mao (tm23477) – HW02 – Radin – (56570)
1
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printout
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have
10
questions.
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001
10.0 points
Find all functions
g
such that
g
′
(
x
) =
4
x
2
+
x
+ 5
√
x
.
1.
g
(
x
) = 2
√
x
(
4
x
2
+
x
+ 5
)
+
C
2.
g
(
x
) = 2
√
x
parenleftbigg
4
5
x
2
+
1
3
x
−
5
parenrightbigg
+
C
3.
g
(
x
) =
√
x
(
4
x
2
+
x
+ 5
)
+
C
4.
g
(
x
) = 2
√
x
parenleftbigg
4
5
x
2
+
1
3
x
+ 5
parenrightbigg
+
C
cor
rect
5.
g
(
x
) = 2
√
x
(
4
x
2
+
x
−
5
)
+
C
6.
g
(
x
) =
√
x
parenleftbigg
4
5
x
2
+
1
3
x
+ 5
parenrightbigg
+
C
Explanation:
After division
g
′
(
x
) = 4
x
3
/
2
+
x
1
/
2
+ 5
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
+
2
3
x
3
/
2
+ 10
x
1
/
2
= 2
√
x
parenleftbigg
4
5
x
2
+
1
3
x
+ 5
parenrightbigg
is an antiderivative of
g
′
. Consequently,
g
(
x
) = 2
√
x
parenleftbigg
4
5
x
2
+
1
3
x
+ 5
parenrightbigg
+
C
with
C
an arbitrary constant.
002
10.0 points
Determine
f
(
t
) when
f
′′
(
t
) = 6(2
t
−
1)
and
f
′
(1) = 6
,
f
(1) = 4
.
1.
f
(
t
) = 6
t
3
+ 6
t
2
−
6
t
−
2
2.
f
(
t
) = 2
t
3
+ 6
t
2
−
6
t
+ 2
3.
f
(
t
) = 2
t
3
−
3
t
2
+ 6
t
−
1
correct
4.
f
(
t
) = 6
t
3
−
6
t
2
+ 6
t
−
2
5.
f
(
t
) = 2
t
3
+ 3
t
2
−
6
t
+ 5
6.
f
(
t
) = 6
t
3
−
3
t
2
+ 6
t
−
5
Explanation:
The most general antiderivative of
f
′′
has
the form
f
′
(
t
) = 6
t
2
−
6
t
+
C
where
C
is an arbitrary constant.
But if
f
′
(1) = 6, then
f
′
(1) =
6
−
6 +
C
= 6
,
i.e.,
C
= 6
.
From this it follows that
f
′
(
t
) = 6
t
2
−
6
t
+ 6
.
The most general antiderivative of
f
′
is thus
f
(
t
) = 2
t
3
−
3
t
2
+ 6
t
+
D ,
where
D
is an arbitrary constant.
But if
f
(1) = 4, then
f
(1) = 2
−
3 + 6 +
D
= 4
,
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mao (tm23477) – HW02 – Radin – (56570)
2
i.e.,
D
=
−
1
.
Consequently,
f
(
t
) = 2
t
3
−
3
t
2
+ 6
t
−
1
.
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 Spring '08
 RAdin
 Derivative, Fundamental Theorem Of Calculus, Cos

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