Fall 2004 Math 151
Exam 3B: Solutions
c 2004, Art Belmonte
Mon, 06/Dec
1. (b) We have f 0 (x) =
get position, resolving constants along the way.
v 0 (t) = a(t) =
v(t) = 3t 2 + C
1 = v(0) = 0 + C
2
1
1
=
.
(2) =
2x + 4
2 (x + 2)
x +2
C
s (t) = v(t)
0
2. (d) The form is 0/0. Apply L’Hospital’s Rule (twice).
2
ex − 1 − x 2
lim
x→0
x3
=
=
L’H
=
e x (2x) − 2x
lim
x→0
3x 2
2
2 ex − 1
lim
3x
x→0
2
x
2 e (2x)
=0
lim
3
x→0
t3 + t + K
0+0+ K
=
=
1
t3 + t + 1
K
s(t)
s(1) =
9. (d) We have f 0 (x) = 1 − p
1+1+1= 3
1
on (−1, 1). Therefore,
1 − x2
f 0 (x) = 0 when x = 0 ∈ (−1, 1). Apply the Closed Interval
Method on [−1, 1]. Compute function values of f at the
critical value and at the endpoints.
x
f (x) = x − sin−1 x
COMMENT
−1 −1 − − π2 ≈ 0.57 absolute maximum
0
0+0=0
—
1
1 − π2 ≈ −0.57
absolute minimum
−∞
ln x
3. (b) As x → 0+ , we have √ → + = −∞.
0
x
(L’Hospital’s Rule is not applicable!)
4. (b) We have cos−1 (cos (3π)) = cos−1 (−1) = π.
5. (b) Let θ = tan−1 (3). Then cos (2θ ) = cos2 θ − sin2 θ
2 2
1 − 9 = − 8 = −4.
= √1
− √3
= 10
10
10
5
10
= 1
= 3t 2 + 1
s(t) =
1 = s(0) =
2
L’H
6t
10. (d) H is increasing where H 0 > 0; that is, on (−1, 1).
10
11. (a) Since H 0 changes sign from − to 0 to + as x increases
through x = −1, the first derivative test says that H has a
local minimum at x = −1.
1/2
10
3
12. (d) Compute f 0 (x) = (1) e x + xe x = (x + 1) e x and
f 00 (x) = (1) e x + (x + 1) e x = (x + 2) e x . Solve f 00 (x) = 0
to obtain x = −2. Since f 00 changes sign as x increases
through −2, the point of inflection is −2, −2e−2 .
θ
1
x2 + 1
= 1 + x −2 . Now compute an
x2
x2 − 1
1
.
antiderivative: F(x) = x − x −1 = x − =
x
x
6. (e) Expand: f (x) =
13. (d) Resolve the absolute value, then differentiate.
f (x) = x − |2x − 1|
=
f (x)
=
2
7. (c) GivenF is an antiderivative
of f , we have F 0 (x) = f (x).
1 0
d
F (3x)
= F (3x) · 3 = F 0 (3x) = f (3x), we
Since
dx
3
3
F (3x)
is an antiderivative of f (3x).
see that
3
0
f (x) =
x 2 − (− (2x − 1)) ,
x 2 − (2x − 1) ,
x 2 + 2x − 1,
x 2 − 2x + 1,
2x + 2,
2x − 2,
x<
x>
x<
x≥
x<
x≥
1
2
1
2
1
2
1
2
1
2
1
2
(Note: f is not differentiable at x = 12 since its graph is sharp
or kinked there. See plot on next page.) Now f 0 (x) = 0 for
x = ±1. Therefore, the critical points of f are x = −1, 12 , 1.
8. (d) Recall that velocity is the derivative of position and
acceleration the derivative of velocity. We antidifferentiate
acceleration to get velocity, then antidifferentiate velocity to
1
2 ln 1 − 1 x −1
3
• Then ln y = 2x ln 1 − 13 x −1 =
,
x −1
which is an indeterminate quotient 0/0 as x → ∞.
Problem 13
1.5
1
0.5
• We may therefore apply L’Hospital’s Rule.
y
0
−0.5
lim ln y
−1
x→∞
=
lim
2 ln 1 − 13 x −1
x→∞
−1.5
2
−2
−2.5
−2 −1.5 −1 −0.5
L’H
0
x
0.5
1
1.5
=
2
=
14. Since the bacterial culture grows at a rate proportional to its
size, it experiences exponential growth. Let y be the number
of bacteria at time t. Then y = y0 ekt = 1,500ekt .
k
y
!
1
1−
1
3x
= e2k
= 2k
=
1
2
17.
ln 28
15
= 1,500e
1
2
ln 28
15 t
= 1,500
• After 6 hours there are y = 1,500
28
15
28
15
(i) The function increases where f 0 (x) =
that is, for x < 0 or (−∞, 0).
t/2
1 −2
3x
2
=− .
3
−4x
x2 + 3
2 > 0;
(ii) Next, f is concave up where
3
f 00 (x) =
≈ 9,756
bacteria.
15.
2
lim
x→∞ −3
1 x −1
3
−x −2
1 2x
is given by
• Hence our limit lim 1 −
x→∞
3x
lim y = lim eln y = elim ln y = e−2/3 .
2,800 = 1,500e2k
ln
x→∞
1−
!
• After 2 hours there are 2,800 bacteria.
28
15
28
15
lim
x −1
1
12x 2 − 12
3 > 0;
x2 + 3
i.e., for |x| > 1 or (−∞, −1) ∪ (1, ∞).
(i) Recursively apply the Chain Rule.
d −1 −x f tan
e
U 0 (x) =
dx
1
−x
= f 0 tan−1 e−x ·
2 · e (−1)
1 + e−x
f 0 tan−1 e−x
= −
e x + e−x
(iii)
• Moreover, since f 0 > 0 for x < 0 and f 0 < 0 for
x > 0, we conclude that f has an absolute
maximum at x = 0 by the first derivative test for
absolute extrema.
• The first derivative of f is defined everywhere
from the statement of the problem. Yet nowhere
does f 0 change sign from − to 0 to +. Therefore,
f has no local minimum.
(ii) Use logarithmic differentiation.
V (x)
=
(2 + cos x) f (x)
ln V (x)
1
V 0 (x)
V (x)
=
f (x) ln (2 + cos x)
=
1
(− sin x)
f 0 (x) ln (2 + cos x) + f (x)
2 + cos x
f (x) sin x
(2 + cos x) f (x) f 0 (x) ln (2 + cos x) −
2 + cos x
V 0 (x)
=
1 2x
16. As x → ∞, the expression 1 −
is seen to be an
3x
indeterminate power, 1∞ .
1
• Let y = 1 −
3x
2x
• Since f 0 changes sign from + to 0 to − as x
increases through x = 0, the first derivative test
says that f has a local maximum at x = 0.
• Since f 0 > 0 for x < 0 and f 0 < 0 for x > 0, we
conclude that f has no absolute minimum. This is
because f is always decreasing as we go further to
the right of x = 0 (and the same is true as we go
further to the left of x = 0).
(iv) The function f has inflection points at x = ±1 since
f 00 changes sign there.
2x
= 1 − 13 x −1 .
18.
2
(i) Let y be the height of the window’s rectangular portion
and r be the radius of the semicircle. Here is a diagram.
πr
Problem 18
70
Area A (square feet)
60
r
y
y
2r
(ii) The perimeter P of the window is 30 feet.
50
40
30
20
10
1
2
P
=
30
(2πr ) + 2r + 2y
(π + 2) r + 2y
y
=
=
30
30
=
1
2
0
(iii-alt)
(30 − (π + 2) r )
The area A of the window is
A
=
A
=
A
=
domain
:
1 πr 2 + 2r y
2
1
2
2 πr + r (30 − (π
30r − 2 + π2 r 2
0≤r ≤
30
.
π +2
+ 2) r )
[Fast track?
30
, we have y = 0 and the
π +2
window is semicircular. There is no rectangular part.]
(iii) Use the Closed Interval Method for Absolute Extrema.
(That’s why we allowed r = 0 above. A window having
width zero has area zero.)
=
r
=
1
2
3
4
Radius r (feet)
5
6
Note that A = 30r − 2 + π2 r 2 is quadratic. Its
graph is therefore a downward opening parabola
(since the coefficient of r 2 is negative). Hence the
absolute maximum value of A occurs at the
parabola’s vertex; i.e., where A0 = 0.
A0
=
r
=
30 − (4 + π) r = 0
30
≈ 4.2 ft
π +4
For this value of r , we find that y is also equal to
30
≈ 4.2 ft.
π +4
See right!%]
[NOTE: When r =
A0
0
30 − (4 + π) r = 0
30
≈ 4.2 ft
π +4
• Crank out function values of A at this interior
r -value and at the endpoints of the domain of A.
r
A
COMMENT
0
0
No width: zero area!
30
450
Maximum area
π +4
π +4
30
450π
Semicircular region
π + 2 (π + 2)2
450π
450
≈ 63.01 and
≈ 53.48,
π +4
(π + 2)2
450
≈ 63 ft2 .
we see that the maximum area is
π +4
30
≈ 4.2 ft. Here is
This occurs when r = y =
π +4
a plot of A versus r .
• Since
3