pokharel (yp624) – HW12 – Radin – (56520)
1
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printout
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have
15
questions.
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before answering.
001
10.0 points
Use Newton’s method to estimate the solu
tion to
x
3
−
x
−
7 = 0
starting with the initial guess
x
0
= 2 and
applying one iteration.
1.
estimate =
9
5
2.
estimate =
21
11
3.
estimate =
5
2
4.
estimate =
23
11
correct
5.
estimate =
3
2
6.
estimate =
11
5
Explanation:
If
x
n
is one estimate of a solution to the
equation
f
(
x
) = 0, then Newton’s method
says that
x
n
+1
=
x
n
−
f
(
x
n
)
f
′
(
x
n
)
will usually be a better estimate.
But when
f
(
x
) =
x
3
−
x
−
7
.
then
f
′
(
x
) = 3
x
2
−
1
,
so Newton’s method gives the iteration for
mula
x
n
+1
=
x
n
−
x
3
n
−
x
n
−
7
3
x
2
n
−
1
.
Consequently, with an initial guess of
x
0
= 2,
x
1
= 2
−
2
3
−
1
·
2
−
7
3
·
2
2
−
1
=
23
11
.
002
10.0 points
A differentiable function
f
has the proper
ties
(i)
f
(
x
)
<
0 for
x < x
0
,
(ii)
f
(
x
)
>
0 for
x > x
0
,
(iii)
f
(3) = 7
,
f
′
(3) = 4.
Use the tangent line at (3
,
7) to the graph of
f
to compute an approximate value for the
value of
x
0
.
1.
x
0
≈
3
2
2.
x
0
≈
3
4
3.
x
0
≈
7
4
4.
x
0
≈
5
4
correct
5.
x
0
≈
1
Explanation:
The tangent line at (3
,
7) to the graph of
f
is given by
y
−
7 =
f
′
(3)(
x
−
3)
,
i
.
e
., y
= 4
x
−
5
.
The basic idea underlying Newton’s Method
is that the
x
intercept of this tangent line pro
vides an approximate value for the
x
intercept
of the graph of
f
. Since the graph of
f
crosses
the
x
axis at
x
=
x
0
,
x
0
≈
5
4
.
003
10.0 points
Find all functions
g
such that
g
′
(
x
) =
4
x
2
+ 3
x
+ 1
√
x
.
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pokharel (yp624) – HW12 – Radin – (56520)
2
1.
g
(
x
) = 2
√
x
(
4
x
2
+ 3
x
+ 1
)
+
C
2.
g
(
x
) =
√
x
(
4
x
2
+ 3
x
+ 1
)
+
C
3.
g
(
x
) =
√
x
parenleftbigg
4
5
x
2
+
x
+ 1
parenrightbigg
+
C
4.
g
(
x
)
=
2
√
x
parenleftbigg
4
5
x
2
+
x
+ 1
parenrightbigg
+
C
cor
rect
5.
g
(
x
) = 2
√
x
parenleftbigg
4
5
x
2
+
x
−
1
parenrightbigg
+
C
6.
g
(
x
) = 2
√
x
(
4
x
2
+ 3
x
−
1
)
+
C
Explanation:
After division
g
′
(
x
) = 4
x
3
/
2
+ 3
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
8
5
x
5
/
2
+ 2
x
3
/
2
+ 2
x
1
/
2
= 2
√
x
parenleftbigg
4
5
x
2
+
x
+ 1
parenrightbigg
is an antiderivative of
g
′
. Consequently,
g
(
x
) = 2
√
x
parenleftbigg
4
5
x
2
+
x
+ 1
parenrightbigg
+
C
with
C
an arbitrary constant.
004
10.0 points
Determine
f
(
t
) when
f
′′
(
t
) = 6(
t
−
1)
and
f
′
(1) = 5
,
f
(1) = 4
.
1.
f
(
t
) =
t
3
−
3
t
2
+ 8
t
−
2
correct
2.
f
(
t
) =
t
3
+ 6
t
2
−
8
t
+ 5
3.
f
(
t
) =
t
3
+ 3
t
2
−
8
t
+ 8
4.
f
(
t
) = 3
t
3
−
6
t
2
+ 8
t
−
1
5.
f
(
t
) = 3
t
3
−
3
t
2
+ 8
t
−
4
6.
f
(
t
) = 3
t
3
+ 6
t
2
−
8
t
+ 3
Explanation:
The most general antiderivative of
f
′′
has
the form
f
′
(
t
) = 3
t
2
−
6
t
+
C
where
C
is an arbitrary constant.
But if
f
′
(1) = 5, then
f
′
(1) =
3
−
6 +
C
= 5
,
i.e.,
C
= 8
.
From this it follows that
f
′
(
t
) = 3
t
2
−
6
t
+ 8
.
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 Spring '10
 TSOI
 Fundamental Theorem Of Calculus, Cos, 3 seconds, 128 ft, 32 ft

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