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9.
Note that
f
9
(
x
)
52
32
x
1
160.
lim
h
→
0
}
f
(3
1
h
h
)
2
f
(3)
}5
f
9
(3)
52
32(3)
1
160
5
64
For graphical support, use the graph shown in the solution
to Exercise 7 and observe that NDER (
f
(
x
), 3)
<
64.
10.
f
9
(
x
)
52
32
x
1
160
f
0
(
x
)
52
32
At
x
5
7 (and, in fact, at any other of
x
),
}
d
dx
2
y
2
}52
32.
Graphical support: the graph of NDER(NDER
f
(
x
)) is
shown.
[
2
1, 9] by [
2
40, 10]
Section 3.4 Exercises
1.
Since
V
5
s
3
, the instantaneous rate of change is
}
d
d
V
s
} 5
3
s
2
.
2. (a)
Displacement =
s
(5)
2
s
(0)
5
12
2
2
5
10 m
(b)
Average velocity
5 }
1
5
0
se
m
c
} 5
2 m/sec
(c)
Velocity
5
s
9
(
t
)
5
2
t
2
3
At
t
5
4, velocity
5
s
9
(4)
5
2(4)
2
3
5
5 m/sec
(d)
Acceleration
5
s
0
(
t
)
5
2 m/sec
2
(e)
The particle changes direction when
s
9
(
t
)
5
2
t
2
3
5
0, so
t
5 }
3
2
}
sec.
(f)
Since the acceleration is always positive, the position
s
is at a minimum when the particle changes direction, at
t
5 }
3
2
}
sec. Its position at this time is
s
1
}
3
2
}
2
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Fall '08 term at University of Florida.
 Fall '08
 GERMAN

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