MTH 121 — Spring — 2008
Essex County College — Division of Mathematics and Physics
Project #3 1 — Printed February 5, 2008
Name:
Signature:
Be neat, orderly, and answer every question. Answers without work will not receive credit.
1. Find all the critical numbers of the function.
g (x) = 4x + sin (4x)
Answer: First, the derivative.
g 0 (x) = 4 + 4 cos (4x)
Now find where g 0 (x) = 0.
g 0 (x) = 4 + 4 cos (4x)
0 = 4 + 4 cos (4x)
−1 = cos (4x)
So,
x=
π (2n + 1)
,
4
n∈Z
2. Find the local and absolute extreme values of the function on the given interval.
f (x) = x3 − 6x2 + 9x + 1,
[2, 4]
Answer: First the derivative.
f 0 (x) = 3x2 − 12x + 9 = 3 (x − 1) (x − 3)
Using the closed interval method, with these values x = 2, x = 3, and x = 4, we get a
maximum at (4, 5) ; and a minimum at (3, 1) .
1
This document was prepared by Ron Bannon using LATEX 2ε . Source and pdf are available by emailing a
request to [email protected].
1
Although not necessary, here’s the graph.
6
5
4
3
2
1
0
1
2
3
4
5
6
7
Figure 1: f (x) = x3 − 6x2 + 9x + 1,
8
9
[2, 4]
3. Find an equation of the line through the point (8, 16) that cuts off the least area from the
first quadrant.
Answer: Drawing a picture may help. Just take a point along the y-axis and x-axis,
(0, y) and (x, 0) respectively. The area is
1
A = xy.
2
Using slope to find a relationship between x and y.
m=−
16
y
=
x
8−x
⇒
y=
16x
x−8
Now we can rewrite A as a function of x.
1
16x
8x2
A (x) = x ·
=
2
x−8
x−8
Taking the derivative.
A0 (x) =
8x (x − 16)
(x − 8)2
Now analyze this derivative (use your head, where x = 0, x = 8 x = 16) we find a
minimization at x = 16. Finally using the relationship between x and y, we find y = 32
and the the slope is −2. Finally, the equation is
y = −2x + 32
2
4. Verify that the function satisfies the three hypotheses of Rolle’s Theorem on the given
interval. Then find all numbers c that satisfy the conclusion of Rolle’s Theorem.
2 2
f (x) = sin 3πx,
− ,
3 3
Answer:
(a) f (x) is continuous everywhere, hence is continuous on − 23 ,
2
3
.
(b) f (x) is differentiable everywhere, hence is differentiable on − 23 ,
(c) f − 32 = sin (−2π) = 0 = f 32 = sin (2π) = 0
The c’s in − 23 , 23 , such that
2
3
.
f 0 (x) = 3π cos 3πx = 0,
are,
1
1 1
1
, − , , −
6
6 2
2
√
5. Estimate the absolute maximum value of the function y = x 3x − x2 to two decimal
places on the interval [0, 3].
3.6
3.2
2.8
2.4
2
1.6
1.2
0.8
0.4
-0.4
0
0.4
0.8
1.2
1.6
2
2.4
2.8
3.2
-0.4
Figure 2: Graph of y = x
3
p
3x − x2 .
3.6
4
4.4
Answer:
p
3x − x2
1/2
3x3 − x4
−1/2
1
3x3 − x4
9x2 − 4x3
2
y = x
y =
y0 =
I’m not listing all the critical values here, mainly because the graph and the derivative
indicate that the maximum is occurring at x = 9/4.
r
r
√
9 27 81
9 27
27 3
y=
−
=
=
≈ 2.92283573777 ≈ 2.92
4 4
16
4 16
16
6. At 4:00 pm a car’s speedometer reads 29 miles per hour (mph). At 4:15 pm it reads 71
mph. At some time between 4:00 and 4:15 the acceleration is exactly x miles per hour per
hour. Find x.
Answer: Using the Mean Value Theorem , we have
x=
∆v
71 − 29
=
mph2 = 168 mph2
∆t
1/4
7. Find f .
f 0 (t) = 2t − 3 sin t,
f (0) = 5
Answer: Working backwards.
f 0 (t) = 2t − 3 sin t
f (t) = t2 + 3 cos t + C
f (0) = 3 + C = 5
f (t) =
⇒
C=2
t2 + 3 cos t + 2
8. Find a cubic function f (x) = ax3 − bx2 + cx − d that has a local maximum value of 112
at 1 and a local minimum value of -1,184 at 7.
Answer:
f (x) = ax3 − bx2 + cx − d
f (1) = a − b + c − d = 112
f (7) = 343a − 49b + 7c − d = −1184
f 0 (x) = 3ax2 − 2bx + c
f 0 (1) = 3a − 2b + c = 0
f 0 (7) = 147a − 14b + c = 0
4
Four equations with four unknowns, use Guassian elimination (matrices were taught in
MTH-119) to solve for a, b, c and d.
f (x) = 12x3 − 144x2 + 252x − 8
250
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
11
12
-250
-500
-750
-1000
-1250
-1500
-1750
Figure 3: Partial graph of f (x) = 12x3 − 144x2 + 252x − 8, with indicated local extrema.
9. Find the inflection points for the function.
f (x) = 8x + 3 − 2 sin x,
0 < x < 3π
Answer: You’ll need to find the second derivative and then find the point on f where the
second derivative changes sign on the interval 0 < x < 3π.
f (x) = 8x + 3 − 2 sin x
f 0 (x) = 8 + −2 cos x
f 00 (x) = 2 sin x
(π, 8π + 3) , (2π, 16π + 3)
5
80
60
40
20
-1
0
1
2
3
4
5
Figure 4: f (x) = 8x + 3 − 2 sin x,
10. Find the limit.
6
7
8
ex − 1
x→0 sin 7x
ex − 1
ex
1
= lim
=
x→0 sin 7x
x→0 7 cos 7x
7
lim
11. Evaluate the limit.
lim
x→0
Answer:
10
0 < x < 3π, with indicated points.
lim
Answer:
9
1 − cos x
x2 + x
1 − cos x
sin x
= lim
= 0
2
x→0 x + x
x→0 2x + 1
lim
12. Find the limit.
lim (1 − 10x)1/x
x→0
Answer: Here I am doing the following first.
y = (1 − 10x)1/x
1
ln (1 − 10x)
ln y =
x
1
y = e x ln(1−10x)
6
Now the limit.
lim (1 − 10x)1/x =
x→0
lim e
ln(1−10x)
x
x→0
lim
= ex→0
ln(1−10x)
x
lim
10
= ex→0 10x−1
=
e−10
13. Consider the following problem: A farmer with 800 feet of fencing wants to enclose a
rectangular area and then divide it into four pens with fencing parallel to one side of the
rectangle. What is the largest possible total area of the four pens?
Answer: Let the length of the outer enclosure be y feet and the width of the outer
enclosure be x feet. You’ll need three dividers, where I am taking them parallel to the side
of length x feet.
800 = 5x + 2y
800 − 5x
= y
2
A = xy
800 − 5x
A (x) = x
2
800x − 5x2
A (x) =
2
800
−
10x
A0 (x) =
2
You should note that x ∈ (0, 160), and the critical number is x = 80. The analysis the
derivative indicates that this maximizes A. So the maximal area is
A (80) =
800 · 80 − 5 · 802
= 16, 000 square feet .
2
14. Find the most general antiderivative of the function.
√
√
f (x) = 8 7 x − 10 9 x
Answer: Here f (x) = F 0 (x).
√
√
f (x) = 8 7 x − 10 9 x
f (x) = 8x1/7 − 10x1/9
F (x) =
7x8/7 − 9x10/9 + C
7
15. A particle is moving with the given data.
v (t) = sin t − cos t,
s (0) = 0
Find the position of the particle.
Answer:
v (t) = sin t − cos t
s0 (t) = sin t − cos t
s (t) = − cos t − sin t + C
s (0) = − cos 0 − sin 0 + C = −1 + C = 0
C = 1
s (t) =
1 − cos t − sin t
16. A rectangular beam will be cut from a cylindrical log of radius 10 inches. Suppose that
the strength of a rectangular beam is proportional to the product of its width and the
square of its depth. Find the dimensions of the strongest beam that can be cut from the
cylindrical log.
Answer: Let w be the width of the beam in inches; let d be the depth of the beam in
inches; let k be the proportionality constant; and let S be the strength of the beam.
400 = d2 + w2
d2 = 400 − w2
S = kwd2
S (w) = kw 400 − w2
S (w) = k 400w − w3
S 0 (w) = k 400 − 3w2
√
We have w ∈ (0, 20), so we only have one critical number to analyze, w = 20/ 3. The
analysis indicates that this value for w maximizes S. So the dimension of the beam is:
√
20
20 2
√ inches wide and √ inches wide.
3
3
8
17. Let P (x) and Q (x) be polynomials. Find
P (x)
x→∞ Q (x)
lim
if the degree of P (x) is 2 and the degree of Q (x) is 14.
Answer: The denominator far exceeds the numerator for large x, and as x → ∞ we have,
P (x)
= 0.
x→∞ Q (x)
lim
18. A painting in an art gallery has height h = 60 cm and is hung so that its lower edge is a
distance d = 17 cm above the eye of an observer (as seen in the figure below). How far from
the wall should the observer stand to get the best view? (In other words, where should
the observer stand so as to maximize the angle θ subtended at his eye by the painting?)
Answer: Let x be the distance between the the wall and the observer.
h+d
d
− arctan
x
x
77
17
= arctan
− arctan
x
x
1
−77
1
−17
=
2 · x2 −
2 · x2
1 + (77/x)
1 + (17/x)
−77
17
=
+
x2 + 772 x2 + 172
θ = arctan
θ0
√
Here, x ∈ (0, ∞), and the only critical number on this interval is x = 77 · 17. The
analysis of the derivative shows that this is in fact the value that maximizes θ. So the
viewer should stand
√
77 · 17 ≈ 36.18 cm
from the wall.
9
19. Graph
f (x) = xe−x
2
using a computer. You also need to indicate all important points2 of this graph.
Answer:
f (x) = xe−x
f 0 (x) = e−x
2
2
1 − 2x2
f 00 (x) = −2xe−x
2
3 − 2x2
0.5
-3
-2
-1
0
1
2
3
-0.5
Figure 5: Partial graph of f (x) with important points indicated.
The important points on f (x) are:
(a) The x and y intercept is (0, 0) .
(b) The global and local maximum is
1
1
√ , √
2
2e
.
1
1
√
√
.
(c) The global and local minimum is −
, −
2
2e
!
r r
3
3
(d) The points-of-inflection are
,
, (0, 0) ,
2
2e3
2
Extrema, intercepts, and points-of-inflection
10
r
−
!
r
3
3
, −
.
2
2e3