6. (d) Given g (x) = sin−1 x, we have
−1/2
1
= 1 − x2
g 0 (x) = p
1 − x2
−3/2
g 00 (x) = − 12 1 − x 2
(−2x) =
Fall 2005 Math 151
Exam 3A: Solutions
c
Mon, 05/Dec
2005,
Art Belmonte
1. (b) In English, sin−1
whose sine is
√
3
2 .”
√ 3
2
is “the angle (between ± π2 )
This angle is
π
3
g 00
radians or 60◦ .
2
G (x) = x 4 + 12 e x is an antiderivative of the stated function
2
g (x) = 4x 3 + xe x .
3. (e) Let y = x x . Use logarithmic differentiation.
=
x ln x
=
(1) ln x + (x)
=
x x (1 + ln x)
x3
L’H
=
=
=
lim
1
x
−1
1−x
+1+x
3x 2
−1 + 1 − x 2
lim
x→0 3x 2 (1 − x)
1
−1
=−
lim
3
x→0 3 (1 − x)
0
−1
−2
−3
Curve
Inflection points
−4
−2
C
dr (t)
= v (t)
dt
= [0, 1]
h
i
= t, t 2 + 1
i
h
1 2 1 3
t
,
t
+
t
+K
=
2
3
r (t)
r (1)
−1
0
x
1
2
8. (b) The Mean Value Theorem (MVT) guarantees that there
is a number c ∈ (1, 3) such that
f 0 (c) =
f (3) − f (1)
8−5
3
f (b) − f (a)
=
=
= .
b−a
3−1
2
2
9. (a) The Extreme Value Theorem (EVT) guarantees that the
continuous polynomial f (x) = x 3 − 12x + 5 does indeed
have absolute maximum and absolute minimum values on the
closed interval [−5, 3].
= [1, 2t]
i
h
= t, t 2 + C
= [0, 0] + C
[−1, 1] = r (0)
K
0
0, ±1
X3A, P7
x→0
[0, 1] = v (0)
r (t)
2
1
5. (c) Recall that velocity is the derivative of position and
acceleration the derivative of velocity. Antidifferentiate
acceleration to get velocity, then antidifferentiate velocity to
get position, resolving constants along the way.
v (t)
3
4
• As x increases through −1, g 00 changes from + to −.
• As x increases through 1, g 00 changes from − to +.
• However, as x increases through 0, g 00 stays negative.
That is, it does NOT change sign.
• Accordingly, the two inflection points of g are x = ±1.
• Here is a graph that shows the inflection points.
[Alternatively, you may use L’Hospital’s Rule three times.]
dv (t)
= a (t)
dt
3/2
y
x→0
ln (1 − x) + x + 12 x 2
1
2
=
40x 2 (x − 1) (x + 1) =
x =
4. (b) Use L’Hospital’s Rule and some algebra.
lim
1
2
3/2
1 − x2
√
1
4
4 3
2
= √ 3 = √ =
9
3 3
3
7. (a) The inflection points of g (x) = 23 2x 6 − 5x 4 are
values of x at which g 00 changes sign. This may occur where
g 00 is either zero or undefined. Since g is a polynomial, g 00
exists for all real numbers. So let’s find where g 00 = 0 to
obtain our candidates.
g 0 (x) = 23 12x 5 − 20x 3
g 00 (x) = 23 60x 4 − 60x 2 = 40x 2 x 2 − 1 = 0
2. (c) G is an
G 0 (x) = g (x).
antiderivative
of g provided
2
2
2
d
1
1
4
x
3
x
Since d x x + 2 e
= 4x + 2 e (2x) = 4x 3 + xe x ,
ln y
1 dy
y dx
dy
dx
x
• Find the critical values of f . These are values of x for
which f 0 is either zero or undefined. The derivative of
f , another polynomial, exists for all real numbers. So
we are looking for values of x for which f 0 (x) = 0.
f 0 (x) = 3x 2 − 12 =
=
3 x2 − 4
= [0, 0] + K
= [−1, 1]
h
i
1 t 2 − 1, 1 t 3 + t + 1
=
2
3
h
i
= − 12 , 73
3 (x − 2) (x + 2) =
x =
0
0
0
−2, 2
Hence the critical numbers of f are x = −2, 2. Both of
these x-values lie in the interval [−5, 3].
1
• f 00 (x) > 0 on (0, 2) indicates where f is concave up.
• Now crank out function values of f at the critical
values along with the endpoints of the closed interval.
• f 00 (x) < 0 on (2, 4) shows where f is concave down.
f (−5) = −125 + 60 + 5 = −60
f (−2) = −8 + 24 + 5 = 21
f (2) = 8 − 24 + 5 = −11
f (3)
These facts enable us to sketch a graph of f .
X3A / P13
= 27 − 36 + 5 = −4
y
The absolute minimum is f (−5) = −60.
• Here is a graph that corroborates this assertion.
X3A / P9 + P10
30
0
20
1
2
x
10
0
14.
y
−10
absolute maximum
−20
dy
dx
absolute minimum
−50
−60
−4
−2
0
dy
=
dx
11. (c) The area under the curve y = f (x) = 1 + x 3 above the
x-axis between x = a = 2 and x = b = 6 may be formulated
as the limit of a Riemann sum that uses subintervals of equal
length and function values at right endpoints. Let’s construct
this limit of a sum one step at a time.
xi
f (x i )
=
=
area =
area =
i=1
n
4i 3
4X
1+ 2+
lim
n→∞ n
n
5x 4
dy
= p
.
dx
2 x5 + 4 x5 + 5
(c) Use logarithmic differentiation.
!
12. (b) Use the law of exponential growth.
9 =
ln 9 =
k =
y
=
y
=
ln y
=
1 dy
y dx
dy
dx
=
=
x + x2
tan x
(tan x) ln x + x 2
1 + 2x
sec2 x ln x + x 2 + (tan x)
x + x2
tan x (1 + 2x) tan x x + x2
sec2 x ln x + x 2 +
x + x2
15. Velocity is the derivative of position and acceleration the
derivative of velocity. Antidifferentiate acceleration to get
velocity, then antidifferentiate velocity to get position,
resolving constants along the way. Deceleration signifies
negative acceleration.
i=1
y =
9000 =
−1/2 1 5
4
x
5x
·
+
4
p
2
2
1+
x5 + 4
1
or
b−a
6−2
4
=
=
n
n
n
4i
a + i 1x = 2 +
n
3
4i
1+ 2+
n
n
X
lim
f (x i ) 1x
n→∞
1
1
1
√ · x −1/2
√ ·
ln 1 + x 1 + x 2
or
10. (d) From the analysis in Problem 9, we see that the absolute
maximum of f on [−5, 3] is f (−2) = 21.
=
=
√ x , we have
1
dy
= √
√ √ .
dx
2 x 1 + x ln 1 + x
p
(b) Given y = tan−1 x 5 + 4, we have
2
x
1x
4
(a) Given y = ln ln 1 +
−30
−40
3
dv
=a =
dt
v =
88 = v (0) =
y0 ekt
1000e2k
e2k
2k
1 ln 9 = ln 91/2 = ln 3
2
t
1000et ln 3 = 1000 eln 3 = 1000 (3)t
C
dr
=v
dt
r
0 = r (0)
After 3 hours, the number of bacteria is 1000 (3)3 = 27,000.
−40t + C
0+C
=
88
=
88 − 40t
= 88t − 20t 2 + K
= 0+K
K =
r (t) =
13. (c) Consider the geometrical meaning conveyed by the
information given about the first and second derivatives.
−40
0
88t − 20t 2 or 4t (22 − 5t)
When the car has stopped, its velocity is zero. Therefore
22
0 = v = 88 − 40t, whence t = 88
40 = 10 = 2.2 s. At this
time the car’s position is r (2.2) = 8.8 (11) = 96.8 ft.
• f 0 (x) > 0 on (1, 3) tells where f is increasing.
• f 0 (x) < 0 on (0, 1) ∪ (3, 4) tells where f is decreasing.
2
18. Let f (x) = x 3 − 3x 2 + 3.
16. Let x be the width of the bottom of the box and h be the
height of the box.
(a) The critical points of f are where f 0 is zero or
undefined. Since f is a polynomial, its derivative is
defined for all values of x. So let’s determine where f 0
is zero. We have
• Here is diagram depicting the situation.
f 0 (x) = 3x 2 − 6x = 3x (x − 2) = 0
h
2x
whence x = 0, 2.
x
(b) Use the First Derivative Test to classify each critical
point as being the location of a local maximum, local
minimum, or neither.
• The volume of the box is
V = (2x) (x) (h) = 2hx 2 = 36,000,
whence h =
18,000x −2,
• As x increases through 0, the sign of f 0 changes
from + to −. This signifies a local maximum at
x = 0.
• As x increases through 2, the sign of f 0 changes
from − to +. This signifies a local minimum at
x = 2.
x > 0.
• The surface area is S = 2x 2 + 2hx + 2 (2x) h or
S = 2x 2 + 6hx = 2x 2 + 108,000x −1, x > 0.
• Thus S 0 = 4x − 108,000x −2 = 0. So x 3 = 27,000
whence x = 30. Hence x = 30 cm is the width of the
base that ostensibly minimizes the total surface area.
(c) To find the inflection points of f , determine where the
second derivative changes sign.
• Let’s verify this assertion. The second derivative of S is
S 00 = 4 + 216,000x −3 > 0 for all x > 0. That is, S is
concave up on its domain. Accordingly, the absolute
minimum of occurs when x = 30 cm. [This minimum
is S (30) = 5,400 cm2 .] The corresponding length is
18,000
= 20 cm.
2x = 60 cm and the height is h =
900
• Here is a graph of S versus x. The problem is readily
solved interactively using the graphing analysis
capabilities of a TI-89. This gives an intuitive and
independent way of checking your [home]work!
• Now 0 = f 00 (x) = 6x − 6, whence x = 1.
• As x increases through 1, the sign of f 00 changes
from − to +. Since the second derivative changes
sign, x = 1 is an inflection point of f .
These assertions are corroborated by the following graph.
X3A, P18
10
5
6000
0
y
Surface area S (cm3)
X3A / P16
6500
5500
5000
20
25
30
Width x (cm)
1/x 2
17. The limit lim (cos 2x)
1∞ .
35
40
−10
−2
involves an indeterminate power,
x→0
2
• Let y = (cos 2x)1/x .
ln (cos 2x)
,
x2
an indeterminate quotient 0/0 as x → 0.
• Therefore,
• Then ln y = (1/x 2 ) (ln (cos 2x)) =
lim
ln (cos 2x)
x2
L’H
=
L’H
=
lim ln y
x→0
=
lim
−2 sin 2x/ cos 2x
−2 tan 2x
= lim
2x
2x
x→0
lim
−4 sec2 2x
2
x→0
x→0
−2.
• Hence lim (cos 2x)1/x = lim y = lim eln y = e−2 .
2
x→0
x→0
−1
0
1
x
2
3
4
• NOTE: Instead of using the First Derivative Test in part
(b), we may use the Second Derivative Test. Observe
that f 00 (x) = 6x − 6 is continuous. Since f 0 (0) = 0
and f 00 (0) = −6 < 0, we conclude that f has a local
maximum at x = 0. Similarly, since f 0 (2) = 0 and
f 0 (2) = 6 > 0, we deduce that f has a local minimum
at x = 2.
We use the four-step procedure from Section 4.8.
x→0
Curve
Local maximum
Local minimum
Inflection point
−5
x→0
3