4
+
2
8
= 15 −
1
2
2
−
4
2 +
8
2 = 15 −
4
8
2
−
8
2 = 15 − 2
+ 4
8
.
Since the lengths and must be positive quantities, we have 0 and 0. For 0, we have 2 0 ⇔
30 − −
1
2 0
⇔ 60 2 + ⇔
60
2 +
. Hence, the domain of is 0
60
2 +
.
73. We can summarize the amount of the fine with a
piecewise defined function.
() =
15(40 − ) if 0 ≤ 40
0 if 40 ≤ ≤ 65
15( − 65) if 65
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CHAPTER 1 FUNCTIONSAND MODELS
74. For the first 1200 kWh, () = 10 + 006.
For usage over 1200 kWh, the cost is
() = 10 + 006(1200)+ 007( − 1200) = 82 + 007( − 1200).
Thus,
() =
10 + 006
if 0 ≤ ≤ 1200
82 + 007( − 1200) if 1200
75. (a) (b) On $14,000, tax is assessed on $4000, and 10%($4000) = $400.
On $26,000, tax is assessed on $16,000, and
10%($10,000) + 15%($6000) = $1000 + $900 = $1900.
(c) As in part (b), there is $1000 tax assessed on $20,000 of income, so
the graph of is a line segment from (10,0000) to (20,0001000).
The tax on $30,000 is $2500, so the graph of for 20,000 is
the ray with initial point (20,0001000) that passes through
(30,0002500).
76. (a) Because an even function is symmetric with respect to the axis, and the point (53) is on the graph of this even function,
the point (−53) must also be on its graph.
(b) Because an odd function is symmetric with respect to the origin, and the point (53) is on the graph of this odd function,
the point (−5−3) must also be on its graph.
77. is an odd function because its graph is symmetric about the origin. is an even function because its graph is symmetric with
respect to the axis.
78. is not an even function since it is not symmetric with respect to the axis. is not an odd function since it is not symmetric
about the origin. Hence, is neither even nor odd. is an even function because its graph is symmetric with respect to the
axis.
79. (a) The graph of an even function is symmetric about the axis. We reflect the
given portion of the graph of about the axis in order to complete it.
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SECTION 1.1 FOURWAYS TO REPRESENTA FUNCTION
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21
(b) For an odd function, (−) = −(). The graph of an odd function is
symmetric about the origin. We rotate the given portion of the graph of
through 180 ◦ about the origin in order to complete it.
80. (a) The graph of an even function is symmetric about the axis. We reflect the
given portion of the graph of about the axis in order to complete it.
(b) The graph of an odd function is symmetric about the origin. We rotate the
given portion of the graph of through 180 ◦ about the origin in order to
complete it.
81. () =
2 + 1 .
(−) =
−
(−) 2 + 1
=
−
2 + 1
= −
2 + 1
= −().
Since (−) = −(), is an odd function.
82. () =
2
4 + 1 .
(−) =
(−) 2
(−) 4 + 1
=
2
4 + 1
= ().
Since (−) = (), is an even function.
83. () =
+ 1 , so (−) =
−
− + 1
=
− 1 .
Since this is neither () nor −(), the function is
neither even nor odd.
84. () = ||.
(−) = (−)|−| = (−)|| = −(||)
= −()
Since (−) = −(), is an odd function.
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