10.
This problem can be solved graphically by using NDER to obtain the graphs shown below.
y
y
9
y
0
[
2
4, 4] by [
2
1, 0.3]
[
2
4, 4] by [
2
0.4, 0.6]
[
2
4, 4] by [
2
0.7, 0.8]
An alternative approach using a combination of algebraic and graphical techniques follows.
Note that the denominator of
y
is always positive because it is equivalent to (
x
1
1)
2
1
2.
y
95
5
y
05
5
5
}
2
(
x
x
3
2
2
1
1
2
8
x
x
1
2
3
1
)
3
2
}
Using graphing techniques, the zeros of 2
x
3
2
18
x
2
12 (and hence of
y
0
) are at
x
<
2
2.584,
x
<
2
0.706, and
x
<
3.290.
(a)
[
2
ˇ
3
w
,
ˇ
3
w
]
(b)
(
2‘
,
2
ˇ
3
w
] and [
ˇ
3
w
,
‘
)
(c)
Approximately (
2
2.584,
2
0.706) and (3.290,
‘
)
(d)
Approximately (
2‘
,
2
2.584) and (
2
0.706, 3.290)
(e)
Local maximum at
1
ˇ
3
w
,
}
ˇ
3
w
4
2
1
}
2
<
(1.732, 0.183);
local minimum at
1
2
ˇ
3
w
,
}
2
ˇ
3
w
4
2
1
}
2
<
(
2
1.732,
2
0.683)
(f)
<
(
2
2.584,
2
0.573), (
2
0.706,
2
0.338), and
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.
 Spring '08
 GERMAN
 Algebra

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