Math 1100-6 Assignment
September 8, 2008
Page 1
Math 1100: Assignment #1
Due: Friday Sept 5, 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.
§9.1
8.
lim f (x) = ∞
lim f (x) = −∞
x→2−
12.
f (x)
1.98019802
1.998001998
1.99980002
DNE
2.00020002
2.002002002
2.02020202
lim f (x) = DN E
x→2
x→2+
f (c) = DN E
x
-0.51
-0.501
-0.5001
From the table, we see that limx→−0.5 f (x) = 2.
-0.5
-0.4999
-0.499
-0.49
18.
lim (2x3 − 12x2 + 5x + 3) = 2 · 33 − 12 · 32 + 5 · 3 + 3
x→3
= −36
22.
x2 − 16
x→−4 x + 4
lim
(x − 4)(x + 4)
x+4
lim (x − 4)
=
lim
x→−4
=
x→−4
= −8
26.
x2 − 8x − 20
x→10 x2 − 11x + 10
lim
(x − 10)(x + 2)
(x − 1)(x − 10)
x+2
= lim
x→10 x − 1
4
=
3
=
lim
x→10
28.
lim f (x) =
x→5−
lim f (x) =
x→5+
∴
lim (7x − 10) = 25
x→5−
lim 25 = 25
x→5+
lim f (x) = 25
x→5
Math 1100-6 Assignment
September 8, 2008
Page 2
30.
x3 − 4
= −4
x→2−
x→2− x − 3
3 − x2
1
lim f (x) = lim
=−
+
+
x
2
x→2
x→2
lim f (x) =
∴
46.
lim
lim f (x) = DN E
x→2
a) limx→2 f (x) = 5 − limx→2 g(x) = −6
b) limx→2 [f (x)]2 − [g(x)]2 = limx→2 (f (x) + g(x))(f (x) − g(x)) = 5(−17) = −85
3g(x)
33
c) limx→2
=
f (x) − g(x)
−17
52.
58.
53
9
+ 10 + 1 =
4
4
9
5
134
+ 10 + =
b) limx→10 S(x) =
10
2
10
a) limx→4+ S(x) =
730000
− 7300 = 29200
100 − 80
b) limp→100− C(p) = ∞
a) limp→80 C(p) =
c) Part b means that the cost to remove 100% of the pollutants increases substantially
as you try to remove more pollutants.
§9.2
4. The function is continuous at x = 3 because there is a hole at x = −3, while the rest of
the function is f (x) = x − 3.
6. Note that f (x) simplifies down to f (x) = x + 2, so limx→2− f (x) = limx→2+ f (x) = 4,
however, f (2) is undefined (using the original equation). Therefore, there is a hole in the
function at x = 2 and the function is not continuous.
8.
lim f (x) =
x→1−
lim f (x) =
x→1+
∴
lim x2 + 1 = 2
x→1−
lim 2x2 − 1 = 1
x→1+
lim f (x) = DN E
x→2
Because the left and right limits are not equal, the function is not continuous.
10. The function is continuous everywhere because the function is defined everywhere.
14. We see that the denominator of the function is always positive, so the function has no
vertical asymptotes or holes.
Math 1100-6 Assignment
September 8, 2008
Page 3
16. Both branches of the function are continuous (polynomials) and we see that
limx→1 f (x) = f (1) = 2.
22. We see that f (x) is undefined at x = 2 and it blows up near x = 2, therefore, the
function is not continuous.
24. Although f (x) → ∞ at x = 2, we see that the function is undefined at x = 2, so the
function is not continuous.
26. limx→−∞
30. limx→∞
4
4
= limx→−∞ 2 = 0. This tells us that the horizontal asymptote is 0.
x2 − 2x
x
4x2 + 5x
4x2
= 4, therefore, the horizontal asymptote is 4
=
lim
x→∞
x2 − 4x
x2
8100p
, the only discontinuity is at p = 100. Note that
100 − p
limp→100− C(p) = ∞, which means that the cost of removing pollutants is very expensive
if all the pollutants must be removed.
46. Given C(p) =
100C
, if cost were not an issue, then we would pay up the wazoo to
8100 + C
remove as many pollutants as possible (i.e. we’d let C → ∞). Note that
100C
limC→∞ p = limC→∞
= 100. This tells us that we can remove all impurities.
C
48. Given p =
§9.3
2. Given f (x) = 6 − x − x2 , we find that
f (2) − f (−1)
= −2
2 − (−1)
f (10) − f (1)
= −12
10 − 1
6. Using f (x) = x2 + 3x + 7, note that
0.69
f (2) − f (1.9)
=
= 6.9
0.1
0.1
f (2) − f (2.1)
−0.071
=
= 7.1
−0.1
−0.1
f (2) − f (1.99)
0.0699
=
= 6.99
0.01
0.01
f (2) − f (2.01)
−0.0701
=
= 7.01
−0.01
−0.01
Therefore, f 0 (2) ≈ 7.
8. Parts a and b ask the same thing... i.e. f 0 (6) = 14. c) f (6) = 141.
10.
− 21 (x + h)2 + 21 x2
h→0
h
−2xh + h2
= lim
h→0
2h
= −x
lim f (x + h) − f (x)h =
h→0
Using this we see that f 0 (2) = −2.
lim
Math 1100-6 Assignment
September 8, 2008
Page 4
12. Points P and A are, (1,1) and (0,-1) respectively. Parts b, c, and d, are all asking for the
same thing... i.e.
slope = f 0 (1) =
31.
1 − (−1)
=2
1
a) Slope is positive in intervals: a, b, d
b) Slope is negative in intervals: c
c) Rate of change is zero at: A, C, E
32.
a) Slope changes from positive to negative at: A, E
b) Slope changes from negative to positive at: C
33. The function is continuous at A, B, C, D, G, H, I, J, while the function is differentiable at
A, C, D, G, I, J.
40. Given S(x) = 20 + 64x − 16x2 , the average velocity in the...
a) ...first two seconds is:
S(2) − S(0)
2−0
20 + 128 − 256 − 20
=
2
= 32f t/s
S =
b) ...next two seconds is:
S(4) − S(2)
4−2
20 + 256 − 256 − 20 − 128 + 256
=
2
= −32f t/s
S =
46. Given R(x) = 36x − 0.01x2
a) Marginal revenue:
R(x + h) − R(x)
h
36(x + h) − 0.01(x + h)2 − 36x + 0.01x2
= lim
h→0
h
36h − 0.02xh − 0.01h2
= lim
h→0
h
= 36 − 0.02x
M R(x) =
lim
h→0
b,c,d) M R(600) = 24 (you’ll get $24 more for every unit produced), M R(2000) = −4
(you’ll loose $4 for every new unit produced), M R(1800) = 0 (there will be no
change in revenues).
Math 1100-6 Assignment
September 8, 2008
Page 5
48. Given E(p) = 10000p − 100p2 , the instantaneous rate of change of E(p) with respect to p
is:
E(p + h) − E(p)
h
10000(p + h) − 100(p + h)2 − 10000p + 100p2
= lim
h→0
h
10000h − 200ph − 100h2
= lim
h→0
h
= 10000 − 200p
E 0 (p) =
lim
h→0
Therefore, E 0 (5) = 9000 and E 0 (20) = 6000.
.