,
,
,
,
,
, SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤ 15
2 + 1
40. The function () = is defined for all values of except those for which 2 + 4 − 21 = 0 ⇔
2 + 4 − 21
( + 7)( − 3) = 0 ⇔ = −7 or = 3. Thus, the domain is { ∈ | 6= −7 3} = (−∞ −7) ∪ (−7 3) ∪ (3 ∞).
√
41. () = 3
2 − 1 is defined for all real numbers. In fact 3 (), where () is a polynomial, is defined for all real numbers.
Thus, the domain is or (−∞ ∞).
√ √
42. () = 3 − − 2 + is defined when 3 − ≥ 0 ⇔ ≤ 3 and 2 + ≥ 0 ⇔ ≥ −2. Thus, the domain is
−2 ≤ ≤ 3, or [−2 3].
√
43. () = 1 4
2 − 5 is defined when 2 − 5 0 ⇔ ( − 5) 0. Note that 2 − 5 6= 0 since that would result in
division by zero. The expression ( − 5) is positive if 0 or 5. (See Appendix A for methods for solving
inequalities.) Thus, the domain is (−∞ 0) ∪ (5 ∞).
+1 1 1
44. () = is defined when + 1 6= 0 [ 6= −1] and 1 + 6= 0. Since 1 + =0 ⇔
1 +1 +1
1+
+1
1
= −1 ⇔ 1 = − − 1 ⇔ = −2, the domain is { | 6= −2, 6= −1} = (−∞ −2) ∪ (−2 −1) ∪ (−1 ∞).
+1
√ √ √ √ √
45. () = 2 − is defined when ≥ 0 and 2 − ≥ 0. Since 2 − ≥ 0 ⇔ 2 ≥ ⇔ ≤2 ⇔
0 ≤ ≤ 4, the domain is [0 4].
√
46. The function () = 2 − 4 − 5 is defined when 2 − 4 − 5 ≥ 0 ⇔ ( + 1)( − 5) ≥ 0. The polynomial
() = 2 − 4 − 5 may change signs only at its zeros, so we test values of on the intervals separated by = −1 and
= 5: (−2) = 7 0, (0) = −5 0, and (6) = 7 0. Thus, the domain of , equivalent to the solution intervals
of () ≥ 0, is { | ≤ −1 or ≥ 5} = (−∞ −1] ∪ [5 ∞).
√ √
47. () = 4 − 2 . Now = 4 − 2 ⇒ 2 = 4 − 2 ⇔ 2 + 2 = 4, so
the graph is the top half of a circle of radius 2 with center at the origin. The domain
is | 4 − 2 ≥ 0 = | 4 ≥ 2 = { | 2 ≥ ||} = [−2 2]. From the graph,
the range is 0 ≤ ≤ 2, or [0 2].
2 − 4
48. The function () = is defined when − 2 6= 0 ⇔ 6= 2, so the
−2
domain is { | 6= 2} = (−∞ 2) ∪ (2 ∞). On its domain,
2 − 4 ( − 2)( + 2)
() = = = + 2. Thus, the graph of is the
−2 −2
line = + 2 with a hole at (2 4).
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
,16 ¤ CHAPTER 1 FUNCTIONS AND MODELS
2 + 2 if 0
49. () =
if ≥ 0
(−3) = (−3) + 2 = 11, (0) = 0, and (2) = 2.
2
5 if 2
50. () = 1
2
−3 if ≥ 2
(−3) = 5, (0) = 5, and (2) = 12 (2) − 3 = −2.
+1 if ≤ −1
51. () = 2
if −1
(−3) = −3 + 1 = −2, (0) = 02 = 0, and (2) = 22 = 4.
−1 if ≤ 1
52. () =
7 − 2 if 1
(−3) = −1, (0) = −1, and (2) = 7 − 2(2) = 3.
if ≥ 0
53. || =
− if 0
2 if ≥ 0
so () = + || =
0 if 0
Graph the line = 2 for ≥ 0 and graph = 0 (the axis) for 0
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
, SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤ 17
+2 if + 2 ≥ 0
54. () = | + 2| =
−( + 2) if + 2 0
+2 if ≥ −2
=
− − 2 if −2
1 − 3 if 1 − 3 ≥ 0
55. () = |1 − 3| =
−(1 − 3) if 1 − 3 0
1 − 3 if ≤ 1
3
=
3 − 1 if 1
3
||
56. () =
The domain of is { | 6= 0} and || = if 0, || = − if 0.
So we can write
−
= −1 if 0
() =
=1 if 0
|| if || ≤ 1
57. To graph () = , graph = || [Figure 16]
1 if || 1
for −1 ≤ ≤ 1 and graph = 1 for 1 and for −1.
1 if −1
− if −1 ≤ 0
We could rewrite f as () = .
if 0 ≤ ≤ 1
1 if 1
|| − 1 if || − 1 ≥ 0
58. () = || − 1 =
−(|| − 1) if || − 1 0
|| − 1 if || ≥ 1
=
− || + 1 if || 1
−1 if || ≥ 1 and ≥ 0 −1 if ≥ 1
− − 1
− − 1
if || ≥ 1 and 0 if ≤ −1
= =
− + 1 if || 1 and ≥ 0
− + 1 if 0 ≤ 1
−(−) + 1 if || 1 and 0 +1 if −1 0
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
,
,
,
,
, SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤ 15
2 + 1
40. The function () = is defined for all values of except those for which 2 + 4 − 21 = 0 ⇔
2 + 4 − 21
( + 7)( − 3) = 0 ⇔ = −7 or = 3. Thus, the domain is { ∈ | 6= −7 3} = (−∞ −7) ∪ (−7 3) ∪ (3 ∞).
√
41. () = 3
2 − 1 is defined for all real numbers. In fact 3 (), where () is a polynomial, is defined for all real numbers.
Thus, the domain is or (−∞ ∞).
√ √
42. () = 3 − − 2 + is defined when 3 − ≥ 0 ⇔ ≤ 3 and 2 + ≥ 0 ⇔ ≥ −2. Thus, the domain is
−2 ≤ ≤ 3, or [−2 3].
√
43. () = 1 4
2 − 5 is defined when 2 − 5 0 ⇔ ( − 5) 0. Note that 2 − 5 6= 0 since that would result in
division by zero. The expression ( − 5) is positive if 0 or 5. (See Appendix A for methods for solving
inequalities.) Thus, the domain is (−∞ 0) ∪ (5 ∞).
+1 1 1
44. () = is defined when + 1 6= 0 [ 6= −1] and 1 + 6= 0. Since 1 + =0 ⇔
1 +1 +1
1+
+1
1
= −1 ⇔ 1 = − − 1 ⇔ = −2, the domain is { | 6= −2, 6= −1} = (−∞ −2) ∪ (−2 −1) ∪ (−1 ∞).
+1
√ √ √ √ √
45. () = 2 − is defined when ≥ 0 and 2 − ≥ 0. Since 2 − ≥ 0 ⇔ 2 ≥ ⇔ ≤2 ⇔
0 ≤ ≤ 4, the domain is [0 4].
√
46. The function () = 2 − 4 − 5 is defined when 2 − 4 − 5 ≥ 0 ⇔ ( + 1)( − 5) ≥ 0. The polynomial
() = 2 − 4 − 5 may change signs only at its zeros, so we test values of on the intervals separated by = −1 and
= 5: (−2) = 7 0, (0) = −5 0, and (6) = 7 0. Thus, the domain of , equivalent to the solution intervals
of () ≥ 0, is { | ≤ −1 or ≥ 5} = (−∞ −1] ∪ [5 ∞).
√ √
47. () = 4 − 2 . Now = 4 − 2 ⇒ 2 = 4 − 2 ⇔ 2 + 2 = 4, so
the graph is the top half of a circle of radius 2 with center at the origin. The domain
is | 4 − 2 ≥ 0 = | 4 ≥ 2 = { | 2 ≥ ||} = [−2 2]. From the graph,
the range is 0 ≤ ≤ 2, or [0 2].
2 − 4
48. The function () = is defined when − 2 6= 0 ⇔ 6= 2, so the
−2
domain is { | 6= 2} = (−∞ 2) ∪ (2 ∞). On its domain,
2 − 4 ( − 2)( + 2)
() = = = + 2. Thus, the graph of is the
−2 −2
line = + 2 with a hole at (2 4).
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
,16 ¤ CHAPTER 1 FUNCTIONS AND MODELS
2 + 2 if 0
49. () =
if ≥ 0
(−3) = (−3) + 2 = 11, (0) = 0, and (2) = 2.
2
5 if 2
50. () = 1
2
−3 if ≥ 2
(−3) = 5, (0) = 5, and (2) = 12 (2) − 3 = −2.
+1 if ≤ −1
51. () = 2
if −1
(−3) = −3 + 1 = −2, (0) = 02 = 0, and (2) = 22 = 4.
−1 if ≤ 1
52. () =
7 − 2 if 1
(−3) = −1, (0) = −1, and (2) = 7 − 2(2) = 3.
if ≥ 0
53. || =
− if 0
2 if ≥ 0
so () = + || =
0 if 0
Graph the line = 2 for ≥ 0 and graph = 0 (the axis) for 0
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
, SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤ 17
+2 if + 2 ≥ 0
54. () = | + 2| =
−( + 2) if + 2 0
+2 if ≥ −2
=
− − 2 if −2
1 − 3 if 1 − 3 ≥ 0
55. () = |1 − 3| =
−(1 − 3) if 1 − 3 0
1 − 3 if ≤ 1
3
=
3 − 1 if 1
3
||
56. () =
The domain of is { | 6= 0} and || = if 0, || = − if 0.
So we can write
−
= −1 if 0
() =
=1 if 0
|| if || ≤ 1
57. To graph () = , graph = || [Figure 16]
1 if || 1
for −1 ≤ ≤ 1 and graph = 1 for 1 and for −1.
1 if −1
− if −1 ≤ 0
We could rewrite f as () = .
if 0 ≤ ≤ 1
1 if 1
|| − 1 if || − 1 ≥ 0
58. () = || − 1 =
−(|| − 1) if || − 1 0
|| − 1 if || ≥ 1
=
− || + 1 if || 1
−1 if || ≥ 1 and ≥ 0 −1 if ≥ 1
− − 1
− − 1
if || ≥ 1 and 0 if ≤ −1
= =
− + 1 if || 1 and ≥ 0
− + 1 if 0 ≤ 1
−(−) + 1 if || 1 and 0 +1 if −1 0
c 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
°
