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Sullivan AP˙Sullivan˙Chapter01 October 8, 2016 17:4
Section 1.3 • Assess Your Understanding 113
⎧
4 − 3x
⎪⎨
2 if x < 0
4 if x = 0 at c = 0
31. f (x) =
⎪⎩ 16 − x 2
if 0 < x < 4
4 − x
⎧√ ⎨ 4 + x if −4 ≤ x ≤ 4
32. f (x) =
⎩ x 2 − 3x − 4
at c = 4
if x > 4
x − 4
In Problems 33–36, each function f has a removable
discontinuity at c. Define f (c) so that f is continuous at c.
33. f (x) = x2 − 4
x − 2 , c = 2
34. f (x) = x2 + x − 12
, c = 3
x − 3
⎧
⎨ 1 + x if x < 1
PAGE
105 35. f (x) = 4 if x = 1 c = 1
⎩
2x if x > 1
⎧
⎨ x 2 + 5x if x < −1
36. f (x) = 0 if x =−1 c =−1
⎩
x − 3 if x > −1
In Problems 37–40, determine if each function f is
continuous on the given interval. If the answer is no,
state the interval, if any, on which f is continuous.
PAGE
107 37. f (x) = x2 − 9
on the interval [−3, 3)
x − 3
38. f (x) = 1 + 1 on the interval [−1, 0)
x
39.
1
f (x) = on the interval [−3, 3]
x 2 − 9
40. f (x) = 9 − x 2 on the interval [−3, 3]
In Problems 41–50, determine where each function f is continuous.
First determine the domain of the function. Then support your decision
using properties of continuity.
41. f (x) = 2x 2 + 5x − 1 42. f (x) = x + 1 + 2x
x
x 2 + 5
43. f (x) = (x − 1)(x 2 + x + 1) 44. f (x) = √ x(x 3 − 5)
PAGE
109 45. f (x) = √ x − 9
46. f (x) = √ x − 4
x − 3 x − 2
PAGE
109 47. f (x) =
x 2 + 1
2 − x
48. f (x) =
4
x 2 − 1
49. f (x) = (2x 2 + 5x − 3) 2/3 50. f (x) = (x + 2) 1/2
In Problems 51–56, use the function
⎧ √
15 − 3x if x < 2
⎪⎨ √
5 if x = 2
f (x) =
9 − x ⎪⎩
2 if 2 < x < 3
x − 2 if 3 ≤ x
51. Is f continuous at 0? Why or why not?
52. Is f continuous at 4? Why or why not?
PAGE
105 53. Is f continuous at 3? Why or why not?
49. Continuous on the set of all real numbers.
50. Continuous on { xx | ≥− 2} .
51. f is continuous at 0 because lim fx ( ) = f (0).
x→0
52. f is discontinuous at 4 because lim fx ( )
x→4
does not exist.
53. f is discontinuous at 3 because lim fx ( ) does
x → 3
not exist.
54. f is discontinuous at 2 because lim fx ( ) does
x→2
not exist.
55. f is continuous at 1 because lim fx ( ) = f (1).
x→1
56. f is continuous at 2.5 because
lim fx ( ) = f (2.5).
x→2.5
54. Is f continuous at 2? Why or why not?
55. Is f continuous at 1? Why or why not?
56. Is f continuous at 2.5? Why or why not?
In Problems 57 and 58:
(a) Use technology to graph f using a suitable scale on each axis.
(b) Based on the graph from (a), determine where f is continuous.
(c) Use the definition of continuity to determine where f is continuous.
(d) What advice would you give a fellow student about using
technology to determine where a function is continuous?
57. f (x) = x3 − 8
x − 2
58. f (x) = x2 − 3x + 2
3x − 6
In Problems 59–64, use the Intermediate Value Theorem to determine
which of the functions must have zeros in the given interval. Indicate
those for which the theorem gives no information. Do not attempt to
locate the zeros.
PAGE
111 59. f (x) = x 3 − 3x on [−2, 2]
60. f (x) = x 4 − 1on[−2, 2]
61.
x
f (x) = − 1 on [10, 20]
(x + 1) 2
62. f (x) = x 3 − 2x 2 − x + 2 on [3, 4]
63. f (x) = x3 − 1
on [0, 2]
x − 1
64. f (x) = x2 + 3x + 2
x 2 on [−3, 0]
− 1
In Problems 65–72, verify that each function has a zero in the indicated
interval. Then use the Intermediate Value Theorem to approximate the
zero correct to three decimal places by repeatedly subdividing the
interval containing the zero into 10 subintervals.
PAGE
112 65. f (x) = x 3 + 3x − 5; interval: [1, 2]
66. f (x) = x 3 − 4x + 2; interval: [1, 2]
67. f (x) = 2x 3 + 3x 2 + 4x − 1; interval: [0, 1]
68. f (x) = x 3 − x 2 − 2x + 1; interval: [0, 1]
69. f (x) = x 3 − 6x − 12; interval: [3, 4]
70. f (x) = 3x 3 + 5x − 40; interval: [2, 3]
71. f (x) = x 4 − 2x 3 + 21x − 23; interval: [1, 2]
72. f (x) = x 4 − x 3 + x − 2; interval: [1, 2]
In Problems 73 and 74,
(a) Use the Intermediate Value Theorem to show that f has a zero in
the given interval.
(b) Use technology to find the zero rounded to three decimal places.
73. f (x) = x 2 + 4x − 2 in [0, 1]
74. f (x) = x 3 − x + 2in[−2, 0]
Applications and Extensions
Heaviside Functions In Problems 75 and 76, determine whether the
given Heaviside function is continuous at c.
0 if t < 1
75. u 1(t) =
1 if t ≥ 1
c = 1
0 if t < 3
76. u 3(t) =
1 if t ≥ 3
c = 3
57. (a)
y
18
14
10
6
2
3 2 1 1 2 3 x
(b) Based on the graph, it appears that the
function is continuous for all real numbers.
(c) f is actually continuous at all real numbers
except x = 2.
(d) Answers will vary. Sample answer:
Conclusions drawn from graphing technology
should always be confirmed using basic
analysis.
58. (a)
23 22 21
y
1.5
1
0.5
21
21.5
1 2 3
(b) Based on the graph, it appears
that the function is continuous on all
real numbers.
(c) f is continuous on { xx | ≠ 2}.
(d) Answers will vary. Sample
answer: Conclusions drawn from
graphing technology should always be
confirmed using analysis.
59. Yes. A zero exists on the given
interval.
60. IVT gives no information.
61. IVT gives no information.
62. IVT gives no information.
63. IVT gives no information.
64. IVT gives no information.
65. 1.154
66. 1.675
67. 0.211
68. 0.445
69. 3.134
70. 2.137
71. 1.157
72. 1.308
73. (a) Since f is continuous on [0,1],
f (0) < 0, and f (1) > 0, the IVT
guarantees that f has a zero on the
interval (0,1).
(b) 0.828
74. (a) Since f is continuous on [ −2,0],
f ( − 2) < 0, and f (2) > 0, the IVT
guarantees that f has a zero on the
interval (0,1).
(b) −1.521
75. Discontinuous at c = 1.
76. Discontinuous at c = 3.
x
Section 1.3 • Assess Your Understanding
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