Math 1100-6 Assignment
September 22, 2008
Page 1
Math 1100: Assignment #2 Solutions
Due: Friday Sept 19, 2008 @ 10:30am
No late assignments will be accepted
Please complete the following questions for the assignment. Please be mindful of “good
presentation” criteria outlined in the syllabus as you’re preparing to hand it in.
§9.4
8.
y = 3x5 − 5x3 − 8x + 8
y 0 = 15x4 − 15x2 − 8
9.
g(x) = 10x9 − 5x5 + 7x3 + 5x − 6
g 0 (x) = 90x8 − 25x4 + 21x2 + 5
14.
R(x) = 16x + x2
R0 (x) = 16 + 2x
Therefore, R0 (1) = 18 (this is the slope and instantaneous rate of change).
19.
4
4
f (x) = 5x− 5 + 2x− 3
9
8 7
f 0 (x) = −4x− 5 − x− 3
3
22.
√
1
7
3
− 3 + 8 x = 7x−7 − 3x−3 + 8x 2
7
x
x
1
49
9
4
0
−8
h (x) = −49x + 9x−4 + 4x− 2 = − 8 + 4 + √
x
x
x
h(x) =
25.
1
x
1
0
f (x) = 8x + 2
x
f (−1/2) = 1 + 2 = 3
f (x) = 4x2 −
@ x=−
1
2
f 0 (−1/2) = −4 + 4 = 0
Therefore, the equation of the tangent line at the desired point is y = 3 (horizontal line).
Math 1100-6 Assignment
September 22, 2008
Page 2
28.
1 3
x − 3x2 − 16x + 8
3
f 0 (x) = x2 − 6x − 16 = (x + 2)(x − 8)
f (x) =
Therefore, the function has horizontal tangents at x = −2, 8.
53. Evaluating the expression at t = 15, we get:
√
W C(s) = 55.174 − 2.15s − 23.18 s
11.59
W C 0 (s) = −2.15 − √
s
W C 0 (25) = −4.468
This means that the temperature drops 4.468o for every mile that the wind increases.
§9.5
2.
s = (t4 + 1)(t3 − 1)
s0 = 4t3 (t3 − 1) + (t4 + 1)3t2
£
¤
= t2 4t4 − 4t + 3t4 + 3
= t2 (7t4 − 4t + 3)
6.
y = (9x9 − 7x7 − 6x)(3x5 − 4x4 + 3x3 − 8)
y 0 = (81x8 − 49x6 − 6)(3x5 − 4x4 + 3x3 − 8) + (9x9 − 7x7 − 6x)(15x4 − 16x3 + 9x2 )
8.
√
√
y = ( 5 x − 2 4 x + 1)(x3 − 5x − 7)
³ 1
´
1
=
x 5 − 2x 4 + 1 (x3 − 5x − 7)


3
³ 1
´
1
1 4 1 −
y 0 =  x− 5 − x 4  (x3 − 5x − 7) + x 5 − 2x 4 + 1 (3x2 − 5)
5
2
15.
x2
1 − x − 2x2
2x(1 − x − 2x2 ) − x2 (−1 − 4x)
= 2x +
(1 − x − 2x2 )2
2x − 2x2 − 4x3 + x2 + 4x3
= 2x +
(1 − x − 2x2 )2
2x − x2
= 2x +
(1 − x − 2x2 )2
z = x2 +
z0
Math 1100-6 Assignment
September 22, 2008
18.
√
1
2x 2 − 1
2 x−1
√ =
3
1 − 4 x3
1 − 4x 2
1
3
1
1
x− 2 (1 − 4x 2 ) − (2x 2 − 1)(−6x 2 )
y =
y0 =
3
(1 − 4x 2 )2
Ã
!
3
1
1
1 − 4x 2
1
+ 6x 2 (2x 2 − 1)
3
1
(1 − 4x 2 )2
x2
³
´
3
1
1
2 + 6x(2x 2 − 1)
1
−
4x
√
3
(1 − 4x 2 )2 x
=
=
3
1 − 6x + 8x 2
√
3
(1 − 4x 2 )2 x
=
41.
R(x) =
R0 (x) =
=
=
∴
R0 (49) =
60x2 + 74x
30x2 + 37x
=
2x + 2
x+1
(60x + 37)(x + 1) − (30x2 + 37)(1)
(x + 1)2
2
60x + 97x + 37 − 30x2 − 37
(x + 1)2
30x2 + 97x
(x + 1)2
30.7132
This means that the revenue will increase by appprox $30.71 with the sale of an
additional unit.
47. We want to find
dR
where
dn
R =
dR
dn
=
=
nr
0<r≤1
1 + (n − 1)r
r(1 + (n − 1)r) − nr(r)
(1 + (n − 1)r)2
r(1 − r)
(1 + (n − 1)r)2
§9.6
2.
y = u4
→
u = x2 + 4x
dy
dx
=
dy
= 4u3
du
du
→
= 2x + 4
dx
dy du
= 4(x2 + 4x)3 (2x + 4)
du dx
Page 3
Math 1100-6 Assignment
September 22, 2008
Page 4
9.
g(x) = (x2 + 4x)−2
dg
= −2(x2 + 4x)−3 (2x + 4)
dx
14.
3
y = (3x3 + 4x + 1)− 2
5
3
y 0 = − (3x3 + 4x + 1)− 2 (9x2 + 4)
2
18.
1
5p
5
1 − x3 = (1 − x3 ) 2
6
6
1
5
=
(1 − x3 )− 2 (−3x2 )
12
15x2
= − √
12 1 − x3
y =
y0
20.
y =
y0 =
√ ¢
1 ¡√
2x − 1 + x
2µ
¶
1
2
1
√
+√
4
x
2x − 1
46.
18
3
−
t + 3 (t + 3)2
3
36
S0 = −
+
2
(t + 3)
(t + 3)3
S 0 (8) = 0.002254
S = 1+
∴
∴
S 0 (10) = −0.00137
The campaign should not be continued after the 10’th week because sales are decreasing
at that time.
§9.7
9.
x3 + 1
1
=x+ 2
2
x
x
2
0
f (x) = 1 − 3
x
f (x) =
Math 1100-6 Assignment
September 22, 2008
14.
y = 3x4 (2x5 + 1)7
£
¤
y 0 = 12x3 (2x5 + 1)7 + 3x4 7(2x5 + 1)6 (10x)
£
¤
= 6x3 (2x5 + 1)6 2(2x5 + 1) + 35x2
= 6x3 (2x5 + 1)6 (4x5 + 35x2 + 2)
18.
y =
y0 =
=
=
(x2 − 3)4
£ x2
¤
x 4(x − 3)3 (2x) − (x2 − 3)4
x2
2
3
¡
¢
(x − 3)
2
2
3
+
8x
−
x
x2
2
(x − 3)3 (7x2 + 3)
x2
27.
f (x) =
=
=
=
=
√
1
3
x2 + 5
(x2 + 5) 3
=
4 − x2 µ 4 − x2
¶
1
1 2
1
− 23
2
2
(x + 5) (2x)(4 − x ) − (x + 5) 3 (−2x)
(4 − x2 )2 3
Ã
!
1
2x
4 − x2
2
+ (x + 5) 3
(4 − x2 )2 3(x2 + 5) 32
¡
¢
2x
4 − x2 + 3(x2 + 5)
2
3(4 − x2 )2 (x2 + 5) 3
2x(19 + 2x2 )
2
3(4 − x2 )2 (x2 + 5) 3
32.
p
1
3
3x3 + 2 = x(3x3 + 2) 3
µ
¶
1
1
3
3
− 23
2
3
= (3x + 2) + x
(3x + 2) (9x )
3
1
3x3
= (3x3 + 2) 3 +
2
(3x3 + 2) 3
6x3 + 2
=
2
(3x3 + 2) 3
R(x) = x
Page 5
Math 1100-6 Assignment
September 22, 2008
40.
q+1
(q + 2)2
¡
¢
1
(1)(q + 2)2 − 2(q + 1)(q + 2)
= 400
4
(q + 2)
1
= 400
(q + 2 − 2(q + 1))
(q + 2)3
q
= −400
(q + 2)3
p = 400
p0
§9.8
14.
1
y = x4 + x 3
dy
1 2
= 4x3 + x− 3
dx
3
d2 y
2 5
= 12x2 − x− 3
2
dx
9
16.
√
1
x − 5 = (x − 5) 2
1
1
f 0 (x) =
(x − 5)− 2
2
3
1
f 00 (x) = − (x − 5)− 2
4
5
3
000
f (x) =
(x − 5)− 2
8
f (x) =
21.
1
y 0 = (4x − 1) 2
1
y 00 = 2(4x − 1)− 2
3
y 000 = −4(4x − 1)− 2
5
y (4) = 24(4x − 1)− 2
38.
1
R = 70x + x2 − 0.001x3
2
R0 = 70 + x − 0.003x2
R00 = 1 − 0.006x
∴ R00 (100) = 1 − 0.6 = 0.4
so the marginal revenue is changing at a rate of 0.4 when x = 100.
Page 6
Math 1100-6 Assignment
September 22, 2008
Page 7
§9.9
17.
C(x) = 40 + x2
C 0 (x) = 2x
C 0 (5) = 10
C(6) − C(5) = 36 − 25 = 11
C 0 (5) = 10 tells us that the cost will increase approximately $10 for an additional unit,
while C(6) − C(5) = 11 tells us that the cost will actually increase by $11.
28.
R(x) = 46x
C(x) = 100 + 30x +
1 2
x
10
P (x) = R(x) − C(x)
1 2
x
10
P (100) = −100 + 1600 − 1000 = −$500
1
P 0 (x) = 16 − x
5
0
P (100) = 16 − 20 = −$4
= −100 + 16x −
P (101) − P (100) = 16 − 1020.1 + 1000 = $4.10
P 0 (100) = $4 tells us that the profit will increase approximately $4 for an additional unit,
while P (101) − P (100) = $4.10 tells us that the profit will actually increase by $4.10.
35. The monthly revenue is R = 300x, while the monthly costs are C = (160 + x)x, so the
monthly profits are:
P
= 300x − (160 + x)x
= 140x − x2
To maximize the profit, we need to set P 0 = 0 and solve for x, i.e.
P 0 = 0 = 140 − 2x
→
x = 70
Therefore, the company should produce 70 units each month.