0 2 ( x  2)( x1) lim x 33 x 22  lim   2

x 0  x  2 x

x ( x  2)

x 0 



negativenegative positivenegative



2 ( x  2)( x 1) lim x 33 x 22  lim  lim x 21  14 , x  2 2

x 2 x  2 x

x 2

x ( x  2)

x 2 x

( x  2)( x 1) lim x 33 x 22  lim  lim x 21  14 , x  2 2 x 2 x  2 x x 2 x ( x  2) x  2 x 2   ( x 2)( x 1) lim x 33 x 22  lim 2  lim x 21  14 , x  2 x 2 x  2 x x 2 x ( x  2) x 2 x 2

(e) lim x 33 x 22  lim x 0 x  2 x

58. (a)

1 1

1 negativenegative

positive positivenegative

1 x

1 21/3

1 positivenegative

positive positivepositive

    0  lim      (b) lim     0  lim  2 2 (c) lim     (d) lim        

55. (a)

  

1 positivenegative

x 0

( x  2)( x 1) x 2 ( x  2)

 



negativenegative positivenegative

2 ( x  2)( x 1) ( x 1) 1 1 lim x 3 3 x  2  lim x ( x  2)( x  2)  lim x ( x  2)  2(4) 8  x 4 x x 2 x  2 x  2

(b)

2 ( x  2)( x 1) ( x 1) lim x 3 3 x  2  lim x ( x  2)( x  2)  lim x ( x  2)      x  4 x x 2 x 2 x 2

(c)

2 ( x  2)( x 1) ( x 1) lim x 3 3 x  2  lim x ( x  2)( x  2)  lim x ( x  2)      x  4 x x 0 x 0 x 0

  

negative negativepositive negative negativepositive

 

100

Chapter 2 Limits and Continuity (d) (e)

( x  2)( x 1)

( x 1)

0 0 lim x 3 3 x  2  lim x ( x  2)( x  2)  lim x ( x  2)  (1)(3) 

x 1

x 4 x



x 1

lim x (xx12)   x 0 

and lim

x 0 



x 1   x ( x  2)



x 1

negative positivepositive

negative negativepositive



so the function has no limit as x  0. 59. (a) 60. (a) 61. (a) (c) 62. (a) (c)

3    lim  2  1/3 t 

(b)

1  7   lim  3/5    t  t 0

(b)

 1 lim  2/3 

(b)

t 0 

x 0   x

 2  ( x 1) 2/3 

 1  lim  2/3  2 2/3     x x  ( 1)   x 1

(d)

 1 lim  1/3 

(b)

x 0   x

 1  ( x 1)4/3 

 1  lim  1/3  1 4/3     x x  ( 1)   x 1

(d)

3  lim  2  1/3 t 

t 0 

1  7    lim  3/5    t  t 0

 1 lim  2/3 

 2  ( x 1)2/3 

x 0   x

 1  lim  2/3  2 2/3     x x  ( 1)   x 1  1 lim  1/3 

x 0   x

 1   ( x 1) 4/3 

 1  lim  1/3  1 4/3     x x  ( 1)   x 1

63. y  x11

64. y  x11

65. y  2 x1 4

66. y  x33

Section 2.6 Limits Involving Infinity; Asymptotes of Graphs 67. y  xx32  1  x 1 2

68. y  x2x1  2  x21

69. Domain  (, );

lim 4  32x  7  horizontal asymptote is y  7 x 1

x 

70. Domain  (, 1)  (1, 1)  (1, ); lim

2x 2

x 1 x 1

101

  and lim

2x 2

x 1 x 1

x 1

lim 22 x  0, lim 22 x  , and lim 22 x  ; x  x 1 x 1 x 1 x 1 x 1

   horizontal asymptote is y  0; vertical asymptotes are x  1,

x x 71. Domain  (, ); lim 8e x  1, and lim 8e x  4  horizontal asymptotes are y  1, y  4

x  2  e

x  2  e

x 2x x 2x 72. Domain  (, ); lim 4ex  e2 x  1, and lim 4ex  e2 x  4  horizontal asymptotes are y  1, y  4

x  e  e

x  e  e

2 73. Domain  (, 0 )  (0,  ) ; if x  0, then 1  x x 4 , lim x 0 

x2  4  1, x

lim

x 0



x2  4  , and lim x x 

x2  4  , and lim x x 

x2  4  1; if x  0, then x

x2 4  1  horizontal asymptotes are y  1, x

y  1; vertical

asymptote is x  0

74. Domain  (, 2 )  (2,  );

x3  , lim x3  , lim x3  1,  horizontal asymptote is 3 3  x3 8 x  8 x x  8 x2 x2

lim



y  1; vertical asymptote is x  2

75. Here is one possibility.

76. Here is one possibility.

102

Chapter 2 Limits and Continuity

77. Here is one possibility.

78. Here is one possibility.

79. Here is one possibility.

80. Here is one possibility.

81. Here is one possibility.

82. Here is one possibility.

f ( x)

83. Yes. If lim g ( x )  2 then the ratio the polynomials’ leading coefficients is 2, so lim g ( x )  2 as well. x x 84. Yes, it can have a horizontal or oblique asymptote. f ( x)

f ( x)

85. If lim g ( x )  L, where L  0, then the ratio of the polynomials’ leading coefficients is L, so lim g ( x )  L x  x  as well. f ( x)

f ( x)

If lim g ( x )  0, then lim g ( x )  0 also. Similarly, if lim g ( x ) DNE, then lim g ( x ) DNE either and the x x x x rational function has no horizontal asymptote. 86.

lim

x 

 x  9  x  4    x 9  x  4   lim ( x 9) ( x  4)  x  9  x  4   xlim   x 9  x  4  x  x 9  x  4   5

 lim

x 

x 5  lim  101  0 x 9  x  4 x  1 9x  1 4x

Section 2.6 Limits Involving Infinity; Asymptotes of Graphs

103

2  2  ( x 2  25)  ( x 2 1) 87. lim  x 2  25  x 2  1   lim  x 2  25  x 2  1    x 2  25  x 2 1   lim  x     x  25  x 1  x  x 2  25  x 2 1 x  

 lim

x 2  25  x 2 1

x 

88.

x  1 252  1 12 x

 101  0

 2  ( x 2 3) ( x 2 ) lim  x 2  3  x   lim  x 2  3  x    x 2 3  x   lim  lim   x 3  x  x  x 2 3  x  x   x  x   3 x2

 lim

x  1 32  x x

89.

26 x

 lim



 3x

lim

x  1 32 1

3 x 2 3  x

 101  0

2   (4 x 2 ) (4 x 2 3 x  2) lim  2 x  4 x 2  3 x  2   lim  2 x  4 x 2  3x  2    2 x  4 x 2 3 x  2   lim     2 x  4 x 3 x  2  x  2 x  4 x 2 3 x  2  x   x  

3 x  2

 lim

x  2 x  4 x 2 3 x  2

3 2x

 lim

x 

3 x  2 x2

2x  x2

3 x  2 x x  2 xx  4  3x  22 x

 lim 2

4  3x  2 x

 3202   34

x  2  4  3x  22 x

90.

 2  (9 x 2  x )  (9 x 2 ) x lim  9 x 2  x  3 x   lim  9 x 2  x  3 x    9 x 2  x 3 x   lim  lim    x    9 x  x 3 x  x  9 x 2  x 3 x x   x  9 x 2  x 3 x  lim

x 

91.

 xx

x 

 x  3xx x2 x2

2  2  ( x 2 3 x ) ( x 2  2 x ) lim  x 2  3 x  x 2  2 x   lim  x 2  3x  x 2  2 x    x 2 3 x  x 2  2 x   lim  x     x 3 x  x  2 x  x  x 2 3 x  x 2  2 x x   5x 5  lim  lim  151  52 2 2 3

x 

92.

1  1   16 9  1x 3 3 3

 lim

9 x2

x 3 x  x  2 x

x 

1 x  1 2x

2  2  ( x 2  x ) ( x 2  x ) 2x x 2  x  x 2  x  lim  x 2  x  x 2  x    x 2  x  x 2  x   lim  lim   x  x  x  x  x  x 2  x  x 2  x x x 2  x  x 2  x x  x   2  lim  121  1 1 1

lim

x  1 x  1 x

93. For any   0, take N  1. Then for all x  N we have that f ( x)  k  k  k  0  . 94. For any   0, take N  1. Then for all y   N we have that f ( x)  k  k  k  0  . 95. For every real number  B  0, we must find a   0 such that for all x, 0  x  0    21   B. x

Now,  12   B  0  12  B  0  x 2  B1  x  1 . Choose   1 , or any smaller positive value, then x

0  x    x  1  21   B so that lim  12  . B

x0

96. For every real number B  0, we must find a   0 such that for all x, 0  x  0    1  B. Now,

x 1  B  0  x  1 . Choose   1 , or any smaller positive value. Then 0  x  0    x  1  1  B so B B B x x 1 that lim  . x0 x

104

Chapter 2 Limits and Continuity

97. For every real number  B  0, we must find a   0 such that for all x, 0  x  3    ( x 3) 2   B  0  2 B0 2 ( x 3) 2 ( x 3) 2

 B1  ( x  3)2  B2  0  X  3 

smaller positive value, then 0  x  3   

2   B. Now, ( x 3) 2

2 . Choose   B

2 , or any B

2   B  0 so that lim 2  . 2 ( x 3) 2 x 3 ( x 3)

98. For every real number B  0, we must find a   0 such that for all x, 0  x  (5)   

1  B. ( x 5) 2

1  B  0  ( x  5)2  B1  x  5  1 . Choose   1 , or any smaller positive value. Then B B ( x  5)2 1 0  x  (5)    x  5  1  1 2  B so that lim  . 2 B ( x 5) x 5 ( x 5)

Now,

99. (a) We say that f ( x) approaches infinity as x approaches x0 from the left, and write lim f ( x)  , x  x0

if for every positive number B, there exists a corresponding number   0 such that for all x, x0    x  x0  f ( x)  B. (b) We say that f ( x) approaches minus infinity as x approaches x0 from the right, and write lim f ( x)  , x  x0

if for every positive number B (or negative number  B) there exists a corresponding number   0 such that for all x, x0  x  x0    f ( x )   B. (c) We say that f ( x) approaches minus infinity as x approaches x0 from the left, and write lim f ( x)  , if x  x0

for every positive number B (or negative number  B) there exists a corresponding number   0 such that for all x, x0    x  x0  f ( x)   B. 100. For B  0, 1x  B  0  x  B1 . Choose   B1 , or any smaller positive value. Then 0  x    0  x  B1  1x  B so that lim 1x  . x 0 

101. For B  0, 1x   B  0   1x  B  0   x  B1   B1  x. Choose   B1 , or any smaller positive value. Then   x  0   B1  x  1x   B so that lim 1x  . x 0 

102. For B  0, x 1 2   B   x 1 2  B  ( x  2)  B1  x  2   B1  x  2  B1 . Choose   B1 , or any smaller positive value. Then 2    x  2    x  2  0   B1  x  2  0  x 1 2   B  0 so that lim x 1 2  . x  2

103. For B  0, x 1 2  B  0  x  2  B1 . Choose   B1 , or any smaller positive value. Then 2  x  2    0  x  2    0  x  2  B1  x 1 2  B  0 so that lim x 1 2  . x  2 1  B  1  x 2  1  (1  x )(1  x )  1 . Now 1 x  1 since x  1. Choose   1 , 2 B B 2B 1 x 2 1 or any smaller positive value. Then 1    x  1    x  1  0  1  x    2 B  (1  x)(1  x)  B1 12x  B1  1 2  B for 0  x  1 and x near 1  lim 1 2  . 1 x x 1 1 x

104. For B  0 and 0  x  1,

 

Section 2.6 Limits Involving Infinity; Asymptotes of Graphs 2

106. y  xx 11  x  1  x21

108. y  2xx 14  12 x  1  2 x3 4

110. y  x 21  x  12

105. y  xx1  x  1  x11

107. y  xx 14  x  1  x31

109. y  x x1  x  1x

111. y 

x 4 x 2

112. y 

1 4 x 2

105

106

Chapter 2 Limits and Continuity

1 113. y  x 2/3  1/3

114. y  sin

  

x 2 1

115. (a) y   (see accompanying graph) (b) y   (see accompanying graph) (c) Cusps at x   1 (see accompanying graph)

116. (a) y  0 and a cusp at x  0 (see the accompanying graph) (b) y  32 (see accompanying graph) (c) A vertical asymptote at x  1 and contains the



point 1,

CHAPTER 2

3 23 4

 (see accompanying graph)

PRACTICE EXERCISES

1. At x  1:

lim f ( x)  lim f ( x)  1

x  1

x  1

 lim f ( x)  1  f (1) x 1

 f is continuous at x  1.

At x  0 :

lim f ( x)  lim f ( x)  0

x 0 

 lim f ( x)  0.

x 0

x 0

But f (0)  1  lim f ( x) x 0

 f is discontinuous at x  0. If we define f (0)  0, then the discontinuity at x  0 is removable.

At x  1:

lim f ( x)  1 and lim f ( x)  1

x 1

x 1

 lim f ( x ) does not exist x 1

 f is discontinuous at x  1.

Chapter 2 Practice Exercises 2. At x  1:

107

lim f ( x)  0 and lim f ( x)  1

x  1

 lim f ( x) does not exist

x 1

x  1

 f is discontinuous at x  1.

At x  0 :

lim f ( x )   and lim f ( x)  

x 0 

x 0 

 lim f ( x) does not exist x 0

 f is discontinuous at x  0.

At x  1:

lim f ( x)  lim f ( x)  1  lim f ( x)  1.

x 1

x 1

x 1

But f (1)  0  lim f ( x) x 1

 f is discontinuous at x  1. If we define f (1)  1, then the discontinuity at x  1 is removable.

3. (a) lim (3 f (t ))  3 lim f (t )  3( 7)  21 t t0

t t0

  (b) lim ( f (t ))   lim f (t )   (7) 2  49 t t0  t t0  (c) lim ( f (t )  g (t ))  lim f (t )  lim g (t )  ( 7)(0)  0 2

t t0

t t0 lim f (t )

f (t )

t t0 lim f (t )

(d) lim g (t ) 7  lim (0g (t ) 7)  lim g (0t )  lim 7  077  1 t t t t

t t

t  t0

  (e) lim cos ( g (t ))  cos  lim g (t )   cos 0  1 t t0 t  t  0 

(f)

lim | f (t )|  lim f (t )  | 7|  7

t t0

(g) lim ( f (t )  g (t ))  lim f (t )  lim g (t )  7  0  7 t t0

(h) lim

t t0

  1 f (t )

t t0

 lim f (t )  17   17 t  t0

4. (a) lim  g ( x)   lim g ( x)   2 x 0

x 0

(b) lim ( g ( x)  f ( x))  lim g ( x)  lim f ( x)  x 0

x 0

 2   12   22

x 0

(d) lim f 1( x )  lim 1f ( x )  11  2 x 0 2

x 0

(e) lim ( x  f ( x))  lim x  lim f ( x)  0  12  12 x 0

(f)

lim

x 0

x 0 x 0 1 (1) lim f ( x ) lim cos x f ( x )cos x 2 x 0 x 0     12 x 1 lim x lim 1 0 1



x0

x 0

5. Since lim x  0 we must have that lim (4  g ( x))  0. Otherwise, if lim (4  g ( x)) is a finite positive number, x 0

x 0

4 g ( x ) 4 g ( x ) we would have lim  x    and lim  x    so the limit could not equal 1 as x  0. Similar   x 0   x 0   reasoning holds if lim (4  g ( x)) is a finite negative number. We conclude that lim g ( x)  4. x 0

x 0

108

Chapter 2 Limits and Continuity

6. 2  lim  x lim g ( x)   lim x  lim  lim g ( x)    4 lim  lim g ( x)    4 lim g ( x) (since lim g ( x) is a x0 x  4  x 0 x  4  x 0 x 0  x  4 x  4  x 0   2 1 constant)  lim g ( x )  4   2 . x 0

7. (a) lim f ( x)  lim x1/3  c1/3  f (c) for every real number c  f is continuous on ( , ). x c

x c

1  h(c ) for every nonzero real number c  h is continuous on ( , 0) and c 2/3

x c

(b) lim g ( x)  lim x3/4  c3/4  g (c) for every nonnegative real number c  g is continuous on [0, ). (c) lim h( x)  lim x 2/3 

(, ). (d) lim k ( x)  lim x 1/6  1/1 6  k (c ) for every positive real number c  k is continuous on (0, ).

8. (a)

   n  12   ,  n  12   , where I  the set of all integers

n I

(b)

 (n , (n  1) ), where I  the set of all integers

n I

( x  2)( x  2)  lim  lim x  2 , x  2; the limit does not exist because 2 x 0 x  5 x 14 x x 0 x ( x  7)( x  2) x 0 x ( x  7) lim x (xx27)   and lim x (xx27)   x 0  x 0  x2  4 x  4 ( x  2)( x  2) 0 0 lim 3 2  lim x ( x  7)( x  2)  lim x (xx27) , x  2, and lim x (xx27)  2(9) x  5 x  14 x x 2 x2 x 2 x 2

9. (a) lim

(b)

x ( x 1) x2  x  lim 3 2  lim 2 x 1  lim 2 1 , x  0 and x  1 5 4 3 x 0 x  2 x  x x 0 x ( x  2 x 1) x 0 x ( x 1)( x 1) x 0 x ( x 1) 2 lim 2 1   and lim 2 1    lim 5 x 4x 3    x ( x 1)  x ( x 1) x 0 x  2 x  x x 0 x 0 x ( x 1) x2  x 1

10. (a) lim

(b)

lim

 lim

x 1 x  2 x  x x 1 x ( x  2 x 1) x 1 x ( x 1) lim 2 1   and lim 2 1  . x 1 x ( x 1) x 1 x ( x 1)

, x  0 and x  1. The limit does not exist because

11. lim 11 xx  lim

1 x  lim 1  12 x 1 (1 x )(1 x ) x 1 1 x

x 1

( x2 a2 )

12. lim x 4  a 4  lim x a x  a

x a ( x  a )( x  a )

1  12 2 2 2a xa x  a

 lim

( x  h)2  x 2 ( x 2  2 hx  h 2 )  x 2  lim  lim (2 x  h)  2 x h h h 0 h 0 h 0

13. lim

( x  h)2  x 2 ( x 2  2 hx  h 2 )  x 2  lim  lim (2 x  h)  h h h x 0 x 0 x 0

14. lim

15.

1

2 (2  x )

1   1 4 x 0 4  2 x

lim 2  xx 2  lim 2 x (2 x )  lim

x 0

(2  x )3 8 ( x3  6 x 2 12 x 8) 8  lim  lim ( x 2  6 x  12)  12 x x x 0 x 0 x 0

16. lim

Chapter 2 Practice Exercises 1/3 2/3 1/3 ( x 1)( x 1) 1 1  lim ( x 1)  ( x  x 1)( x 1)  lim  lim 2/3 x 1/3  11  2 2/3 1/3 2/3 1/3 x 1 x 1 ( x 1) ( x 1)( x  x 1) x 1 ( x 1)( x  x 1) x 1 x  x 1 111 3

1/3

17. lim x x 1

18.

2/3 ( x1/3  4)( x1/3  4) ( x1/3  4)( x1/3  4) ( x 2/3  4 x1/3 16)( x 8) lim x 16  lim  lim  2/3 1/3

x 8

x 64

x 8 ( x 8)( x  4 x 16) ( x1/3  4) ( x 8) (4  4) (88) 8  lim  lim 2/3 1/3  161616  3 2/3 1/3 x 64 ( x 64) ( x  4 x 16) x 64 x  4 x 16 x 8 x 64 ( x 64) ( x1/3  4) ( x 8)

x 64

cos  x

tan 2 x

sin 2 x   lim 19. lim tan  x  lim cos 2 x sin  x x 0

20.

x 0

cos  x x 2 x  1 1 1  2  2  sin2 x2 x  cos 2 x   sin  x    x   

lim csc x  lim sin1 x   x  

x  









 

21. lim sin 2x  sin x  sin 2  sin   sin 2  1 x 

22. lim cos 2 ( x  tan x)  cos 2 (  tan  )  cos 2 ( )  (1)2  1 x 

23. lim 3sin8 xx  x  lim

8  8 4 sin x x 0 3 x 1 3(1) 1

x 0

 sin 2 x 2 x 1  lim cos 2 x 1  lim  cossin2xx1  cos cos 2 x 1  x 0 sin x (cos 2 x 1) x 0 sin x (cos 2 x 1) x 0 2

24. lim  cossin2 xx 1  lim x 0

4sin x cos 2 x 4(0)(1) 2  0 11 x 0 cos 2 x 1

 lim

25. Let x = t  3  lim ln(t  3)  lim ln x   t 3



x 0 



26. lim t 2 ln 2  t  ln1  0 t 1

 

27. 1  cos   1  e1  ecos( / )  e1   e 1   ecos( / )   e  lim

 0 

 ecos( / )  0 by the

Sandwich Theorem 28.

lim

2e1/ z

z 0 e1/ z  1

 lim

1/ z

z  1 e

 120  2 1/3

29. 30.

  lim [4 g ( x)]1/3  2   lim 4 g ( x)  x 0   x 0   1

lim

x 5 x  g ( x)

 2  lim 4 g ( x)  8, since 23  8. Then lim g ( x )  2. x 0 

x 0 

 2  lim ( x  g ( x))  12  5  lim g ( x)  12  lim g ( x)  12  5 x 5

x 5

31. lim 3gx( x)1    lim g ( x)  0 since lim (3x 2  1)  4 x 1

32.

lim

x 2

x 1

5 x 2  0  lim g ( x )   since lim (5  x 2 )  1 g ( x) x 2 x 2

109

110

Chapter 2 Limits and Continuity

33. (a) f (1)  1 and f (2)  5  f has a root between  1 and 2 by the Intermediate Value Theorem. (b), (c) root is 1.32471795724 34. (a) f (2)  2 and f (0)  2  f has a root between  2 and 0 by the Intermediate Value Theorem. (b), (c) root is 1.76929235424 35. At x  1:

lim



f ( x)  lim

x 1 x ( x 2 1)

 lim

x 2 1

x 1

lim

x 1

x ( x 2 1) 

x 1

| x 2 1|

 lim x  1, and x 1

f ( x)  lim

x ( x 2 1) 2

x 1 | x 1|

lim

x 1

x 1 ( x 1)

 lim ( x)  (1)  1. Since lim x 1

x ( x 2 1)

 lim

x 1

f ( x) 

f ( x)  lim f ( x) does not exist, the x1

function f cannot be extended to a continuous function at x  1. At x  1:

lim f ( x)  lim

x 1

x ( x 2 1) 2

x 1 | x 1| x ( x 2 1)

 lim

x 1

x 2 1

 lim

x ( x 2 1) 2

x 1  ( x 1)

 lim ( x)  1, and lim f ( x)  lim x 1

x 1

x ( x 2 1)

2 x 1 | x 1|

 lim x  1 x 1

Again lim f ( x) does not exist so f cannot be extended to a continuous function at x  1 either. x 1

36. The discontinuity at x  0 of f ( x)  sin

lim sin 1x does not exist.  1x  is nonremovable because x 0

37. Yes, f does have a continuous extension at a  1: define f (1)  lim x 41  43 . x 1 x  x

38. Yes, g does have a continuous extension at a  2 :

    5 cos 

g 2  lim 4  2   54 . 

Chapter 2 Practice Exercises 39. From the graph we see that lim h(t )  lim h(t ) t 0 

t 0 

so h cannot be extended to a continuous function at a  0.

40. From the graph we see that lim k ( x)  lim k ( x) x 0 

x 0 

so k cannot be extended to a continuous function at a  0.

41.

43.

lim 52xx73  lim

2  3x

 5200  52 7

42.

x  5 x

x 



x 2  4 x 8  lim 1  4  8 3 2 3 x3 x  3 x x  3 x 3 x

lim

2  32 20 x 2 x 2 3  lim  5  0  52 2 7 x  5 x  7 x  5  2

lim

  000  0

44.

2 1  lim 7x 1  100 0  0 2 1   x  x 7 x 1 x  x 2

lim

45.

x 2 7 x  lim x 7   1 x  x 1 x  1

lim

46.

x 4  x3  lim x 1   3 128 12 x  x 128 x  12 3

lim

47. 48.

49.

50.

lim sin x  lim 1  0 since x   x  x   x 

x 

lim cos 1  lim 2  0  lim cos 1  0

 

x2 x lim x sin  lim x sin x

x 

1 sinx x  2

x 

1 sinx x





 1100 0  1

x 2/3  x 1  lim  1 x 5/3   1 0  1 x  cos 2 x x   1 cos2 x  1 0 x 

lim

2/3



51.

sin x  x    as x    lim  x   0



x 2/3



lim e1/ x cos 1x  e0  cos(0)  1 1  1

x 

52.

 

lim ln 1  1t  ln1  0

x

111

112 53.

Chapter 2 Limits and Continuity lim tan 1 x   2

54.

x 



lim e3t sin 1 1t  0  sin 1 (0)  0  0  0

t 

55. (a) y  xx 34 is undefined at x  3 : lim xx 34   and lim xx 34   , thus x  3 is a vertical asymptote. 



x 3

x 3 x 2  x  2   and lim x 2  x  2  , thus x  1 is a vertical 2 2 x 1 x  2 x 1 x 1 x  2 x 1

(b) y  x2  x  2 is undefined at x  1: lim x  2 x 1

asymptote. 2

x  2 x 8

x2

x 4

x 2  x 6 

 x 2  2 x 8

lim

x 4

x 3  ;

 x4

x 2  x 6  lim x 3  . Thus x  4 is a vertical asymptote. x4  x 2  2 x 8 x 4 x 4

lim

1 1 2

x 1 x  x 1

x  1 2

x  x 1

y  12 x : lim 12 x  lim x 1  11  1 and lim 12 x  lim

56. (a)

1 1

x  1 2

 11  1, thus y  1 is a

horizontal asymptote. y

x 4 : x4

x2  4 : x

(b)

lim

1 4 x 4 x  lim  1 0 x4 1 0 x  1 4x

lim

x2  4  lim x x 

x 

1 42 x

 lim

x 

x x



 1, thus y  1 is a horizontal asymptote.

1 0  1 and lim 1 x 

 lim

x 

1 42 x 1



x2 4  lim x x 

1 42 x

x x2

1 0  11  1, 1

thus y  1 and y  1 are horizontal asymptotes. x 2 9 : 9 x 2 1

(d) y 

lim

x 

x 2 9  lim 9 x 2 1 x 

1  92

x 9  12 x



1 0  1 and lim 9 0 3 x 

1 92

x 2 9  lim 9 x 2 1 x 

9  12 x

thus y  13 is a horizontal asymptote.

57. Domain  [4, 2)  (2, 4]; y in range and y  16  x x2

lim

x 2



x 

x  lim x b  x 

1. (a) x x



16  x 2  , and x 2

x  x  x b

a  42  lim x

0.01

0.001

a  42

ax 2  4 x  x b

lim

a  42   a  horizontal asymptote is y   a x

ADDITIONAL AND ADVANCED EXERCISES 0.1

x2

2 x ax 2  4 4    vertical asymptote is x  b; lim ax  lim x b  x b  x b x  x 



 lim x x b a  42  a  horizontal asymptote is y  a ,

CHAPTER 2

lim

 ,  range  (, )

58. Since lim x b

16 x 2 , if x  4, then y  0, x 2

0.0001 0.00001

0.7943 0.9550 0.9931 0.9991 0.9999

Apparently, lim x x  1 x 0 

 1900  13 ,

Chapter 2 Additional and Advanced Exercises

113

(b)

2. (a) x

 1x 

1/(ln x )

1000

0.3679 0.3679 0.3679

1 x x

Apparently, lim (b)

100

1/(ln x )

2 lim L  lim L0 1  v 2  L0 c v c  v c 

 0.3678  1e

1

lim v 2

v c 2

 L0 1  c 2  0 c

The left-hand limit was needed because the function L is undefined if v  c (the rocket cannot move faster than the speed of light). 4. (a)

x  1  0.2  0.2  2x  1  0.2  0.8  2x  1.2  1.6  2

x  2.4  2.56  x  5.76

(b)

x  1  0.1  0.1  2x  1  0.1  0.9  2x  1.1  1.8  2

x  2.2  3.24  x  4.84

5. |10  (t  70)  104  10|  0.0005  |(t  70)  104 |  0.0005  0.0005  (t  70)  104  0.0005  5  t  70  5  65  t  75  Within 5 F 6. We want to know in what interval to hold values of h to make V satisfy the inequality |V  1000|  |36 h  1000|  10. To find out, we solve the inequality: 990  h  1010  8.8  h  8.9 |36 h  1000|  10  10  36 h  1000  10  990  36 h  1010  36  36 where 8.8 was rounded up, to be safe, and 8.9 was rounded down, to be safe. The interval in which we should hold h is about 8.9  8.8  0.1 cm wide (1 mm). With stripes 1 mm wide, we can expect to measure a liter of water with an accuracy of 1%, which is more than enough accuracy for cooking. Copyright  2023 Pearson Education, Inc.

114

Chapter 2 Limits and Continuity

7. Show lim f ( x)  lim ( x 2  7)  6  f (1). x 1

x 1

Step 1: |( x  7)  6|      x 2  1    1    x 2  1    1    x  1   Step 2: | x  1|      x  1      1  x    1. Then   1  1   or   1  1   . Choose   min 1  1   , 1    1 , or any smaller positive value, then 0  | x  1|    |( x 2  7)  6|   and





lim f ( x)  6. By the continuity test, f ( x) is continuous at x  1.

x 1

x  14



1 2g 1 . 4 x  14 2 x

8. Show lim g ( x)  lim

Step 1: 21x  2      21x  2    2    21x  2    412  x  412 Step 2: X  14      x  14      14  x    14 . Then   14  412    14  412  4(2 ) , or

  14  412    412  14  4(2 ) . Choose   4(2 ) , the smaller of the two values, or any smaller positive value. Then 0  x  14    21x  2   and lim 21x  2. By the continuity test, g ( x) is continuous at x  14 . x  14

9. Show lim h( x)  lim 2 x  3  1  h(2). x 2

x2

(1 ) 2  3 (1 ) 2  3 x 2 2 (1 ) 2 3 (1 ) 2 3 1(1 )2   2  Step 2: | x  2|      x  2   or    2  x    2. Then   2  2 2 2 2 2 2 2 2 2 (1   )  3 (1   )  3 (1   )  1       2 , or   2    2     2 . Choose     2 , the smaller of the two 2 2 2

Step 1:

2 x  3  1      2 x  3  1    1    2 x  3  1   

values, or any smaller positive value. Then, 0  | x  2|   

2 x  3  1  , so lim 2 x  3  1. By the x2

continuity test, h( x) is continuous at x  2. 10. Show lim F ( x)  lim 9  x  2  F (5). x 5

Step 1:

x 5

9  x  2      9  x  2    9  (2   ) 2  x  9  (2   )2

Step 2: 0  | x  5|      x  5      5  x    5. Then   5  9  (2   )2    (2   )2  4   2  2, or   5  9  (2   )2    4  (2   ) 2   2  2. Choose    2  2, the smaller of the two values,

or any smaller positive value. Then, 0  | x  5|    F ( x) is continuous at x  5.

9  x  2  , so lim 9  x  2. By the continuity test, x 5

11. Suppose L1 and L2 are two different limits. Without loss of generality assume L2  L1. Let   13 ( L2  L1 ). Since lim f ( x)  L1 there is a 1  0 such that 0  | x  x0 |  1  | f ( x )  L1 |      f ( x )  L1    x  x0  13 ( L2  L1 )  L1 

f ( x)  13 ( L2  L1 )  L1  4 L1  L2  3 f ( x)  2 L1  L2 . Likewise, lim f ( x)  L2 so there

x  x0 1 is a  2 such that 0  | x  x0 |   2  | f ( x)  L2 |      f ( x)  L2     3 ( L2  L1 )  L2  f ( x)  13 ( L2  L1 )  L2  2 L2  L1  3 f ( x)  4 L2  L1  L1  4 L2  3 f ( x)  2 L2  L1. If   min{1 ,  2 } both

4 L1  L2  3 f ( x)  2 L1  L2    5( L1  L2 )  0  L1  L2 . That L1  4 L2  3 f ( x)  2 L2  L1  is, L1  L2  0 and L1  L2  0, a contradiction.

inequalities must hold for 0  | x  x0 |   :

Chapter 2 Additional and Advanced Exercises

115

12. Suppose lim f ( x)  L. If k  0, then lim k f ( x)  lim 0  0  0  lim f ( x) and we are done. If k  0, then given x c

x c

any   0, there is a   0 so that 0  | x  c |     | f ( x)  L |  | k |  | k || f ( x)  L |    | k ( f ( x)  L)|  

 |(kf ( x))  (kL)|  . Thus lim k f ( x)  kL  k  lim f ( x)  . x c  x c 

13. (a) Since x  0 , 0  x3  x  1  ( x3  x)  0  lim f ( x3  x)  lim f ( y )  B where y  x3  x. 

x 0 



y 0

(b) Since x  0 , 1  x  x  0  ( x  x)  0  lim f ( x  x)  lim f ( y )  A where y  x3  x. 3



x 0 



y 0



x 0  4

y 0 2



(d) Since x  0 , 1  x  0  0  x  x  1  ( x  x )  0  lim f ( x  x 4 )  A as in part (c). x 0 

14. (a) True, because if lim ( f ( x)  g ( x)) exists then lim ( f ( x)  g ( x))  lim f ( x)  lim [( f ( x)  g ( x))  f ( x)]  x a

lim g ( x) exists, contrary to assumption.

x a

(b) False; for example take f ( x) 





x a

1 and g ( x )   1 . Then neither lim f ( x ) nor lim g ( x ) exists, but x x x 0 x 0

lim ( f ( x)  g ( x))  lim 1x  1x  lim 0  0 exists. x 0 x 0 (c) True, because g ( x) | x | is continuous  g ( f ( x))  | f ( x)| is continuous (it is the composite of continuous functions). 1, x  0 (d) False; for example let f ( x)    f ( x) is discontinuous at x  0. However | f ( x)| 1 is  1, x  0 continuous at x  0. x 0

15. Show lim f ( x )  lim xx 11  lim x 1

x 1

( x 1)( x 1)  2, x  1. ( x 1)

 x 2 1 , x  1 . We now prove the limit of f ( x) as x  1 Define the continuous extension of f ( x) as F ( x )   x 1  2 , x  1

exists and has the correct value. 2

Step 1: xx 11  (2)     

( x 1)( x 1)  2      ( x  1)  2  , x  1    1  x    1 ( x 1)

Step 2: | x  (1)|     x  1      1  x    1. Then   1    1    , or   1    1 2

   . Choose   , or any smaller positive value. Then 0  | x  (1)|    xx 11  (2)    lim F ( x)  2. Since the conditions of the continuity test are met by F ( x), then f ( x ) has a continuous

x 1

extension to F ( x) at x  1. 2

16. Show lim g ( x)  lim x 2x2x63  lim x 3

x 3

( x 3)( x 1)  2, x  3. 2( x 3)

 x 2  2 x 3 , x  3 Define the continuous extension of g ( x) as G ( x)   2 x 6 . We now prove the limit of g ( x) as x  3 , x3 2 exists and has the correct value. 2

Step 1: x 2x2x63  2     

( x 3)( x 1)  2      x21  2  , x  3  3  2  x  3  2 2( x 3)

Step 2: | x  3|      x  3    3    x    3. Then, 3    3  2    2, or   3  3  2 2

   2. Choose   2, or any smaller positive value. Then 0  | x  3|    x 2x2x63  2   

lim

x 3

( x 3)( x 1)  2. 2( x 3)

Since the conditions of the continuity test hold for G ( x), g ( x) can be continuously

116

Chapter 2 Limits and Continuity

17. (a) Let   0 be given. If x is rational, then f ( x)  x  | f ( x)  0|  | x  0|   | x  0|  ; i.e., choose   . Then | x  0|    | f ( x)  0|   for x rational. If x is irrational, then f ( x)  0  | f ( x)  0|    0   which is true no matter how close irrational x is to 0, so again we can choose   . In either case, given   0 there is a     0 such that 0  | x  0|    | f ( x)  0|  . Therefore, f is continuous at x  0. (b) Choose x  c  0. Then within any interval (c   , c   ) there are both rational and irrational numbers. If c is rational, pick   2c . No matter how small we choose   0 there is an irrational number x in (c   , c   )  | f ( x)  f (c)|  |0  c |  c  2c  . That is, f is not continuous at any rational c  0. On the other hand, suppose c is irrational  f (c)  0. Again pick   2c . No matter how small we choose   0 there is a rational number x in (c   , c   ) with | x  c |  2c    2c  x  32c . Then | f ( x)  f (c)|  | x  0|  | x |  2c    f is not continuous at any irrational c  0. |c|

If x  c  0, repeat the argument picking   2  2c . Therefore f fails to be continuous at any nonzero value x  c. 18. (a) Let c  mn be a rational number in [0, 1] reduced to lowest terms  f (c)  1n . Pick   21n . No matter how small   0 is taken, there is an irrational number x in the interval (c   , c   )  | f ( x)  f (c)|  0  1n  1n  21n  . Therefore f is discontinuous at x  c, a rational number.

(b) Now suppose c is an irrational number  f (c)  0. Let   0 be given. Notice that 12 is the only rational number reduced to lowest terms with denominator 2 and belonging to [0, 1]; 13 and 23 the only rationals with denominator 3 belonging to [0, 1]; 14 and 34 with denominator 4 in [0, 1]; 15 , 52 , 53 and 54 with denominator 5 in [0, 1]; etc. In general, choose N so that N1    there exist only finitely many rationals in [0, 1] having

denominator  N , say r1 , r2 , , rp . Let   min {| c  ri |: i  1,  , p}. Then the interval (c   , c   ) contains no rational numbers with denominator  N . Thus, 0  | x  c |    | f ( x)  f (c)|  | f ( x )  0|  | f ( x)|  N1    f is continuous at x  c irrational. (c) The graph looks like the markings on a typical ruler when the points ( x, f ( x)) on the graph of f ( x ) are connected to the x-axis with vertical lines.

19. Yes. Let R be the radius of the equator (earth) and suppose at a fixed instant of time we label noon as the zero point, 0, on the equator  0   R represents the midnight point (at the same exact time). Suppose x1 is a point on the equator “just after” noon  x1   R is simultaneously “just after” midnight. It seems reasonable that the temperature T at a point just after noon is hotter than it would be at the diametrically opposite point just after midnight: That is, T ( x1 )  T ( x1   R)  0. At exactly the same moment in time pick x2 to be a point just before midnight  x2   R is just before noon. Then T ( x2 )  T ( x2   R )  0. Assuming the temperature function T is continuous along the equator (which is reasonable), the Intermediate Value Theorem says there is a point c Copyright  2023 Pearson Education, Inc.

Chapter 2 Additional and Advanced Exercises

117

between 0 (noon) and  R (simultaneously midnight) such that T (c)  T (c   R )  0; i.e., there is always a pair of antipodal points on the earth’s equator where the temperatures are the same. 2 2  20. lim f ( x ) g ( x)  lim 14  ( f ( x)  g ( x))2  ( f ( x)  g ( x))2   14  lim ( f ( x )  g ( x))    lim ( f ( x)  g ( x))    x c x c    x c    xc  14 (32  (1)2 )  2

a 0

(b)







1(1 a ) 1 1 a 1 1 a  lim  1  12 a 1 1 0 1 1 a a 0 a ( 1  1 a ) a 0 1(1 a )  a  1  lim  1 At x  1: lim r (a )  lim 1 0 a 1 a 1 a ( 1 1 a ) a 1 a ( 1  1 a ) 1 (1 a ) a At x  0: lim r (a )  lim 1 a1 a  lim 1 a1 a 1 1 a  lim  lim     a ( 1 1 a ) 1 1 a    a 0 a 0 a 0 a 0 a 0 a ( 1 1 a ) 1  lim   (because the denominator is always negative); a 0 1 1 a 1 lim r (a) lim   (because the denominator is always positive). a 0  a 0 1 1 a

21. (a) At x  0: lim r (a )  lim 1 a1 a  lim a 0







Therefore, lim r (a) does not exist. a 0

At x  1: lim r (a )  lim 

(c)

a 1



a 1

1 1 a  a

lim

1

a 1 1 1 a

1

(d)

22. f ( x)  x  2 cos x  f (0)  0  2 cos 0  2  0 and f ( )    2 cos( )    2  0. Since f ( x ) is continuous on [ , 0], by the Intermediate Value Theorem, f ( x ) must take on every value between [  2, 2]. Thus there is some number c in [ , 0] such that f (c)  0; i.e., c is a solution to x  2 cos x  0. 23. (a) The function f is bounded on D if f ( x)  M and f ( x)  N for all x in D. This means M  f ( x)  N for all x in D. Choose B to be max {| M |, | N |}. Then | f ( x)|  B. On the other hand, if | f ( x)|  B, then Copyright  2023 Pearson Education, Inc.

118

Chapter 2 Limits and Continuity  B  f ( x)  B  f ( x )   B and f ( x)  B  f ( x) is bounded on D with N  B an upper bound and M   B a lower bound. (b) Assume f ( x)  N for all x and that L  N . Let   L 2N . Since lim f ( x)  L there is a   0 such that x  x0

0  | x  x0 |    | f ( x)  L |    L    f ( x )  L    L  L 2N  f ( x)  L  L 2N  L 2N  f ( x)

 3 L2 N . But L  N  L 2N  N  N  f ( x) contrary to the boundedness assumption f ( x)  N . This contradiction proves L  N . (c) Assume M  f ( x) for all x and that L  M . Let   M2 L . As in part (b), 0  | x  x0 |    L  M2 L  f ( x)  L  M2 L  3L 2 M  f ( x)  M2 L  M , a contradiction.

| a b | a  b a b  2  2  22a  a. 2 |a b| If a  b, then a  b  0  | a  b |  (a  b)  b  a  max {a, b}  a 2 b  2  a 2 b  b 2 a  22b  b. | a b | Let min {a, b}  a 2 b  2 .

24. (a) If a  b, then a  b  0  | a  b |  a  b  max {a, b}  a 2 b  (b) 25.

lim 

x 0

sin(1 cos x ) sin(1 cos x ) x  1 cos x  lim sin(1cos x )  lim 1cos 2 x  lim 1cos x  1cos 1 cos x x 0 1cos x x x x 0 x 0 x (1 cos x )



x  lim sin x  sin x  1  0  0  1  lim x (1sin x 1 cos x 2  cos x ) x 0

26.

27.

28.

x 0

lim sin x  lim sinx x  x  x  1  lim x sin x x 0 sin x x 0  x 0 lim

x 0



1 sin x x

 lim

 x 0



x  1 1  0  0

sin(sin x ) sin(sin x ) sin(sin x )  lim sin x  sinx x  lim sin x  lim sinx x  1 1  1 x x 0 x 0 x 0

sin( x 2  x ) sin( x 2  x ) sin( x 2  x )  lim  ( x  1)  lim  lim ( x  1)  1 1  1 2 2 x x 0 x 0 x  x x 0 x  x x 0

lim

sin( x 2  4) sin( x 2  4) sin( x 2  4)  lim  ( x  2)  lim  lim ( x  2)  1  4  4 2 2 2 x  x 2 x 2 x  4 x2 x 4 x2

29. lim

sin( x 3) sin( x 3) sin( x 3)  lim  1  lim  lim x 9  x 3 x 3 x 3   x 9 x 9 x 9 x 9

30. lim

 

1  1 1  1 6 6 x 3

 

31. lim sin2 x  lim  sinx x  (2 x11)   (1) 11  1  x 0 2 x  x x 0  

x 0

sin r  lim 33. lim tan 2r

r 0

r0

34. lim



lim  cos17 x  sin7 x7 x  1   32  1 1  72   2  3x  sin 7 x   32  x x 0 2 x 2 x cos 7 x 0 

7 x  lim 32. lim 3 x 2tan x

sin(sin  )

 0





2 7



cos 2 r  1 (1) 1  1  sinr r  tan2r2r  12    12  (1) rlim 0    2  1 2

 lim

 0

sin 2 r 2r



sin(sin  ) sin 

    lim sin 

sin(sin  )  lim sinx x  1.  0 sin  x 0



 0

sin (sin  ) . Let x  sin  . Then x  0 as   0 sin 

 lim

Chapter 2 Additional and Advanced Exercises 35.

 4 1  1   tan   2 (4  0  0) tan 2   4 tan   tan   1 lim  lim   (1 0)  4 2   1 5  tan  5      2  2  tan 2  

36.

 1 2   2  (0  2)  cot   1 2 cot 2   lim    52 2  0 5cot  7 cot  8  0  5 cot7   82  (50 0) cot   

119

 

lim

x  lim x sin x  lim 37. lim 2x2sin cos x 2(1  cos x ) x 0

38. lim

 0

x 0

1  cos 

2

 lim

   lim  sin 2   sin 2   1   (1)(1) 1  1  2 2 2  0   2   2  2 



1  sin x x x 0 cos x

x0



2sin 2 2

 0

39. lim tanx x  lim

 xx   x   x  lim  22 2 x  sinx x   lim  2 x  2 x  sinx x   (1)(1)(1)  1 sin sin sin 0 0 2   x   2   x   2  

x sin x 2 x0 2 2sin 2x

tan(tan x ) g ( x )  lim tan x  lim tan   1.   1; let   tan x    0 as x  0  xlim 0 x0  0

Therefore, to make g continuous at the origin, define g (0)  1.

tan(tan x ) tan(tan x ) sin x  1   1  lim sin x ; let   sin x    0 as x  0  40. lim f ( x)  lim sin(sin x )  lim  tan x  sin(sin x ) cos x   x 0 x0 x0 x0 sin(sin x ) sin x  lim   1. Therefore, to make f continuous at the origin, define f (0)  1. lim sin(sin x) sin 

 0

x0

41. Since the highest power of x in the numerator is 1 more than the highest power of x in the denominator, there is 3/ 2 an oblique asymptote. y  2 x  2 x 3  2 x  3 , thus the oblique asymptote is y  2 x. x 1



x 1

 



   x; thus

42. As x  , 1x  0  sin 1x  0  1  sin 1x  1, thus as x  , y  x  x sin 1x  x 1  sin 1x the oblique asymptote is y  x.

43. As x   , x 2  1  x 2  x 2  1  x 2 ; as x  , x 2   x, and as x   , x 2  x; thus the oblique asymptotes are y  x and y   x. 44. As x   , x  2  x  x 2  2 x  x( x  2)  x 2 ; as x  , x 2   x, and as x   , x 2  x; asymptotes are y  x and y   x. 45. Assume 1  a  b and ax  x  x 1b  a ( x  b)  x 2 ( x  b)  x  f ( x)  a( x  b)  x 2 ( x  b)  x  0; f is continuous for all x-values and f (0)  ab  0, f (a  b)  a 2  (a  b)2 a  (a  b)  a2 a  (a  b) 2 a  b  0.    ()

()

Thus, by the Intermediate Value Theorem there is at least one number c, 0  c  a  b, so that f (c)  0  a(c  b)  c 2 (c  b)  c  0  ac  c  c 1b .

46. (a)

lim

x 0

a  bx 1 " a 1"  0 , so x

a  1  0  a  1, then lim

x 0

(1 bx ) 1 1 bx 1 1 bx 1   lim x 1 bx 1 x 0 x ( 1 bx 1)

b  b 2b4 x 0 1bx 1 2

 lim

120

Chapter 2 Limits and Continuity (b)

lim

x 1

tan( ax  a ) b  2 "tan 0 b  2" "b  2" tan( ax  a ) tan a ( x 1)   0 , so b  2  0  b  2, then lim x 1  lim a  a ( x 1) x 1 0 x 1 x 1

 lim cos aa( x 1)  x 1

sin a ( x 1)  cosa 0 1  a  3 a ( x 1)

47.

1/6 4 4 1  lim ( x ) 1 3 1/6 3 x 1 1 x x 1 1 ( x )

48.

lim

lim x

x 0



2/3

1/2

3x 4  x 4  x

lim

x 0



 lim

( x1/6 1)( x1/6 1)(( x1/6 )2 1) 1/6

x 1 (1 x

(3 x  4)  x  4 x

1/6

)(1 ( x

1/6 2

)( x

) )

 lim

x 1

( x1/6 1)( x1/3 1)

 lim 22x  2; assume x  43  x 0





(2)(2) 4  3 3

lim

3x  4  x 4  x

1 x1/6  x1/3

x 0

3x  4  x 4 lim 4xx  4  lim does not exist. x  x 0 x 0



lim

x 0



(3 x  4) (  x )  4 x

CHAPTER 3 DERIVATIVES 3.1

TANGENT LINES AND THE DERIVATIVE AT A POINT

1. P1: m1  1, P2 : m2  5

2. P1: m1  2, P2 : m2  0

3. P1: m1  52 , P2 : m2   12

4. P1: m1  3, P2 : m2  3

 (1 2 h  h 2 ) 1 [4  ( 1 h ) 2 ](4 ( 1)2 )  lim h h h 0 h 0 h (2  h )  lim h  2; at (1, 3): y  3  2( x  ( 1)) h 0

5. m  lim

 y  2 x  5, tangent line

2 [(1 h  1) 2 1]  [(1  1) 2  1]  lim hh  lim h  0; h h 0 h 0 h 0

6. m  lim

at (1,1) : y  1  0( x  1)  y  1, tangent line

7. m  lim 2 1 hh  2 1  lim 2 1hh  2  2 1 h  2

2 1 h  2

h 0

h 0 4(1 h )  4

 lim





h0 2 h 1 h 1

2  1; h 0 1 h 1

 lim

at (1, 2): y  2  1( x  1)  y  x  1, tangent line

8. m  lim

h 0

 lim

( 1 h )2



h 2

 ( 2 h  h )

2 h0 h ( 1 h )

( 1)2

 lim

1( 1 h )2

h0 h ( 1 h )

2  h  2; 2 h0 ( 1 h )

 lim

at (1,1):

y  1  2( x  (1))  y  2 x  3, tangent line

121

122

Chapter 3 Derivatives 2 3 ( 2  h )3 ( 2)3  lim 812 h 6hh  h 8 h h 0 h 0 2

9. m  lim

 lim (12  6h  h )  12; h 0

at (2, 8): y  8  12( x  (2))  y  12 x  16, tangent line

10. m  lim



( 2  h )3

( 2)3

h h 0 (12 h 6 h 2  h3 )

 lim

8 ( 2  h )3

h  0  8 h ( 2  h ) 2  lim 126h  h3 h 0 8( 2  h )

3 h 0 8h ( 2 h ) 3 12  8( 8)   16 ; 3 ( x  ( 2)) at 2,  18 : y   18  16 3 x  1 , tangent line  y   16 2





[(2  h ) 2 1]5 (5 4 h  h 2 ) 5 h (4  h )  lim lim  4; at (2,5): y  5  4( x  2), tangent line h h h 0 h 0 h 0 h

11. m  lim

h ( 3 2 h ) [(1 h )  2(1 h )2 ]( 1) (1 h  2  4 h  2 h 2 ) 1  lim  lim  3; at (1, 1) : y  1  3( x  1), tangent line h h h h 0 h 0 h 0

12. m  lim

3 h

13. m  lim (3 h )h 2

3

h 0

14. m  lim

h 0

(2  h )2

2

 lim

h 0

 lim

(3 h ) 3( h 1)  lim h(h2h1)  2; at (3,3): y  3  2( x  3), tangent line h ( h 1) h 0 8  2(2  h )2

h 0 h (2 h )

 lim

h 0

8 2(4 4 h  h 2 ) h (2  h )2

 lim

2 h (4  h )

h 0 h (2  h )

 48  2; at (2, 2): y  2  2( x  2)

(2 h )3 8 (812 h  6 h 2  h3 ) 8 h (12  6 h  h 2 )  lim  lim  12; at (2,8): y  8  12(t  2), tangent line h h h h 0 h 0 h 0

15. m  lim

[(1 h )3 3(1 h )] 4 (1 3h 3h 2  h3 33h )  4 h (6 3h  h 2 )  lim  lim  6; at (1, 4): y  4  6(t  1), tangent line h h h h 0 h 0 h 0

16. m  lim 17. m  lim

h 0

(4  h )  4 4 h  2  lim 4hh  2  4 h  2  lim h 4  h  2 h 0 h 4  h  2 h 0





 lim

h 0 h

 4 h  2 



1  1 ; at (4, 2): 4 2 4

y  2  14 ( x  4), tangent line (8 h ) 1 3 (9 h ) 9  lim 9hh 3  9 h 3  lim  lim h 9  h  3 h 0 h 9  h  3 h 0 h 0 h h 0

18. m  lim





 9  h  3



1  1 ; at (8,3): 9 3 6

y  3  16 ( x  8), tangent line 2 5(1 h ) 3(1 h )2  2 h ( 13h )2  lim  h h3h  lim  1 h h h 0 h0 h 0

19. At x  1, y  2  m  lim

Section 3.1 Tangent Lines and the Derivative at a Point

123

[( 2  h )3  2( 2 h ) 7]3 h ( h 2 6h 10)  lim  10 h h h 0 h0

20. At x  2, y  3  m  lim

1

2 (2  h )

h   14 , slope 21. At x  3, y  12  m  lim (3 h )h1 2  lim 2h (2 h)  lim 2h(2  h) h 0

h 0

h 1  ( 1)

22. At x  0, y  1  m  lim h 1 h h 0

 lim

h 0

( h 1)  ( h 1)  lim h (2hh1)  2 h ( h 1) h 0

23. (a) It is the rate of change of the number of cells when t  5. The units are the number of cells per hour. (b) P (3) because the slope of the curve is greater there. (c)

6.10(5  h ) 2  9.28(5  h )  16.43  [6.10(5) 2  9.28(5)  16.43] 61.0h  6.10h 2  9.28h  lim  h 0 h 0 h h lim 51.72  6.10h  51.72  52 cells/hr

P (5)  lim h 0

24. (a) From t  0 to t  3, the derivative is positive. (b) At t  3, the derivative appears to be 0. From t  2 to t  3, the derivative is positive but decreasing. [( x  h )2  4( x  h) 1]( x 2  4 x 1) h h 0

25. At a horizontal tangent line the slope m  0  0  m  lim

( x 2  2 xh  h2  4 x  4 h 1) ( x 2  4 x 1) (2 xh  h 2  4 h)  lim  lim (2 x  h  4)  2 x  4; 2 x  4  0 h h h 0 h 0 h 0

 lim

 x  2. Then f (2)  4  8  1  5  (2, 5) is the point on the graph where there is a horizontal tangent line. 2 2 3 [( x  h )3 3( x  h )]( x3 3 x ) ( x3  3 x 2 h  3 xh 2  h3 3 x 3h )  ( x3 3 x )  lim 3 x h 3 xhh  h 3h   lim h h h 0 h 0 h 0

26. 0  m  lim

lim (3x 2  3xh  h 2  3)  3 x 2  3; 3 x 2  3  0  x  1 or x  1. Then f (1)  2 and f (1)  2  (1, 2)

h0

and (1, 2) are the points on the graph where a horizontal tangent line exists. 1

 1

( x 1) ( x  h 1)

27. 1  m  lim ( x  h ) h1 x 1  lim h ( x 1)( x  h 1)  lim h( x 1)( hx  h 1)   h 0

h 0

1  ( x  1) 2  1  x 2  2 x  0 ( x 1)2

 x ( x  2)  0  x  0 or x  2. If x  0, then y  1 and m  1  y  1  ( x  0)  ( x  1). If x  2, then y  1 and m  1  y  1  ( x  2)  ( x  3).

28.

( x  h)  x h 1  m  lim x  h  x  lim x  h  x  x  h  x  lim  lim  1 . h h 4 2 x x  h  x h 0 h x  h  x h 0 h 0 h 0 h x  h  x x 1 1 1  x  2  x  4  y  2. The tangent line is y  2  4 ( x  4)  4  1. Thus, 4  2 x









f (2  h )  f (2) (100 4.9(2  h ) 2 )  (100 4.9(2) 2 ) 4.9(4 4 h  h 2 )  4.9(4)  lim  lim  lim (19.6  4.9h)  19.6. h h h h 0 h 0 h 0 h 0

29. lim

The minus sign indicates the object is falling downward at a speed of 19.6 m/sec.

f (10  h )  f (10) 3(10  h )2 3(10) 2  lim h h h 0 h 0

30. lim

3(20 h  h 2 )  60 ft/sec h h 0

 lim

f (3 h )  f (3)  (3 h)2  (3)2 )  [9 6 h  h2 9]  lim  lim  lim  (6  h)  6 h h h h 0 h 0 h 0 h 0

31. lim

124

Chapter 3 Derivatives 4

4

(2  h )  3 (2) f (2  h )  f (2)  lim 3 h h h 0 h 0

32. lim

4

 lim 3 h 0

[12 h  6 h 2  h3 ]  lim 43 [12  6h  h 2 ]  16 h h 0

33. At ( x0 , mx0  b) the slope of the tangent line is lim

h 0

( m ( x0  h ) b ) ( mx0 b ) mh  lim h  lim m  m. ( x0  h )  x0 h 0 h 0

The equation of the tangent line is y  (mx0  b)  m( x  x0 )  y  mx  b.





 41 h  12 2 4 h  2 4  h  lim  lim  2 4 h  2 4 h   h    lim h  2 4 4 h 2 h 4  h h 0 h 0  2 h 4  h 2  4  h  h 0   h0       4 (4 h ) h 1 1 1   16  lim   lim   lim     2 4  2 4  h 0  2 h 4  h  2 4 h   h 0  2 h 4  h  2 4 h   h 0  2 4  h  2  4  h  

34. At x  4, y  1  12 and m  lim

1 1 4 h 2

 1   lim h sin 1  0  yes, f ( x) does have a tangent line at the

h sin h f (0  h )  f (0)  lim h h h 0 h 0

35. Slope at origin  lim

h

h 0

origin with slope 0.

 1   lim sin 1 . Since lim sin 1 does not exist, f ( x) has no tangent line at the origin.

h sin g (0 h )  g (0)  lim h h h h 0 h 0

36. lim

h 0

h0

37. No. This graph has no tangent line at x = 0 since the graph is not continuous at x = 0. 38.

lim

h 0



U (0  h ) U (0)  h

lim 0h1  , and lim

h 0



h 0



U (0  h ) U (0)  h

lim 1h1  0  no, the graph of f does not have a

h 0 

vertical tangent line at (0, 1) because the limit does not exist.

39. (a) The graph appears to have a cusp at x  0.

(b)

lim

h 0



2/5 f (0  h )  f (0) 1   and lim 1    limit does not exist  the graph  lim h h 0  lim 3/5 3/5 h  h 0 h 0  h h 0  h 2/5

of y  x

does not have a vertical tangent line at x  0.

40. (a) The graph appears to have a cusp at x  0.

(b)

lim

h 0



f (0  h )  f (0)  h

lim

h 0



h 4/5 0  h

1    and lim 1    limit does not exist  y  x 4/5 1/5 1/5 h h 0 h 0  h

lim



does not have a vertical tangent line at x  0.

Section 3.1 Tangent Lines and the Derivative at a Point

125

41. (a) The graph appears to have a vertical tangent line at x  0.

1/5 f (0  h )  f (0) 1    y  x1/5 has a vertical tangent line at x  0.  lim h h0  lim 4/5 h h 0 h 0 h

(b) lim

h 0

42. (a) The graph appears to have a vertical tangent line at x  0.

h3/5  0 f (0  h )  f (0)  lim  lim 21 5    the graph of y  x3/5 has a vertical tangent line at x  0. h h h 0 h 0 h 0 h

(b) lim

43. (a) The graph appears to have a cusp at x  0.

(b)

lim

h 0



f (0 h )  f (0)  h

lim



h 0 2/5

 the graph of y  4 x

4 h 2/5  2 h  h

lim

h 0 



4 2 h3/5

   and lim  h 0 

4 2 h3/5

    limit does not exist

 2 x does not have a vertical tangent line at x  0.

44. (a) The graph appears to have a cusp at x  0.

(b)

5/3 2/3 f (0 h )  f (0)  lim h h5h  lim h h 0 h 0 h 0

lim

 h    0  lim 2/3

5 h1/3

1/3

h 0 h

does not exist  the graph of

y  x5/3  5 x 2/3 does not have a vertical tangent line at x  0.

126

Chapter 3 Derivatives

45. (a) The graph appears to have a vertical tangent line at x  1 and a cusp at x  0.

(b) x  1:

(1 h )2/3  (1 h 1)1/3 1 (1 h ) 2/3  h1/3 1  lim    y  x 2/3  ( x  1)1/3 has a vertical tangent line h h h 0 h 0

lim

at x  1; x  0:

f (0  h )  f (0) h 2/3 ( h 1)1/3 ( 1)1/3  1 ( h 1)1/3 1   lim  lim  h  h  does not exist 1/3 h h h 0 h 0 h 0  h  2/3 1/3

lim

yx

 ( x  1)

does not have a vertical tangent line at x  0.

46. (a) The graph appears to have vertical tangent lines at x  0 and x  1.

(b) x  0: x  0; x  1: x  1.

f (0  h )  f (0) h1/3  ( h 1)1/3  ( 1)1/3  lim    y  x1/3  ( x  1)1/3 has a vertical tangent line at h h h 0 h 0

lim

f (1 h )  f (1) (1 h )1/3  (1 h 1)1/3 1  lim    y  x1/3  ( x  1)1/3 has a vertical tangent line at h h h 0 h 0

lim

47. (a) The graph appears to have a vertical tangent line at x  0.

(b)

 |h| 0  |h| f (0  h )  f (0) f (0  h )  f (0)  lim hh0  lim 1  ; lim  lim  lim |h| h h h     h h 0 x 0 h 0 h 0 h 0 h 0  lim 1    y has a vertical tangent line at x  0. h 0 |h|

lim



Section 3.2 The Derivative as a Function 48. (a) The graph appears to have a cusp at x  4.

(b)

49–52.

|4 (4  h )| 0 |4 (4  h )| |h| f (4  h )  f (4) lim  lim h  lim 1  ; lim  lim h h h h 0  h 0  h 0  h 0  h 0  h |h|  lim | h |  lim 1    y  4  x does not have a vertical tangent line at x  4. h 0  | h | h 0 

lim

h 0 

f (4  h )  f (4)  h

Example CAS commands:

Maple: f : x - x^3  2*x;x0 : 0; plot( f (x), x  x0-1/2..x0  3, color  black,

# part (a)

title "Section 3.1, #49(a)" ); q : unapply( (f (x0  h)-f (x0))/h, h );

# part (b)

L : limit( q(h), h  0 ); sec_lines : seq( f(x0)  q(h)*(x-x0), h 1..3 ); tan_ line : f(x0)  L*(x-x0); plot( [f(x),tan_line,sec_lines], x  x0-1/2..x0  3, color  black,

# part (c) # part (d)

linestyle [1,2,5,6,7], title "Section 3.1, #49(d)", legend ["y  f(x)","Tangent line at x  0","Secant line (h 1)", "Secant line (h  2)","Secant line (h 3)"] );

Mathematica: (function and value for x0 may change) Clear[f , m, x, h] x0  p; f[x_ ]:  Cos[x]  4Sin[2x] Plot[f [x],{x, x0  1, x0  3}] dq[h_ ]:  (f [x0  h]  f [x0])/h m  Limit[dq[h], h  0] ytan:  f [x0]  m(x  x0) y1:  f [x0]  dq[1](x  x0) y2:  f [x0]  dq[2](x  x0) y3:  f [x0]  dq[3](x  x0) Plot[{f [x], ytan, y1, y2, y3}, {x, x0  1, x0  3}] 3.2

THE DERIVATIVE AS A FUNCTION

1. Step 1: Step 2: Step 3:

f ( x)  4  x 2 and f ( x  h)  4  ( x  h) 2

2 f ( x  h )  f ( x ) [4 ( x  h )2 ](4  x 2 ) (4  x 2  2 xh  h 2 )  4  x 2 h ( 2 x  h )    2 xhh h   2 x  h h h h h

f ( x)  lim (2 x  h)  2 x; f (3)  6, f (0)  0, f (1)  2 h 0

127

128

Chapter 3 Derivatives [( x  h 1)2 1][( x 1)2 1] h h 0

2. F ( x)  ( x  1)2  1 and F ( x  h)  ( x  h  1)2  1  F ( x )  lim

2 ( x 2  2 xh  h 2  2 x  2 h 11) ( x 2  2 x 11)  lim 2 xh  hh  2h  lim (2 x  h  2)  2( x  1); h h0 h 0 h 0

 lim

F (1)  4, F (0)  2, F (2)  2 g (t )  12 and g (t  h) 

1 (t  h ) 2  t 2  ( t  h )2  1 1  ( t  h )2 t 2  2 2 2 g (t  h )  g ( t ) t 2 (t 2  2th  h 2 ) h ( 2t  h )  (t  h )h t   h    2th 2 h2   2t 2h2 2 2 h (t  h ) t h (t  h ) 2 t 2 h (t  h ) t h (t  h ) t g (t )  lim 2t 2h 2  2 2t2  32 ; g (1)  2, g (2)   14 , g  3   2 3 3 t t t h 0 ( t  h ) t

3. Step 1:

Step 2:

 

Step 3:





1 ( z  h )

 12zz 2 2( z  h ) 1( z  h ) (1 z  h ) z (1 z )( z  h ) zh  z  h  z 2  zh k ( z )  12zz and k ( z  h)  2( z  h)  k ( z )  lim  lim  lim z  z 2( z  h ) zh h 2( z  h ) zh h 0 h 0 h 0  h  1  1 1 1 1     lim 2( z  h) zh  lim 2( z  h ) z  2 ; k (1)   2 , k (1)   2 , k 2 4 2z h 0 h 0

 

p ( )  3 and p (  h)  3(  h)

5. Step 1:

3(  h )  3  h

p (  h )  p ( )  h

Step 2:



3 3  3h  3



 3 3h  3 

3  h0 3 3h  3

p ( )  lim

Step 3:

 3 3h  3    3 3h  3   (3 3h)3 h  3 3h  3  h 3 3h  3 

   2 32

3  3 ; p (1)  3 , p (3)  12 , p  32 3  3 2 3 2 3 2 s  2 h 1  2 s 1 h

6. r ( s )  2 s  1 and r ( s  h)  2( s  h)  1  r ( s )  lim

h 0

 2s  h1 2 s 1   2s  2h1 2s 1   lim (2 s  2h1)(2s 1) h h 0  2s  2h1 2s 1  h0 h 2s  2h1 2s 1 

 lim

h 0 h

 2s  2h1 2s 1 

2  h 0 2 s  2 h 1  2 s 1

2  2  2 s 1  2 s 1 2 2 s 1

 lim

 

1 ; 2 s 1

r (0)  1, r (1)  1 , r  12  1 3

2( x  h )3  2 x3 2( x3 3 x 2 h 3 xh 2  h3 )  2 x3  lim h h h 0 h 0

7. y  f ( x)  2 x3 and f ( x  h)  2( x  h)3  dx  lim 2

h (6 x 2  6 xh  2 h 2 )  lim (6 x 2  6 xh  2h 2 )  6 x 2 h h 0 h 0

 lim 6 x h  6hxh  2 h  lim h 0

3 2 2 3 2 2 3 2 (( s  h )3  2( s  h )2 3) ( s 3  2 s 2 3)  lim s 3s h 3sh  h  2 s h 4 sh  2h 3 s  2 s 3 h h 0 h 0

 lim 8. r  s 3  2 s 2  3  dr ds 2

h (3s 2 3sh  h 2  4 s  2 h )  lim (3s 2  3sh  h 2  4 s  2h)  3s 2  4s h h 0 h 0

 lim 3s h 3sh hh  4 sh  2h  lim h 0

9. s  r (t )  2tt1 and r (t  h)  2(tthh) 1  ds  lim dt



    lim 

t h  2tt1 2( t  h ) 1

( t  h )(2 t 1)  t (2t  2 h 1) (2t  2 h 1)(2 t 1)



h 0 h 0 (t  h )(2t 1) t (2t  2 h 1) h 2t 2 t  2 ht  h  2t 2  2 ht t  lim 1  lim  lim (2t  2 h 1)(2t 1) h (2t  2 h 1)(2t 1) h h 0 h 0 (2t  2 h 1)(2t 1) h h 0 (2t  2 h 1)(2t 1) h 0 1  1 2  (2t 1)(2 t 1) (2t1)

 lim

Section 3.2 The Derivative as a Function



h (t  h )t t  (t  h ) (t  h )t

10.

 (t  h )  1   (t  1 ) h 1 1 t h  t dv  lim   lim t h h t  lim h dt h 0 h 0 h0

11.

dp ( q  h )3/2  q 3/2 ( q  h )( q  h )1/2  qq1/2  lim  lim  lim dq h0 h h h 0 h 0



1/2



1/2



  lim ht h t h  lim t ht 1  t 1  1  1 2

q [( q  h )1/2  q1/2 ] h



1/2

 h ( qhh )



h 0 (t  h )t

h 0 h (t  h )t

129







1/2

][( q  h )  q ] lim q[( qh )h[(qqh )1/2  ( q  h )1/2  lim h[(qq[(qh)1/2h )qq1/2] ]  ( q  h )1/2  lim ( qh )1/2q q1/2  ( q  h )1/2  q2  q1/2  23 q1/2  q1/2 ] h 0

1 ( w h )2 1

dz  lim dw h0

12.



w2 1  ( w h )2 1

 lim

w2 1 ( w2  2 wh  h 2 1)

lim

h0 h ( w h )2 1 w2 1



w2 1  ( w h )2 1



 lim



2 h ( x 2  xh 9) 9 ;  xx (xxh x( x  h)h  h)

k ( x  h)  k ( x )  lim h h 0 h 0 h 1 1 ; k (2)   1  lim h (2 x )(2  lim   x  h ) h 0 (2  x )(2  x  h ) 16 (2  x ) 2 h 0

14. k ( x)  21 x and k ( x  h)  2 ( 1x  h)  k ( x)  lim





w ( w2 1)3/2

9  x 9  1  9 ; f ( x)  lim xx (xxh 2 2  h) x

h 0

m  f (3)  0

( x  h )  9    x  9  x ( x h)   f ( x h) f ( x) x ( x  h )2 9 x  x 2 ( x  h ) 9( x  h )    h h x( x  h)h

9 h   x  2 x h  xh x( 9x x h)xh  x h 9 x 9h  x xh(xxh  h) h

15.

2 w  h h0 h ( w  h )2 1 w2 1 w2 1  ( w h )2 1

 lim

13. f ( x )  x  9x and f ( x  h)  ( x  h)  ( x 9 h)  3

h 0

 w 1  ( wh) 1 w 1  ( wh) 1  h0 h ( w h ) 1 w 1 h 0 h ( w h ) 1 w 1 w 1  ( w h ) 1 

1 w2 1

h 0

 2 1x  h  21 x   lim (2 x)(2 x  h)

ds  lim [(t  h ) (t  h ) ](t t )  lim (t 3t h 3th  h ) (t  2th  h ) t t h dt h h 0 h 0

h 0 h (2 x )(2  x  h )

 lim 3t h 3th hh  2th  h

h 0

h (3t 2  3th  h 2  2t  h )  lim (3t 2  3th  h 2  2t  h)  3t 2  2t ; m  ds 5 h dt t1 h 0 h 0

 lim

( x  h)3

16.

x 3

 dy  lim 1 ( x  hh) 1 x  lim dx h0 h 0

 lim (1 x  h4)(1 x )  h 0

17. f ( x) 

( x  h  3)(1 x )  ( x  3)(1 x  h ) (1 x  h )(1 x )

4 ; (1 x )2

8 and f ( x  h)  x2

4h 3 x  x 3 x 3 x  xh 3h  lim  lim x  h 3 x  xh h (1 x  h )(1 x ) h (1 x  h )(1 x ) h 0

h 0

dy  4 2  94 dx x2 (3) f ( x  h)  f ( x ) 8   h ( x  h) 2

8  8 x 2 ( x  h)2



 x2  x  h2    x2  x  h2  h x  h2 x 2  x 2  x  h2 

8[( x  2) ( x  h  2)] 8h 8   f ( x)  lim h x  h2 x 2 x 2  x  h2 h x  h2 x2 x 2  x  h2 h 0 x  h  2 x  2 x  2  x  h  2 8 4   ; m  f (6)  4   12  the equation of the tangent line at (6, 4) is ( x  2) x  2 4 4 x2 x 2 x 2  x 2 1 1 1 y  4   2 ( x  6)  y   2 x  3  4  y   2 x  7.





18. g ( z )  lim

h0











(1 4  ( z  h ) )  1 4  z

h

  lim  4 z h  4 z    4 z h  4 z   lim (4 z h)(4 z ) h h 0  4  z  h  4  z  h 0 h  4  z  h  4  z 

1 ; m  g (3)  1   1  the equation 2 2 4 z 2 4 3 3 1 1 1 of the tangent line at (3, 2) is w  2   2 ( z  3)  w   2 z  2  2  w   2 z  72 .

 lim

h 0 h

 4 z  h  4  z 

 lim

h 0

1

 4 z  h  4 z 





130

Chapter 3 Derivatives f (t  h )  f (t ) h

19. s  f (t )  1  3t 2 and f (t  h)  1  3(t  h)2  1  3t 2  6th  3h 2  ds  lim dt

h 0

(13t 2 6th 3h 2 ) (13t 2 )  lim (6t  3h)  6t  ds 6 h dt t 1 h 0 h 0

 lim



    lim 1x  x 1 h

1 x 1 h  1 1x f ( x  h) f ( x)  lim h h h 0 h 0

20. y  f ( x)  1  1x and f ( x  h)  1  x 1 h  dx  lim

h 0

dy  lim x ( x h h) h  lim x ( x1 h)  12  dx  13 x h 0 h 0 x 3

2  2 f (  h )  f ( ) 4   h 4  2 and f (  h)  2  lim  ddr  lim  lim 2 4  2 4  h h h 4 4 (  h ) h 0 h 0 h 0 h 4  4  h 4(4  )  4(4   h ) 2 4   2 4  h 2 4   2 4   h  lim   lim 2 4   2 4   h h 0 2 h 4  4   h 4   4   h h0 h 4  4  h

21. r  f ( ) 

 lim

h 0 4  4   h

 

 

 4  4 h 



2 (4  ) 2 4



22. w  f ( z )  z  z and f ( z  h)  ( z  h)    lim h  z hh  z  lim 1  h 0 h 0  dw 5  dz z 4  4

zh  z  h







1  ddr 1 (4  ) 4  0 8







z  h  z  h ( z  z ) f ( z  h) f ( z )  lim h h h 0 h 0

z  h  dw  lim dz

 z  h  z    1  lim ( z  h) z  1  lim 1  1  1  2 z h 0 h z  h  z  h 0 z  h  z  z  h  z  

1  1 f ( z ) f ( x) ( x  2) ( z  2) z2 x2 z 1  lim  lim ( z  x )( z  2)( x  2)  lim ( z  x )( zx2)(  lim  1 2 zx x  2) z  x ( z  2)( x  2) ( x  2) zx z x zx zx zx

23. f ( x)  lim

2 2 2 2 f ( z ) f ( x) ( z  x )( z  x ) 3( z  x ) ( z 2 3 z  4) ( x 2 3 x  4)  lim z 3 zz  xx 3 x  lim z  x z 3x z 3 x  lim  lim z  x zx z  x zx zx zx zx zx ( z  x ) ( z  x ) 3  lim  lim ( z  x)  3  2 x  3 zx zx zx

24. f ( x)  lim





 g ( z ) g ( x) z ( x 1)  x ( z 1) x 1  lim  1 2  lim z z1  xx 1  lim ( z  x )( z 1)( x 1)  lim ( z  x )(zz1)( zx x 1) z  x ( z 1)( x 1) ( x 1) zx zx zx zx

25. g ( x)  lim

g ( z ) g ( x) (1 z ) (1 x )  lim  lim zx z  x zx zx zx

26. g ( x)  lim

z x zx  z  x  lim  lim zx z x zx z  x ( z  x )( z  x )

1  1 z x 2 x

27. Note that as x increases, the slope of the tangent line to the curve is first negative, then zero (when x  0), then positive  the slope is always increasing which matches (b). 28. Note that the slope of the tangent line is never negative. For x negative, f 2 ( x) is positive but decreasing as x increases. When x  0, the slope of the tangent line to x is 0. For x  0, f 2 ( x) is positive and increasing. This graph matches (a). 29. f3 ( x) is an oscillating function like the cosine. Everywhere that the graph of f3 has a horizontal tangent line we expect f3 to be zero, and (d) matches this condition. 30. The graph matches with (c).

Section 3.2 The Derivative as a Function

131

31. (a) f  is not defined at x  0, 1, 4. At these points, the left-hand and right-hand derivatives do not agree. For f ( x )  f (0) f ( x )  f (0)  slope of line joining (4, 0) and (0, 2)  12 but lim  slope of line example, lim x 0 x 0 x 0 

x 0  f ( x )  f (0) joining (0, 2) and (1,  2)  4. Since these values are not equal, f (0)  lim does not exist. x 0 x 0

(b)

(b) Shift the graph in (a) down 3 units

32. (a)

33.

y’ 2 1 6

1 2 3 4 5

34. (a)

(b) The fastest is between the 20th and 30th days; slowest is between the 40th and 50th days.

35. Answers may vary. In each case, draw a tangent line and estimate its slope. ii) slope  2.86  dT (a) i) slope  1.54  dT  1.54 °F  2.86 °F dt dt hr hr  0 °F iii) slope  0  dT dt hr

iv) slope  3.75  dT  3.75 °F dt hr

(b) The tangent line with the steepest positive slope appears to occur at t  6  12 p.m. and slope  7.27  dT  7.27 °F . The tangent line with the steepest negative slope appears to occur at t  12  6 p.m. and dt hr  8.00 °F slope  8.00  dT dt hr

132

Chapter 3 Derivatives (c)

36. (a) Decrease: 2006–2012, increase: 2012–2015 (b) i) $300,000 ii) $190,000 iii) $280,000 (c) i) $35,000/yr ii) $0/yr iii) $8,000/yr (d) During 2008 at $90,000/yr (e) During 2013 at $68,500/yr (f)



f (0  h )  f (0) h



f (0  h )  f (0) h

37. Left-hand derivative: For h  0, f (0  h)  f (h)  h 2 (using y  x 2 curve)  lim h 0

2  lim h h0  lim h  0  h0 h 0 

Right-hand derivative: For h  0, f (0  h)  f (h)  h (using y  x curve)  lim f (0  h )  f (0)  h 

 lim h h 0  lim 1  1; Then lim h 0 

h 0 

h 0

h 0 f (0  h )  f (0)  the derivative f (0) does not exist. h 

lim

h 0

f (1 h )  f (1)  h

lim 2h 2  lim 0  0 h 0  h 0 f (1 h )  f (1) (2  2 h )  2 Right-hand derivative: When h  0, 1  h  1  f (1  h)  2(1  h)  2  2h  lim  lim h h h 0  h 0   lim 2hh  lim 2  2 h 0 h 0 f (1 h )  f (1) f (1 h )  f (1) Then lim  lim  the derivative f (1) does not exist. h h   h 0 h 0

38. Left-hand derivative: When h  0, 1  h  1  f (1  h)  2  lim

h 0

39. Left-hand derivative: When h  0,1  h  1  f (1  h)  1  h  lim

 1 h 1   1 h 1  lim (1 h)1  lim 1  1 h  1 h 1 h0 h 1 h 1 h0 1 h 1 2 h 0

h 0 

f (1 h )  f (1)  h

 lim



lim

h 0 

1 h 1 h

Section 3.2 The Derivative as a Function f (1 h )  f (1) h

Right-hand derivative: When h  0,1  h  1  f (1  h)  2(1  h)  1  2h  1  lim (2 h 1) 1  h 

 lim

h 0



lim 2  2

h 0 f (1 h )  f (1) f (1 h )  f (1) Then lim  lim  the derivative f (1) does not exist. h h h 0  h 0  h 0

f (1 h )  f (1)  h

40. Left-hand derivative: lim

h 0 

lim

h 0 

(1 h ) 1  h





lim 1  1

h 0



1 (1 h )



1 1 1 h f (1 h )  f (1) 1 h Right-hand derivative: lim  lim  lim  lim h(1hh ) h h h    h 0 h 0 h 0 h 0  f (1 h )  f (1) f (1 h )  f (1) Then lim  lim  the derivative f (1) does not exist. h h h 0  h 0 

 lim 11h  1 h 0 

41. f is not continuous at x  0 since lim f ( x)  does not exist and f (0)  1 x 0

g ( h )  g (0)  h

42. Left-hand derivative: lim h 0



Then lim

h 0 

g ( h )  g (0)  h



43. Left-hand derivative: lim

f ( h )  f (0) h

Right-hand derivative: lim

f ( h )  f (0) h

h 0



h 0 h

2/3

lim h h 0  lim

h 0



 

1/3

h 0 h

 

g ( h )  g (0)    the derivative g (0) does not exist. h

lim

h 0



h 0

g ( h )  g (0)  h

Right-hand derivative: lim h 0

1/3

lim h h0  lim

h 0

 lim hh  lim h  0 h 0



h 0



 lim 2 h htan h  lim 2  sinh h  cos1 h h 0 

h 0

  2  (1)  (1)  3

Then the derivative f (0) does not exist.

h 0

h  h11  ( 1) h 

g ( h )  g (0) h 

 lim

g ( h )  g (0) h0

 lim

44. Left-hand derivative: lim

h 0

h ( h  1)  1  ( h  1) h 1

 lim 1h  h 0



 lim 1h  h h  21h h0



h ( h  2)  lim 1h  h  1  2 h 0

Right-hand derivative: lim h 0



h 0



2 h  h3  1  ( 1) h

 lim h 0



h (2  h 2 ) h

 2 and g is continuous at

x  0 since lim g ( x)  g (0)  1 h 0

Then the derivative g (0)  2. 45. (a) The function is differentiable on its domain 3  x  2 (it is smooth). (b) none (c) none 46. (a) The function is differentiable on its domain 2  x  3 (it is smooth). (b) none (c) none 47. (a) The function is differentiable on 3  x  0 and 0  x  3. (b) none (c) The function is neither continuous nor differentiable at x  0 since lim f ( x)  lim f ( x). h 0 

h 0 

133

134

Chapter 3 Derivatives

48. (a) f is differentiable on 2  x  1, 1  x  0, 0  x  2, and 2  x  3 (b) f is continuous but not differentiable at x  1: lim f ( x)  0 exists but there is a corner at x  1 since f ( 1 h )  f ( 1)  3 and h 

lim

h 0

x 1 f ( 1 h )  f ( 1)  3  f (1) does not exist h 

lim

h 0

(c) f is neither continuous nor differentiable at x  0 and x  2: at x  0, lim f ( x)  3 but lim f ( x)  0  lim f ( x ) does not exist; x 0 

x 0 

x 2

x 0

at x  2, lim f ( x ) exists but lim f ( x )  f (2)

49. (a) f is differentiable on 1  x  0 and 0  x  2 (b) f is continuous but not differentiable at x  0: lim f ( x)  0 exists but there is a cusp at x  0, f (0  h )  f (0) does not exist h h 0

x 0

so f (0)  lim (c) None

50. (a) f is differentiable on 3  x  2, 2  x  2, and 2  x  3 (b) f is continuous but not differentiable at x  2 and x  2: there are corners at those points (c) None 2 2 2 f ( x  h) f ( x) ( x  h)2 (  x 2 )  lim  lim  x  2 xhh h  x  lim (2 x  h)  2 x h h h 0 h 0 h 0 h 0

51. (a) f ( x)  lim (b)

(c) y   2 x is positive for x  0, y  is zero when x  0, y  is negative when x  0 (d) y   x 2 is increasing for   x  0 and decreasing for 0  x  ; the function is increasing on intervals where y   0 and decreasing on intervals where y   0 f ( x  h)  f ( x )  lim h h 0 h 0

52. (a) f ( x)  lim (b)

 x1h  x1   lim  x( x  h)  lim h

h 0 x ( x  h ) h

1  12 x h 0 x ( x  h )

(c) y  is positive for all x  0, y  is never 0, y  is never negative (d) y   1x is increasing for   x  0 and 0  x  

53. (a)

 z 3  x3   3 3  3 3 f ( z ) f ( x)    lim  lim 3(z zxx ) Using the alternate formula for calculating derivatives: f ( x)  lim zx zx zx zx zx 2 ( z  x )( z 2  zx  x 2 )  x 2  x 2  f ( x )  x 2  lim z  zx 3 3( ) z  x zx zx

 lim

Section 3.2 The Derivative as a Function

135

(b)

(c) y  is positive for all x  0, and y   0 when x  0; y  is never negative 3 (d) y  x3 is increasing for all x  0 (the graph is horizontal at x  0 ) because y is increasing where y   0; y is never decreasing

54. (a)

 z 4  x4   4 4  f ( z ) f ( x)   Using the alternate form for calculating derivatives: f ( x)  lim  lim zx zx zx zx 4

3 2 2 3 ( z  x )( z 3  xz 2  x 2 z  x3 )  lim z  xz 4 x z  x  x3  f ( x)  x3 4( ) z  x zx zx

z  x  lim  lim 4( z  x) zx

(b)

(c) y  is positive for x  0, y  is zero for x  0, y  is negative for x  0 4 (d) y  x4 is increasing on 0  x   and decreasing on   x  0 2 2 2 2 (2( x  h )2 13( x  h ) 5) (2 x 2 13 x 5)  lim 2 x  4 xh  2h 13 xh13h 5 2 x 13 x 5  lim 4 xh  2hh 13h h h 0 h 0 h 0

55. y   lim

 lim (4 x  2h  13)  4 x  13, slope at x. The slope is 1 when 4 x  13  1  4 x  12  x  3 h 0

 y  2  32  13  3  5  16. Thus the tangent line is y  16  (1)( x  3)  y   x  13 and the point of tangency is (3, 16).

 x  h  x    x  h  x   lim ( x  h) x  lim 1  1 . h h 0  x  h  x  h0  x  h  x h h0 x  h  x 2 x

56. For the curve y  x , we have y   lim





Suppose a, a is the point of tangency of such a line and (1, 0) is the point on the line where it crosses the 0 x-axis. Then the slope of the line is a a( 1)  a a1 which must also equal

1 ; using the derivative formula at 2 a

x  a  a a1 

1  2a  a  1  a  1. Thus such a line does exist: its point of tangency is (1, 1), its slope is 2 a 1  1 ; and an equation of the line is y  1  1 ( x  1)  y  1 x  1 . 2 2 2 2 2 a

57. Yes; the derivative of  f is  f  so that f ( x0 ) exists   f ( x0 ) exists as well. 58. Yes; the derivative of 3g is 3 g  so that g (7) exists  3 g (7) exists as well. g (t )

59. Yes, lim h(t ) can exist but it need not equal zero. For example, let g (t )  mt and h(t )  t. Then g (0)  h(0)  0, t0 g (t )

but lim h (t )  lim mt  lim m  m, which need not be zero. t t0

t0

t 0

136

Chapter 3 Derivatives

60. (a) Suppose | f ( x)|  x 2 for 1  x  1. Then | f (0)|  02  f (0)  0. Then f (0)  lim

h 0

 lim

h 0

f (0 h )  f (0) h

f (h) f ( h) f (h)0 f ( h)  lim h . For | h |  1, h 2  f (h)  h 2   h  h  h  f (0)  lim h  0 by the h h 0 h0

Sandwich Theorem for limits. (b) Note that for x  0, | f ( x)| | x 2 sin 1x | | x 2 ||sin 1x |  | x 2 | 1  x 2 (since  1  sin x  1). By part (a), f is differentiable at x  0 and f (0)  0.

1 is the derivative of the function y  x so 2 x xh  x gets closer to y  1 as h gets smaller and h 2 x

61. The graphs are shown below for h  1, 0.5, 0.1 The function y  that

1  lim x  h  x . The graphs reveal that y  h 2 x h 0

smaller.

62. The graphs are shown below for h  2,1, 0.5. The function y  3 x 2 is the derivative of the function y  x3 so that ( x  h )3  x 3 ( x  h )3  x 3 . The graphs reveal that y  gets closer to y  3 x 2 as h gets smaller and smaller. h h h 0

3 x 2  lim

63. The graphs are the same. So we know that for | x| f ( x)  | x |, we have f ( x)  x .