jiwani (amj566) – Homework12 – Fouli – (58395)
1
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printout
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have
22
questions.
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before answering.
001
10.0 points
Find all functions
g
such that
g
′
(
x
) =
2
x
2
+ 4
x
+ 1
√
x
.
1.
g
(
x
) = 2
√
x
parenleftbigg
2
5
x
2
+
4
3
x
+ 1
parenrightbigg
+
C
cor
rect
2.
g
(
x
) = 2
√
x
(
2
x
2
+ 4
x
−
1
)
+
C
3.
g
(
x
) = 2
√
x
(
2
x
2
+ 4
x
+ 1
)
+
C
4.
g
(
x
) =
√
x
parenleftbigg
2
5
x
2
+
4
3
x
+ 1
parenrightbigg
+
C
5.
g
(
x
) = 2
√
x
parenleftbigg
2
5
x
2
+
4
3
x
−
1
parenrightbigg
+
C
6.
g
(
x
) =
√
x
(
2
x
2
+ 4
x
+ 1
)
+
C
Explanation:
After division
g
′
(
x
) = 2
x
3
/
2
+ 4
x
1
/
2
+
x
−
1
/
2
,
so we can now find an antiderivative of each
term separately. But
d
dx
parenleftbigg
ax
r
r
parenrightbigg
=
ax
r
−
1
for all
a
and all
r
negationslash
= 0. Thus
4
5
x
5
/
2
+
8
3
x
3
/
2
+ 2
x
1
/
2
= 2
√
x
parenleftbigg
2
5
x
2
+
4
3
x
+ 1
parenrightbigg
is an antiderivative of
g
′
. Consequently,
g
(
x
) = 2
√
x
parenleftbigg
2
5
x
2
+
4
3
x
+ 1
parenrightbigg
+
C
with
C
an arbitrary constant.
002
10.0 points
Determine
f
(
t
) when
f
′′
(
t
) = 2(3
t
−
2)
and
f
′
(1) = 1
,
f
(1) = 2
.
1.
f
(
t
) =
t
3
−
2
t
2
+ 2
t
+ 1
correct
2.
f
(
t
) =
t
3
+ 2
t
2
−
2
t
+ 1
3.
f
(
t
) =
t
3
+ 4
t
2
−
2
t
−
1
4.
f
(
t
) = 3
t
3
−
2
t
2
+ 2
t
−
1
5.
f
(
t
) = 3
t
3
−
4
t
2
+ 2
t
+ 1
6.
f
(
t
) = 3
t
3
+ 4
t
2
−
2
t
−
3
Explanation:
The most general antiderivative of
f
′′
has
the form
f
′
(
t
) = 3
t
2
−
4
t
+
C
where
C
is an arbitrary constant.
But if
f
′
(1) = 1, then
f
′
(1) =
3
−
4 +
C
= 1
,
i.e.,
C
= 2
.
From this it follows that
f
′
(
t
) = 3
t
2
−
4
t
+ 2
.
The most general antiderivative of
f
is thus
f
(
t
) =
t
3
−
2
t
2
+ 2
t
+
D ,
where
D
is an arbitrary constant.
But if
f
(1) = 2, then
f
(1) = 1
−
2 + 2 +
D
= 2
,
i.e.,
D
= 1
.
Consequently,
f
(
t
) =
t
3
−
2
t
2
+ 2
t
+ 1
.
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jiwani (amj566) – Homework12 – Fouli – (58395)
2
003
10.0 points
Find the value of
f
(0) when
f
′
(
t
) = sin 2
t ,
f
parenleftBig
π
2
parenrightBig
= 3
.
1.
f
(0) = 2
correct
2.
f
(0) = 0
3.
f
(0) =
−
1
4.
f
(0) = 3
5.
f
(0) = 1
Explanation:
Since
d
dx
cos
mt
=
−
m
sin
mt ,
for all
m
negationslash
= 0, we see that
f
(
t
) =
−
1
2
cos 2
t
+
C
where the arbitrary constant
C
is determined
by the condition
f
(
π/
2) = 3. But
cos 2
t
vextendsingle
vextendsingle
vextendsingle
t
=
π/
2
= cos
π
=
−
1
.
Thus
f
parenleftBig
π
2
parenrightBig
=
1
2
+
C
= 3
,
and so
f
(
t
) =
−
1
2
cos 2
t
+
5
2
.
Consequently,
f
(0) = 2
.
004
10.0 points
Consider the following functions:
(
A
)
F
1
(
x
) =
cos
2
x
2
,
(
B
)
F
2
(
x
) = sin
2
x ,
(
C
)
F
3
(
x
) =
−
cos 2
x
4
.
Which are antiderivatives of
f
(
x
) = sin
x
cos
x
?
1.
F
1
only
2.
none of them
3.
F
2
only
4.
F
1
and
F
2
only
5.
all of them
6.
F
3
only
correct
7.
F
2
and
F
3
only
8.
F
1
and
F
3
only
Explanation:
By trig identities,
cos 2
x
= 2 cos
2
x
−
1 = 1
−
2 sin
2
x ,
while
sin 2
x
= 2 sin
x
cos
x .
But
d
dx
sin
x
= cos
x,
d
dx
cos
x
=
−
sin
x .
Consequently, by the Chain Rule,
(
A
)
Not antiderivative.
(
B
)
Not antiderivative.
(
C
)
Antiderivative.
005
10.0 points
Find
f
(
π/
2) when
f
′
(
t
) = cos
1
3
t
−
4 sin
2
3
t
and
f
(0) =
−
1.
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 Spring '11
 Jensen
 Inverse, Inverse function, F3

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