with slope 2 passing through (30); that is, − 0 = 2( − 3), or = 2 − 6. So the function is
() =
− + 3 if 0 ≤ ≤ 3
2 − 6 if 3 ≤ 5
64. For −4 ≤ ≤ −2, the graph is the line with slope − 3
2
passing through (−20); that is, − 0 = − 3
2 [ − (−2)], or
= − 3
2 −3. For −2 2, the graph is the top half of the circle with center (00) and radius 2. An equation of the circle
is 2 + 2 = 4, so an equation of the top half is =
√ 4 − 2 . For 2 ≤ ≤ 4, the graph is the line with slope
3
2
passing
through (20); that is, − 0 =
3
2 ( − 2), or =
3
2 − 3. So the function is
() =
− 3
2 − 3
if −4 ≤ ≤ −2
√ 4 − 2
if −2 2
3
2 − 3
if 2 ≤ ≤ 4
65. Let the length and width of the rectangle be and . Then the perimeter is 2 + 2 = 20 and the area is = .
Solving the first equation for in terms of gives =
20 − 2
2
= 10−. Thus, () = (10−) = 10− 2 . Since
lengths are positive, the domain of is 0 10. If we further restrict to be larger than , then 5 10 would be
the domain.
66. Let the length and width of the rectangle be and . Then the area is = 16, so that = 16. The perimeter is
= 2 + 2, so () = 2 + 2(16) = 2 + 32, and the domain of is 0, since lengths must be positive
quantities. If we further restrict to be larger than , then 4 would be the domain.
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SECTION 1.1 FOURWAYS TO REPRESENTA FUNCTION
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19
67. Let the length of a side of the equilateral triangle be . Then by the Pythagorean Theorem, the height of the triangle satisfies
2 +
1
2
2
= 2 , so that 2 = 2 −
1
4
2
=
3
4
2
and =
√ 3
2
. Using the formula for the area of a triangle,
=
1
2 (base)(height), we obtain () =
1
2 ()
√
3
2
=
√ 3
4
2 , with domain 0.
68. Let the length, width, and height of the closed rectangular box be denoted by , , and , respectively. The length is twice
the width, so = 2. The volume of the box is given by = . Since = 8, we have 8 = (2) ⇒
8 = 2 2 ⇒ =
8
2 2
=
4
2
, and so = () =
4
2
.
69. Let each side of the base of the box have length , and let the height of the box be . Since the volume is 2, we know that
2 = 2 , so that = 2 2 , and the surface area is = 2 + 4. Thus, () = 2 + 4(2 2 ) = 2 + (8), with
domain 0.
70. Let and denote the radius and the height of the right circular cylinder, respectively. Then the volume is given by
= 2 , and for this particular cylinder we have 2 = 25 ⇔ 2 =
25
. Solving for and rejecting the negative
solution gives =
5
√ , so = () =
5
√ in.
71. The height of the box is and the length and width are = 20 − 2, = 12 − 2. Then = and so
() = (20 − 2)(12 − 2)() = 4(10 − )(6 − )() = 4(60 − 16 + 2 ) = 4 3 − 64 2 + 240.
The sides , , and must be positive. Thus, 0 ⇔ 20 − 2 0 ⇔ 10;
0 ⇔ 12 − 2 0 ⇔ 6; and 0. Combining these restrictions gives us the domain 0 6.
72. The area of the window is = +
1
2
1
2
2
= +
2
8
, where is the height of the rectangular portion of the window.
The perimeter is = 2 + +
1
2 = 30
⇔ 2 = 30 − −
1
2
⇔ =
1
4 (60 − 2 − ). Thus,
() =
60 − 2 −