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(1)
Use Newton’s method to ﬁnd critical points of the function
y
=
e
x

2
x
2
.
Solution:
The critical points are located at x values for which
y
0
=
f
(
x
)=
e
x

4
x
=0.If
there exists any root for
f
(
x
) = 0, there should normally be two. Let’s start from
x
0
=0
to ﬁnd the smaller root.
x
1
=
x
0

f
(
x
0
)
/f
0
(
x
0
)=0
.
333333,
x
2
=
x
1

f
(
x
1
)
/f
0
(
x
1
.
357246,
x
3
=
x
2

f
(
x
2
)
/f
0
(
x
2
.
357402,
x
4
=
x
3

f
(
x
3
)
/f
0
(
x
3
.
357402.
Let start from
x
0
= 5 to ﬁnd the larger root.
x
1
=
x
0

f
(
x
0
)
/f
0
(
x
0
)=4
.
110793,
x
2
=
x
1

f
(
x
1
)
/f
0
(
x
1
)=3
.
329113,
x
3
=
x
2

f
(
x
2
)
/f
0
(
x
2
)=2
.
718701,
x
4
=
x
3

f
(
x
3
)
/f
0
(
x
3
.
334689,
x
5
=2
.
178594,
x
6
.
153872,
x
7
.
153292,
x
8
.
153292.
(2)
Solution:
(3)
Use Newton’s method to ﬁnd an approximate value for
√
8. (Hint: First think of a
function,
f
(
x
), such that
f
(
x
) = 0 has the solution
x
=
√
8).
Solution:
x
=
√
8 is a solution of the equation
f
(
x
x
2

8 = 0. Thus,
f
0
(
x
x
.W
e
know also that
√
9=3and
√
8 must be close to this value. We start from
x
0
=3.
x
1
=
x
0

f
(
x
0
)
/f
0
(
x
0

(9

8)
/
2
×
3=3

1
/
6=2
5
6
≈
2
.
83
x
2
=
x
1

f
(
x
1
)
/f
0
(
x
1
.
83

(2
.
83
2

8)
/
2
×
2
.
83
x
2
.
83

0
.
0089
/
5
.
66 = 2
.
83

0
.
0016
≈
2
.
83
Therefore,
x
=
√
8
≈
2
.
83.
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This note was uploaded on 01/12/2010 for the course STAT 100 taught by Professor Denissjerve during the Winter '06 term at San Jose State University .
 Winter '06
 denissjerve

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