obiahu (vo879) – HW12 – Schultz – (56395)
1
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printout
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have
15
questions.
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001
10.0 points
Find all functions
g
such that
g
′
(
x
) =
4
x
2
+ 5
x
+ 3
√
x
.
1.
g
(
x
) = 2
√
x
(
4
x
2
+ 5
x
+ 3
)
+
C
2.
g
(
x
) = 2
√
x
parenleftbigg
4
5
x
2
+
5
3
x
−
3
parenrightbigg
+
C
3.
g
(
x
) =
√
x
parenleftbigg
4
5
x
2
+
5
3
x
+ 3
parenrightbigg
+
C
4.
g
(
x
) = 2
√
x
parenleftbigg
4
5
x
2
+
5
3
x
+ 3
parenrightbigg
+
C
cor
rect
5.
g
(
x
) = 2
√
x
(
4
x
2
+ 5
x
−
3
)
+
C
6.
g
(
x
) =
√
x
(
4
x
2
+ 5
x
+ 3
)
+
C
Explanation:
After division
g
′
(
x
) = 4
x
3
/
2
+ 5
x
1
/
2
+ 3
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
8
5
x
5
/
2
+
10
3
x
3
/
2
+ 6
x
1
/
2
= 2
√
x
parenleftbigg
4
5
x
2
+
5
3
x
+ 3
parenrightbigg
is an antiderivative of
g
′
. Consequently,
g
(
x
) = 2
√
x
parenleftbigg
4
5
x
2
+
5
3
x
+ 3
parenrightbigg
+
C
with
C
an arbitrary constant.
002
10.0 points
Find the value of
f
(0) when
f
′′
(
t
) = 2(3
t
+ 1)
and
f
′
(1) = 2
,
f
(1) = 3
.
1.
f
(0) = 6
2.
f
(0) = 5
3.
f
(0) = 4
correct
4.
f
(0) = 8
5.
f
(0) = 7
Explanation:
The most general antiderivative of
f
′′
has
the form
f
′
(
t
) = 3
t
2
+ 2
t
+
C
where
C
is an arbitrary constant.
But if
f
′
(1) = 2, then
f
′
(1) =
3 + 2 +
C
= 2
,
i.e.,
C
=
−
3
.
From this it follows that
f
′
(
t
) = 3
t
2
+ 2
t
−
3
,
and the most general antiderivative of the
latter is
f
(
t
) =
t
3
+
t
2
−
3
t
+
D ,
where
D
is an arbitrary constant.
But if
f
(1) = 3, then
f
(1) = 1 + 1
−
3 +
D
= 3
,
i.e.,
D
= 4
.
Consequently,
f
(
t
) =
t
3
+
t
2
−
3
t
+ 4
.
At
x
= 0, therefore,
f
(0) = 4
.
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obiahu (vo879) – HW12 – Schultz – (56395)
2
003
10.0 points
Find the value of
f
(0) when
f
′
(
t
) = 5 sin 2
t ,
f
parenleftBig
π
2
parenrightBig
= 1
.
1.
f
(0) =
−
5
2.
f
(0) =
−
3
3.
f
(0) =
−
7
4.
f
(0) =
−
6
5.
f
(0) =
−
4
correct
Explanation:
Since
d
dx
cos
mt
=
−
m
sin
mt ,
for all
m
negationslash
= 0, we see that
f
(
t
) =
−
5
2
cos 2
t
+
C
where the arbitrary constant
C
is determined
by the condition
f
(
π/
2) = 1. But
cos 2
t
vextendsingle
vextendsingle
vextendsingle
t
=
π/
2
= cos
π
=
−
1
.
Thus
f
parenleftBig
π
2
parenrightBig
=
5
2
+
C
= 1
,
and so
f
(
t
) =
−
5
2
cos 2
t
−
3
2
.
Consequently,
f
(0) =
−
4
.
004
10.0 points
Consider the following functions:
(
A
)
F
1
(
x
) =
cos 2
x
4
,
(
B
)
F
2
(
x
) =
cos
2
x
2
,
(
C
)
F
3
(
x
) =
sin
2
x
2
.
Which are antiderivatives of
f
(
x
) = sin
x
cos
x
?
1.
F
2
only
2.
F
3
only
correct
3.
all of them
4.
F
1
and
F
3
only
5.
F
1
only
6.
F
2
and
F
3
only
7.
none of them
8.
F
1
and
F
2
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 the value of
f
(
π
) when
f
′
(
t
) =
2
3
cos
1
3
t
−
8 sin
2
3
t
and
f
(
π
2
) = 4.
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 Spring '10
 schultz
 Calculus, Derivative, Fundamental Theorem Of Calculus, Cos

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