Curve Sketching Summary
October 6, 2008
Math 1100: Assignment #3 Solutions
Due: Friday Oct 3, 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.
§10.5
6.
3x − 1
x+5
3x − 1
lim
x→−∞ x + 5
lim
=
x→∞
lim
x→∞
3x
=3
x
= 3
and we see that y is undefined when x = −5, so there’s a VA at x = −5
8.
4x
9 − x2
4x
lim
x→−∞ 9 − x2
lim
=
x→∞
lim
x→∞
4x
=0
−x2
= 0
and we see that y is undefined when x = ±3, so there’s a VA at x = ±3
10.
6x3
x→∞ 4x2 + 9
6x3
lim
x→−∞ 4x2 + 9
=
lim
lim
x→∞
6x
=∞
4
= −∞
and we see that y is defined for all x, so there are no VA.
19. Step 2: HA & VA
lim y =
x→±∞
lim
x→±∞
x
=0
x2
and note that there is a VA at x = 1, so note that, because y > 0 for x > 0
lim y = lim y = ∞
x→1−
x→1+
Therefore, the HA is at x = 0 and the VA is at x = −1 with y → ∞ at the VA.
Page 1
Curve Sketching Summary
October 6, 2008
Page 2
Step 3: CP & Max/Min
y 0 = 0 when x = −1, and y 00 is undefined at x = 1. Therefore, we note the following:
x
x < −1 x = −1 −1 < x < 1 x = 1 1 < x
0
sign of y
−
0
+
DN E
−
y
↓
− 14
↑
DN E
↓
This tells us that there is a minimum at x = −1 and a corner at x = 1.
Step 4: POI & Curvature
y 00 = 0 when x = −2, and y 0 is undefined at x = 1. Therefore, we note the following:
x
x < −2 x = −2 −2 < x < 1 x = 1 1 < x
00
sign of y
−
0
+
DN E
+
y
CD
− 29
CU
DN E CU
Step 5: Sketch
f(x) of Question 19
10
f(x)
5
0
−5
−3
−2
−1
0
x
1
2
3
20. Step 2: HA & VA
lim y =
x→±∞
x2
=1
x→±∞ x2
lim
and note that there is a VA at x = 0, so note that, because y > 0 for all x
lim y = lim y = ∞
x→0−
x→0+
Therefore, the HA is at x = 1 and the VA is at x = 0 with y → ∞ at the VA.
Step 3: CP & Max/Min
y 0 = 0 when x = 1, and y 0 is undefined at x = 0. Therefore, we note the following:
x
x<0 x=0 0<x<1 x=1 1<x
sign of y 0
+
DN E
−
0
+
y
↑
DN E
↓
0
↑
This tells us that there is a minimum at x = 1 and a corner at x = 0.
Curve Sketching Summary
October 6, 2008
Page 3
Step 4: POI & Curvature
y 00 = 0 when x = 32 , and y 00 is undefined at x = 0. Therefore, we note the following:
x
x<0 x=0 0<x<
sign of y 00
+
DN E
+
y
CU DN E
CU
3
2
x=
0
3
2
1
9
3
2
<x
−
CD
Step 5: Sketch
f(x) of Question 20
10
f(x)
5
0
−5
−10
−8
−6
−4
−2
0
x
2
4
6
8
10
22. Step 2: HA & VA
lim y =
x→±∞
√
lim 3 3 x = ±∞
x→±∞
and note that there is a VA at x = 0, so note that
lim y = −∞
x→0−
lim y = ∞
x→0+
Therefore, there is no HA and the VA is at x = 0.
Step 3: CP & Max/Min
y 0 = 0 when x = ±1, and y 0 is undefined at x = 0. Therefore, we note the following:
x
x < −1 x = −1 −1 < x < 0 x = 0 0 < x < 1 x = 1 1 < x
sign of y 0
+
0
−
DN E
−
0
+
y
↑
−4
↓
DN E
↓
4
↑
This tells us that there is a minimum at x = 1, a maximum at x = −1, and a corner at
x = 0.
Step 4: POI & Curvature
3
y 00 = 0 when x = ±3 4 , and y 00 is undefined at x = 0. Therefore, we note the following:
3
x
x < −3 4
00
sign of y
+
y
CU
3
x = −3 4
0
3
3
−3 4 < x < 0 x = 0 0 < x < 3 4
−
DN E
+
CD
DN E
CU
3
x = 34
0
3
34 < x
−
CD
Curve Sketching Summary
October 6, 2008
Page 4
Step 5: Sketch
f(x) of Question 22
10
f(x)
5
0
−5
−10
−20
−15
−10
−5
0
x
5
10
15
20
23. Step 2: HA & VA
2
9x 3
lim y =
lim
=0
x→±∞
x→±∞ x2
and note that there is a VA at x = 0, so note that, because y > 0 for all x
lim y = lim y = ∞
x→0−
x→0+
Therefore, there is a HA at x = 0 and the VA is at x = 0 with y → ∞.
Step 3: CP & Max/Min
y 0 = 0 when x = 3, and y 0 is undefined at x = 0, 2. Therefore, we note the following:
x
x<0 x=0 0<x<2 x=2 2<x<3 x=3 3<x
0
sign of y
+
DN E
−
DN E
+
0
−
y
↑
DN E
↓
0
↑
1
↓
This tells us that there is a minimum at x = 2, a maximum at x = 0, 2.
Step 4: POI & Curvature
y 00 = 0 when x = 4.134, 1.866, and y 00 is undefined at x = 0, 2. Therefore, we note the
following:
x
x < 0 x = 0 0 < x < 1.866 x = 1.866 1.866 < x < 2 x = 2 2 < x < 4.134
sign of y 00
+
DN E
+
0
−
DN E
−
y
CU DN E
CU
CD
0
CD
x
x = 4.134 4.134 < x
sign of y 00
0
+
y
CU
Step 5: Sketch
f(x) of Question 23
20
f(x)
15
10
5
0
−5
−20
−15
−10
−5
0
x
5
10
15
20
Curve Sketching Summary
October 6, 2008
Page 5
24. Step 2: HA & VA
2
lim y =
x→±∞
2x 3
=0
x→±∞ x
lim
and note that there is a VA at x = −1, so note that
lim y = −∞
x→−1−
lim y = ∞
x→−1+
Therefore, there is a HA at x = 0 and the VA is at x = −1.
Step 3: CP & Max/Min
y 0 = 0 when x = 2, and y 0 is undefined at x = 0, −1. Therefore, we note the following:
x
x < −1 x = −1 −1 < x < 0 x = 0 0 < x < 2 x = 2 2 < x
0
sign of y
−
DN E
−
DN E
+
0
−
2
3
y
↓
DN E
↓
0
↑
2
↓
This tells us that there is a minimum at x = 0, a maximum at x = 2.
Step 4: POI & Curvature
y 00 = 0 when x = 4.12132, −0.12132, and y 00 is undefined at x = 0, −1. Therefore, we note
the following:
x
x < −1 x = −1 −1 < x < −0.12 x = −0.12 −0.12 < x < 0 x = 0 0 < x < 4.1
00
sign of y
−
DN E
+
0
−
DN E
−
y
CD
DN E
CU
CD
0
CD
x
x = 4.1 4.1 < x
sign of y 00
0
+
y
CU
Step 5: Sketch
f(x) of Question 24
20
f(x)
10
0
−10
−20
−10
−8
−6
−4
−2
0
x
2
4
6
8
10
Curve Sketching Summary
October 6, 2008
Page 6
§10.1
8.
y = x3 − 3x2 + 6x + 1
y0 = 0
→
y 0 = 3x2 − 6x + 6 = 3(x2 − 2x + 2)
√
¢
1¡
x = − 4 ± −12
2
There is no solution, so there are no critical points.
10. Because y 0 6= 0, then y 0 is either strictly positive or strictly negative. Note that
y 0 (0) = 6 > 0, therefore, y 0 > 0 for all x. This tells us that the function is increasing for
all x.
11.
a) The graph has one max at x = −1 and one min at x = 1 and no horizontal POI.
b) (and part c) We see that y 0 = 3x2 − 3, so the critical points are when 0 = 3x2 − 3.
This occurs when x = ±1.
c) The critical points agree with what we observed in part a.
14.
a) The graph has one horizontal POI at x = 2 and no CP’s.
b) (and part c) We see that y 0 = 3x2 − 12x + 12 = 3(x − 2)2 , so the critical points are
when 0 = x − 2. This occurs when x = 2.
c) The horizontal POI agree with what we observed in part a.
16.
y = x2 + 4x
y 0 = 2x + 4
a)
b)
y0 = 0 = x + 2
c)
y(−2) = −4
→
x = −2
d) Note that y 0 < 0 (y is decreasing) when x < −2 and y 0 > 0 (y is increasing) when
x > −2.
e) From part d, we know that x = −2 is a minimum and there are no HPOI.
18.
x4 x3
−
−2
4
3
= x3 − x2
y =
a)
b)
c)
y0
y 0 = 0 = x2 (x − 1)
→
25
y(0) = −2, y(1) = −
12
x = 0, 1
d) Note that y 0 < 0 (y is decreasing) when x < 1 and y 0 > 0 (y is increasing) when x > 1.
e) From part d, we know that x = 1 is a minimum and there is a HPOI at x = 0.
Curve Sketching Summary
October 6, 2008
Page 7
20.
2
a)
b)
c)
y = −(x − 3) 3
1
2
y 0 = − (x − 3)− 3
3
1
0
y = 0 = (x − 3)− 3
→
No solution, but y 0 DNE at x = 0
y(0) = 0
d) Note that y 0 > 0 (y is increasing) when x < 3 and y 0 < 0 (y is decreasing) when x > 3.
e) From part d, we know that x = 1 is a minimum and there are no HPOI (note that
there is a “corner” or vertex at x = 3).
32.
f (x) = (x2 − 4)2
f 0 (x) = 4x(x − 2)(x + 2)
Set f 0 (x) = 0 = 4x(x2 − 4)
√
x = 0, ± 2
→
x
x < −2 x = −2 −2 < x < 0 x = 0 0 < x < 2 x = 2 2 < x
sign of y 0
−
0
+
0
−
0
+
y
↓
0
↑
16
↓
0
↑
√
From this we see that x = ± 2 are minmums and x = 0 is a maximum.
33.
x3 (x − 5)2
27
2 (x − 3)(x − 5)
5x
f 0 (x) =
27
Set f 0 (x) = 0 = x2 (x − 3)(x − 5)
f (x) =
→
x = 0, 3, 5
x
x<0 x=0 0<x<3 x=3 3<x<5 x=5 5<x
sign of y 0
+
0
+
0
−
0
+
y
↑
0
↑
4
↓
0
↑
From this we see that x = 3 is a maximum, x = 5 is a minimum, and x = 0 is a HPOI.
36.
2
f (x) = x − 3x 3
1
0
f (x) =
x3 − 2
1
x3
1
0
Set f (x) = 0 = x 3 − 2
→
x=8
Curve Sketching Summary
October 6, 2008
Page 8
Note also that f 0 (x) is undefined at x = 0
x
x<0 x=0 0<x<8 x=8 8<x
0
sign of y
+
0
−
0
+
y
↑
0
↓
−4
↑
From this we see that x = 0 is a maximum, x = 8 is a minimum.
54.
1278900
x
1278900
a)
C 0 (x) = 2900 −
x2
441
Set C 0 (x) = 0 = 1 − 2
x = 21
x
C(x) = 2900x +
b) C 0 (x) < 0 for 0 < x < 21
c) C 0 (x) > 0 for x > 21
d) 21 Machines are needed to minimize the cost
e) The minimum average cost is: C(21) = $58, 000
§10.2
21.
f (x) = 3x5 − 20x3 = x3 (3x2 − 20) f 0 (x) = 15x2 (x − 2)(x + 2) f 00 (x) = 60x(x2 − 2)
Step 3: CP & max/min
Note that f 0 (x) = 0 when x = 0, ±2. Therefore, we note the following:
x
x < −2 x = −2 −2 < x < 0 x = 0 0 < x < 2 x = 2 2 < x
sign of f 0 (x)
+
0
−
0
−
0
+
f (x)
↑
64
↓
0
↓
−64
↑
This tells us that there is a minimum at x = 2, maximum at x = −2, and a HPOI at
x = 0.
Step 4: POI & Curvature
√
Note that f 00 (x) = 0 when x = 0, ± 2. Therefore, we note the following:
√
√
√
√
√ √
x
x<− 2 x=− 2 − 2<x<0 x=0 0<x< 2 x= 2
2<x
sign of f 00 (x)
−
0
+
0
−
0
+
f (x)
CD
CU
0
CD
CU
Curve Sketching Summary
October 6, 2008
Page 9
Step 5: Sketch
f(x) of Question 21
60
40
f(x)
20
0
−20
−40
−60
−5
−4
−3
−2
−1
0
x
1
2
3
4
5
22.
f (x) = x5 − 5x4 = x4 (x − 5)
f 0 (x) = 5x3 (x − 4)
f 00 (x) = 20x2 (x − 3)
Step 3: CP & max/min
Note that f 0 (x) = 0 when x = 0, 4. Therefore, we note the following:
x
x<0 x=0 0<x<4 x=4 4<x
0
sign of f (x)
+
0
−
0
+
f (x)
↑
0
↓
−256
↑
This tells us that there is a minimum at x = 4 and maximum at x = 0.
Step 4: POI & Curvature
Note that f 00 (x) = 0 when x = 0, 3. Therefore, we note the following:
x
x<0 x=0 0<x<3 x=3 3<x
00
sign of f (x)
−
0
−
0
+
f (x)
CD
0
CD
−162 CU
Step 5: Sketch
f(x) of Question 22
300
200
f(x)
100
0
−100
−200
−300
−5
−4
−3
−2
−1
0
x
1
2
3
4
5
Curve Sketching Summary
October 6, 2008
Page 10
23.
1
f 0 (x) =
f (x) = x 3 (x − 4)
4(x − 1)
3x
f 00 (x) =
2
3
4(x + 2)
5
9x 3
Step 3: CP & max/min
Note that f 0 (x) = 0 when x = 1 and is undefined at x = 0. Therefore, we note the
following:
x
x<0 x=0 0<x<1 x=1 1<x
sign of f 0 (x)
−
DN E
−
0
+
f (x)
↓
0
↓
−3
↑
This tells us that there is a minimum at x = 1.
Step 4: POI & Curvature
Note that f 00 (x) = 0 when x = −2 and is undefined at x = 0. Therefore, we note the
following:
x
x < −2 x = −2 −2 < x < 0 x = 0 0 < x
00
sign of f (x)
+
0
−
DN E
+
f (x)
CU
CD
0
CU
Step 5: Sketch
f(x) of Question 23
15
10
f(x)
5
0
−5
−10
−15
−5
−4
−3
−2
−1
0
x
1
2
3
4
5
Curve Sketching Summary
October 6, 2008
Page 11
24.
1
4
f 0 (x) =
f (x) = x 3 (x − 7)
7x 3 (x − 4)
3
f 00 (x) =
28(x − 1)
2
9x 3
Step 3: CP & max/min
Note that f 0 (x) = 0 when x = 0, 4. Therefore, we note the following:
x
x<0 x=0 0<x<4 x=4 4<x
0
sign of f (x)
+
0
−
0
+
f (x)
↑
0
↓
↑
This tells us that there is a minimum at x = 4 and a maximum at x = 0.
Step 4: POI & Curvature
Note that f 00 (x) = 0 when x = 1 and is undefined at x = 0. Therefore, we note the
following:
x
x<0 x=0 0<x<1 x=1 1<x
sign of f 00 (x)
−
DN E
−
0
+
f (x)
CD
0
CD
−6
CU
Step 5: Sketch
f(x) of Question 24
30
20
f(x)
10
0
−10
−20
−30
−8
−6
−4
−2
0
x
33. a) The rate of change of P with respect to time is:
dP
reaches its maximum
dt
c) The upper limit of production is point C
b) Point B is when
2
dP
dt
4
6
8
Curve Sketching Summary
October 6, 2008
Page 12
39. Given
S(t) = 3(t + 3)−1 − 18(t + 3)−2 + 1
dS
= 0, i.e. when
dt
a) The sales volume is maximized when
dS
= −3(t + 3)−2 + 36(t + 3)−3 = 0
dt
0 = −3(t + 3) + 36
→
∴
t = 9
b) Find the point of diminishing returns
d2 S
dt2
Set
= 6(t + 3)−3 − 108(t + 3)−4
d2 S
= 0 = t + 3 − 18
dt2
→
t = 15
The point of diminishing returns is at t = 15.
§10.3
2.
f (x) = x3 − 3x + 3
0
x ² [−3, 1.5]
2
f (x) = 3x − 3
∴
0
f (x) = 0
→
x = ±1
x f (x)
−3 −15
−1
5
1
1
1.5 1.875
Therefore, x = −3 is the absolute minimum and x = −1 is the absolute maximum.
10. We want to maximize revenue. We know that
x = # of $10 increases in rent
0≤x≤9
# of units sold = 50 − x
Monthly rent = 360 + 10x
R(x) = # of units sold × Monthly rent
= (50 − x)(360 + 10x)
To maximize the revenue, set R0 (x) = 0 and solve for x
R0 (x) = 0 = −(360 + 10x) + 10(50 − x)
140
→
x =
=7
20
Therefore the rent should be $430.
Curve Sketching Summary
October 6, 2008
Page 13
14. The average revenue is
R=
R(x)
x
= 2800 + 8x − x2
0
a) To maximize the avg revenue, we set R (x) = 0 and solve for x:
0
R (x) = 0 = 8 − 2x
→
x=4
b) Note that
M R(x) = R0 (x) = 2800 + 16x − 3x2
Set R(x) = M R(x) to get:
2800 + 8x − x2 = 2800 + 16x − 3x2
2x2 − 8x = 0
→
→
x = 0, 4
Therefore, we see that R(x) attains it’s maximum when R(x) = M R(x).
16.
C(x) = 300 + 10x + 0.03x2
C(x)
300
→
C(x) =
=
+ 10 + 0.03x
x
x
0
Set C (x) = 0 = −300x−2 + 0.03
→
x = 100
C(100) = 3 + 10 + 3 = 16
26.
1
P (x) = 6400x − 18x2 − x3 − 40000
3
Set P 0 (x) = 0 = 6400 − 36x − x2 = (x + 100)(x − 64)
→
x = 64
P (64) = $208, 490.67
28.
P (x) = R(x) − C(x) = −800 + 600, 000x − 150x2 − x3
Set P 0 (x) = 0 = 200000 − 100x − x2 = (500 + x)(400 − x)
→
x = 400
P (400) = $151, 999, 200
32. a) At point C, the profit will be largest
b) At point C, the slope of the tangents are equal
c) The marginal cost is equal and opposite to the marginal revenue at the maximum
profit
Curve Sketching Summary
October 6, 2008
Page 14
35.
C(x) = 45000 + 100x + x2
R(x) = 1600x
P (x) = −45000 + 1500x − x2
Set P 0 (x) = 0 = 1500 − 2x
→
x = 750
but x ≤ 600, so x = 600 is the max allowed
P (600) = 495000
46. a) The trade deficit reached a minimum around Feb ’92 and was approx $3.5B
b) The average trade deficit reached a minimum around Feb ’92 and was approx $5.5B
c) The trade deficit reached a maximum around July ’93 and was approx $12B
d) The average trade deficit reached a maximum around Dec ’93 and was approx $9.75B
§10.4
5.
1
y = 70t + t2 − t3
2
y 0 = 70 + t − 3t2
√
−1 ± 1 + 12 · 70
→
t =
−6
−1 ± 29
=
−6
14
= − ,5
3
25
− 125 = 237.5
y(5) = 350 +
2
y(0) = 0
Set y 0 = 0
y(8) = 560 + 32 − 512 = 80
Therefore, after 5 hrs the hourly units will be maximized at 237.5 units/hr.
8.
870000
x
870000
C 0 = 25000 −
x2
870000
x =
= 34.8
25000
C = 25000x +
→
34.8 units are required to minimize the cost.
Curve Sketching Summary
October 6, 2008
Page 15
12.
50t
+ 36
50(t2 + 36) − 50t(2t)
R0 (t) =
(t2 + 36)2
−50t2 + 1800
=
(t2 + 36)2
→
t = 6
R(t) =
Set R0 (t) = 0
t2
R(6) = 4.167
After 6 weeks the revenue will be maximized at $4.167M
15. The total amount of area enclosed and the total amount of fence is given by
A = 2xy = 2400
P
→
y=
= 2x + 2y + 2y + x = 3x + 4y
1200
x
We want to minimize the perimeter with respect to the dimensions of the rectangles.
Using A to eliminate y from the equation for P gives:
P (x) = 3x +
4800
x
To minimize the perimeter, we set P 0 (x) = 0 and solve for x, i.e.
P 0 (x) = 0 = 3 −
→
x = 40
4800
x2
Therefore, minimum amount of fence required is:
P (40) = 120 + 120 = 240
18. Similar to the previous question, we are given:
13500
x
C = $5(2x + 4y) + $2(x)
A = 13500 = xy
y=
= $12x + $20y
270000
= 12x +
x
270000
0
Set C (x) = 0 = 12 −
x2
→
x = 150
→
y = 90
Curve Sketching Summary
October 6, 2008
Page 16
24. Let
x = # of items produced for every production run
n = # of production runs
60000
xn = 60000
n=
x
3
C = 400n + 4 · 60000 + x
8
24000000
3
=
+ 240000 + x
x
8
24000000 3
0
Set C (x) = 0 = −
+
x2
8
→
x = 8000
Therefore, each run will produce 8000 units to minimize the cost.
30. We know that
V = 4000 = x2 y
→
y=
A = x2 + 4xy
16000
= x2 +
x
16000
0
Set A = 0 = 2x −
x2
→
x = 20
→
4000
x2
y = 10
a) The dimensions that will minimize the material is x = 20, y = 10.
b) Doing the same as above, but with 4000 = k instead gives us:
V = k = x2 y
→
A = x2 + 4xy
4k
= x2 +
x
4k
Set A0 = 0 = 2x − 2
x
→
x = (2k)
1
3
→
y=
k
x2
µ ¶1
k 3
y=
2 =
4
(2k) 3
k