CHAPTER 1 ADDITIONAL AND ADVANCED EXERCISES
2.6 CONTINUITY
59. yœtan xx"# 60. yœ "x tan x
61. yœ x 62. yœ
4 x 4 x
È È"
# #
63. yœx#Î$x""Î$ 64. yœsinˆx#11‰
3. Continuous on [ 1 3] ß
4. No, discontinuous at xœ1, 1.5œ lim k(x)Á lim k(x)œ !
xÄ "c xÄ "b
5. (a) Yes (b) Yes, lim f(x) 0
xÄ "b œ
(c) Yes (d) Yes
6. (a) Yes, f(1)œ1 (b) Yes, lim f(x)œ2
xÄ1
(c) No (d) No
7. (a) No (b) No
8. ["ß ! !ß " "ß # #ß $) ( ) ( ) ( )
9. f(2)œ0, since lim f(x)œ 2(2) œ œ4 0 lim f(x)
xÄ #c xÄ #b
10. f(1) should be changed to 2œ lim f(x)
xÄ1
11. Nonremovable discontinuity at xœ1 because lim f(x) fails to exist ( lim f(x)œ1 and lim f(x)œ0).
xÄ1 xÄ "c xÄ "b
Removable discontinuity at xœ0 by assigning the number lim f(x)œ0 to be the value of f(0) rather than
xÄ !
f(0)œ1.
12. Nonremovable discontinuity at xœ1 because lim f(x) fails to exist ( lim f(x)œ2 and lim f(x)œ1).
xÄ1 xÄ "c xÄ "b
Removable discontinuity at xœ2 by assigning the number lim f(x)œ1 to be the value of f(2) rather than
xÄ #
f(2)œ2.
13. Discontinuous only when x œ2 0 Ê xœ2 14. Discontinuous only when (x2)#œ0 Ê xœ 2 15. Discontinuous only when x# % $ œ ! Êx (x3)(x1)œ0 Ê xœ3 or xœ1
16. Discontinuous only when x#3x10œ0 Ê (x5)(x2)œ0 Ê xœ5 or xœ 2
17. Continuous everywhere. ( xk 1k sin x defined for all x; limits exist and are equal to function values.) 18. Continuous everywhere. ( xk k " Á0 for all x; limits exist and are equal to function values.)
19. Discontinuous only at xœ0
20. Discontinuous at odd integer multiples of , i.e., x = (2n1# ") , n an integer, but continuous at all other x.1# 21. Discontinuous when 2x is an integer multiple of , i.e., 2x1 œn , n an integer 1 Ê xœ n#1, n an integer, but
continuous at all other x.
22. Discontinuous when 1#x is an odd integer multiple of , i.e., 1# 1#x œ(2n1) , n an integer 1# Ê xœ2n1, n an integer (i.e., x is an odd integer). Continuous everywhere else.
23. Discontinuous at odd integer multiples of , i.e., x = (2n1# 1) , n an integer, but continuous at all other x.1# 24. Continuous everywhere since x% 1 1 and " Ÿsin xŸ1 Ê 0Ÿsin x# Ÿ1 Ê 1 sin x# 1; limits exist
and are equal to the function values.
25. Discontinuous when 2x 3 0 or x Ê3# continuous on the interval ß _3# ‰. 26. Discontinuous when 3x 1 0 or x "3 Ê continuous on the interval "3ß _‰.
27. Continuous everywhere: (2x1)"Î$ is defined for all x; limits exist and are equal to function values.
28. Continuous everywhere: (2x)"Î& is defined for all x; limits exist and are equal to function values.
29. lim sin (xxÄ1 sin x)œsin (1sin )1 œsin (10)œsin 1œ0, and function continuous at xœ1.
30. lim sin cos (tan t) sin cos (tan (0)) sin cos (0) sin 1, and function continuous at t .
tÄ ! ˆ1# ‰œ ˆ1# ‰œ ˆ1# ‰œ ˆ ‰1# œ œ !
31. lim sec y sec y tan y 1 lim sec y sec y sec y lim sec (y 1) sec y sec ( ) sec 1
yÄ1 a # # bœyÄ1 a # # bœyÄ1 a # bœ a" " # b sec 0 1, , and function continuous at y .
œ œ œ "
32. lim tan cos sin x tan cos (sin(0)) tan cos (0) tan 1, and function continuous at x .
xÄ ! 14 ˆ "Î$‰‘œ 14 ‘œ ˆ14 ‰œ ˆ ‰14 œ œ !
33. lim cos cos cos cos , and function continuous at t .
tÄ ! ’È1913 sec 2t“œ ’È1913 sec 0“œ È116 œ 14 œÈ#2 œ !
34. lim csc x 5 3 tan x csc 5 3 tan 4 5 3 9 3, and function continuous at
xÄ1' É # È œÉ #ˆ ‰16 È ˆ ‰16 œÊ È ŠÈ"3‹œÈ œ
xœ 1'.
35. g(x)œ xx#39 œ (x(x3)(x3)3)œ x 3, xÁ3 Ê g(3)œ lim (x3)œ6
xÄ $
36. h(t)œt# t3t #10 œ(tt5)(t #2) œ t 5, tÁ # Ê h(2)œ lim (t5)œ7
tÄ #
37. f(s)œss$# 1 œ s(ss1)(s1 (s1)1) œs#s s1 , sÁ1 Ê f(1)œ lim s#s s11 œ 3
" # "
a b #
sÄ1Š ‹
38. g(x)œ x#x#3x164 œ(x(x 4)(x4)(x 1)4) œxx41, xÁ4 Ê g(4)œ lim xx41 œ 85
xÄ %ˆ ‰
39. As defined, lim f(x) (3) 1 8 and lim (2a)(3) 6a. For f(x) to be continuous we must have
xÄ $c œ # œ xÄ $b œ
6aœ8 Ê œa 43.
40. As defined, lim g(x) 2 and lim g(x) b( 2) 4b. For g(x) to be continuous we must have
xÄ #c œ xÄ #b œ #œ
4bœ Ê2 bœ "#.
41. The function can be extended: f(0)¸2.3. 42. The function cannot be extended to be continuous at xœ0. 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.
43. The function cannot be extended to be continuous 44. The function can be extended: f(0)¸7.39.
at xœ0. If f(0)œ1, it will be continuous from the right. Or if f(0)œ 1, it will be continuous from the left.
45. f(x) is continuous on [!ß "] and f(0)0, f(1)0 by the Intermediate Value Theorem f(x) takes Ê
on every value between f(0) and f(1) Ê the equation f(x)œ0 has at least one solution between xœ0 and xœ1.
46. cos xœx Ê (cos x) œx 0. If xœ 1#, cosˆ1#‰ ˆ 1#‰0. If xœ 1#, cosˆ ‰1# 1# 0. Thus cos x œx 0 for some x between 1# and according to the Intermediate Value Theorem.1#
47. Let f(x)œx$15x1 which is continuous on [ 4 4]. Then f( 4) ß œ 3, f( 1) œ15, f(1)œ 13, and f(4)œ5.
By the Intermediate Value Theorem, f(x)œ0 for some x in each of the intervals % " x 1, x 1, and x 4. That is, x 15x 1 0 has three solutions in [ 4]. Since a polynomial of degree 3 can have at most 3
" $ œ %ß
solutions, these are the only solutions.
48. Without loss of generality, assume that ab. Then F(x)œ(xa) (x# b)#x is continuous for all values of x, so it is continuous on the interval [a b]. Moreover F(a)ß œa and F(b)œb. By the Intermediate Value Theorem, since aa#b b, there is a number c between a and b such that F(x)œ a#b.
49. Answers may vary. Note that f is continuous for every value of x.
(a) f(0)œ10, f(1)œ1$8(1)10œ3. Since $ 1 10, by the Intermediate Value Theorem, there exists a c so that ! c 1 and f(c)œ1.
(b) f(0)œ10, f( 4) œ ( 4)$ 8( 4) 10œ 22. Since 22 È310, by the Intermediate Value Theorem, there exists a c so that 4 c 0 and f(c)œ È3.
(c) f(0)œ10, f(1000)œ(1000)$8(1000)10œ999,992,010. Since 105,000,000999,992,010, by the Intermediate Value Theorem, there exists a c so that ! c 1000 and f(c)œ5,000,000.
50. All five statements ask for the same information because of the intermediate value property of continuous functions.
(a) A root of f(x)œx$3x1 is a point c where f(c)œ0.
(b) The points where yœx crosses y$ œ3x1 have the same y-coordinate, or yœx$œ3x1 Ê f(x)œx$3x œ1 0.
(c) x$3xœ1 Ê x$3x œ1 0. The solutions to the equation are the roots of f(x)œx$3x1.
(d) The points where yœx$3x crosses yœ1 have common y-coordinates, or yœx$3xœ1 Ê f(x)œx$3x œ !1 .
(e) The solutions of x$3x œ1 0 are those points where f(x)œx$3x1 has value 0.
51. Answers may vary. For example, f(x)œ sin (xx22) is discontinuous at xœ2 because it is not defined there.
However, the discontinuity can be removed because f has a limit (namely 1) as x Ä 2.
52. Answers may vary. For example, g(x)œ x"1 has a discontinuity at xœ 1 because lim g(x) does not exist.
xÄ "
lim g(x) and lim g(x) .
Šx ‹
Ä "c œ _ xÄ "b œ _
53. (a) Suppose x is rational ! Ê f(x )! œ1. Choose %œ "#. For any $0 there is an irrational number x (actually infinitely many) in the interval (x! ß $ x! $) Ê f(x)œ0. Then 0 kx x!k$ but f(x)k f(x )! k
1 , so lim f(x) fails to exist f is discontinuous at x rational.
œ œ"# % xÄx! Ê !
On the other hand, x irrational ! Ê f(x )! œ0 and there is a rational number x in (x! ß $ x! $) Ê f(x) 1. Again lim f(x) fails to exist f is discontinuous at x irrational. That is, f is discontinuous at
œ xÄx! Ê !
every point.
(b) f is neither right-continuous nor left-continuous at any point x because in every interval (x! ! ß$ x ) or! (x x!ß ! $) there exist both rational and irrational real numbers. Thus neither limits lim f(x) andxÄx
!
lim f(x) exist by the same arguments used in part (a).
xÄx!
54. Yes. Both f(x)œx and g(x)œ x "# are continuous on [!ß "]. However g(x)f(x) is undefined at xœ "# since gˆ ‰"# œ0 Ê g(x)f(x) is discontinuous at xœ "#.
55. No. For instance, if f(x)œ0, g(x)œ Ü Ýx , then h(x)œ0aÜ Ý œx b 0 is continuous at xœ0 and g(x) is not.
56. Let f(x)œ x"1 and g(x)œ x 1. Both functions are continuous at xœ0. The composition f‰ œg f(g(x)) is discontinuous at x 0, since it is not defined there. Theorem 10 requires that f(x) be
œ (x "1) 1 œ "x œ
continuous at g(0), which is not the case here since g(0)œ1 and f is undefined at 1.
57. 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].ß
58. Let f(x) be the new position of point x and let d(x)œf(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)œ0 for some point in between. That is, f(x)œx for some point x, which is then in its original position.
59. If f(0)œ0 or f(1)œ1, we are done (i.e., cœ0 or cœ1 in those cases). Then let f(0)œ a 0 and f(1)œ b 1 because 0Ÿf(x)Ÿ1. Define g(x)œf(x) Êx g is continuous on [0 1]. Moreover, g(0)ß œf(0) œ 0 a 0 and g(1)œf(1) œ 1 b 1 0 Ê by the Intermediate Value Theorem there is a number c in (!ß ") such that
g(c)œ0 Ê f(c) œc 0 or f(c)œc.
60. Let %œ kf(c)#k 0. Since f is continuous at xœc there is a $0 such that xk Êck $ f(x)k f(c)k% Ê f(c) % f(x)f(c)%.
If f(c)0, then %œ "#f(c) Ê f(c)"# f(x) #3f(c) Ê f(x)0 on the interval (c ß $ c $).
If f(c)0, then %œ "#f(c) Ê f(c)#3 f(x) "#f(c) Ê f(x)0 on the interval (c ß $ c $).
61. By Exercises 52 in Section 2.3, we have lim f xx c L lim f c h L.
h 0
Ä a bœ Í Ä a bœ
Thus, f x is continuous at xa b œ Íc x lim f xc a bœf ca bÍ lim f ca hbœf c .a b
h 0
Ä Ä
62. By Exercise 61, it suffices to show that lim sin c h sin c and lim cos c h cos c.
hÄ0 a bœ hÄ0 a bœ
Now lim sin c h lim sin c cos h cos c sin h sin c lim cos h cos c lim sin h
hÄ0 a bœhÄ0a ba ba ba b‘œa bŠhÄ0 ‹a bŠhÄ0 ‹
By Example 6 Section 2.2, lim cos h and lim sin h . So lim sin c h sin c and thus f x sin x is
hÄ0 œ " hÄ0 œ ! hÄ0 a bœ a bœ continuous at xœc. Similarly,
lim cos c h lim cos c cos h sin c sin h cos c lim cos h sin c lim sin h cos c.
hÄ0 a bœhÄ0a ba ba ba b‘œa bŠhÄ0 ‹a bŠhÄ0 ‹œ
Thus, g xa bœcos x is continuous at xœc.
63. x¸1.8794, 1.5321, 0.3473 64. x¸1.4516, 0.8547, 0.4030
65. x¸1.7549 66. x¸1.5596
67. x¸3.5156 68. x¸ 3.9058, 3.8392, 0.0667
69. x¸0.7391 70. x¸ 1.8955, 0, 1.8955