50. () =
5
if 2
1
2 − 3
if ≥ 2
(−3) = 5, (0) = 5, and (2) =
1
2 (2) − 3 = −2.
51. () =
+ 1
if ≤ −1
2 if −1
(−3) = −3 + 1 = −2, (0) = 0 2 = 0, and (2) = 2 2 = 4.
52. () =
−1
if ≤ 1
7 − 2 if 1
(−3) = −1, (0) = −1, and (2) = 7 − 2(2) = 3.
53. || =
if ≥ 0
− if 0
so () = + || =
2
if ≥ 0
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 FOURWAYS TO REPRESENTA FUNCTION
¤
17
54. () = | + 2| =
+ 2
if + 2 ≥ 0
−( + 2) if + 2 0
=
+ 2
if ≥ −2
− − 2 if −2
55. () = |1 − 3| =
1 − 3
if 1 − 3 ≥ 0
−(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
57. To graph () =
||
if || ≤ 1
1 if || 1
, graph = || [Figure 16]
for −1 ≤ ≤ 1 and graph = 1 for 1 and for −1.
We could rewrite f as () =
1 if −1
− if −1 ≤ 0
if 0 ≤ ≤ 1
1 if 1
.
58. () =
|| − 1
=
|| − 1
if || − 1 ≥ 0
−(|| − 1) if || − 1 0
=
|| − 1
if || ≥ 1
−|| + 1 if || 1
=
− 1 if || ≥ 1 and ≥ 0
− − 1 if || ≥ 1 and 0
− + 1 if || 1 and ≥ 0
−(−) + 1 if || 1 and 0
=
− 1 if ≥ 1
− − 1 if ≤ −1
− + 1 if 0 ≤ 1
+ 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.
18
¤
CHAPTER 1 FUNCTIONSAND MODELS
59. Recall that the slope of a line between the two points ( 1 1 ) and ( 2 2 ) is =
2 − 1
2 − 1
and an equation of the line
connecting those two points is − 1 = ( − 1 ). The slope of the line segment joining the points (1−3) and (57) is
7 − (−3)
5 − 1
=
5
2 , so an equation is − (−3) =
5
2 ( − 1). The function is () =
5
2 −
11
2
, 1 ≤ ≤ 5.
60. The slope of the line segment joining the points (−510) and (7−10) is
−10 − 10
7 − (−5)
= − 5
3 , so an equation is
− 10 = − 5
3 [ − (−5)]. The function is () = −
5
3 +
5
3 , −5 ≤ ≤ 7.
61. We need to solve the given equation for . + ( − 1) 2 = 0 ⇔ ( − 1) 2 = − ⇔ − 1 = ± √ − ⇔
= 1 ±
√ −. The expression with the positive radical represents the top half of the parabola, and the one with the negative
radical represents the bottom half. Hence, we want () = 1 −
√ −. Note that the domain is ≤ 0.
62. 2 + ( − 2) 2 = 4 ⇔ ( − 2) 2 = 4 − 2 ⇔ − 2 = ± √ 4 − 2 ⇔ = 2 ±
√ 4 − 2 . The top half is given by
the function () = 2 +
√ 4 − 2 , −2 ≤ ≤ 2.
63. For 0 ≤ ≤ 3, the graph is the line with slope −1 and intercept 3, that is, = − + 3. For 3 ≤ 5, the graph is the line