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(e)
h
(
x
)
5
(1
1
x
)
1/2
h
9
(
x
)
5 }
1
2
}
(1
1
x
)
2
1/2
h
0
(
x
)
52}
1
4
}
(1
1
x
)
2
3/2
Since
h
(0)
5
1,
h
9
(1)
5 }
1
2
}
, and
h
0
(1)
1
4
}
, the
coefficients are
b
0
5
1,
b
1
5 }
1
2
}
, and
b
2
55
2
}
1
8
}
.
The quadratic approximation is
Q
(
x
)
5
1
1 }
2
x
}
2 }
x
8
2
}
.
[
2
1.35, 3.35] by [
2
1.25, 3.25]
As one zooms in, the two graphs quickly become
indistinguishable. They appear to be identical.
(f)
The linearization of any differentiable function
u
(
x
) at
x
5
a
is
L
(
x
)
5
u
(
a
)
1
u
9
(
a
)(
x
2
a
)
5
b
0
1
b
1
(
x
2
a
),
where
b
0
and
b
1
are the coefficients of the constant and
linear terms of the quadratic approximation. Thus, the
linearization for
f
(
x
) at
x
5
0 is 1
1
x
; the linearization
for
g
(
x
) at
x
5
1 is 1
2
(
x
2
1) or 2
2
x
; and the
linearization for
h
(
x
) at
x
5
0 is 1
1 }
2
x
}
.
53.
Just multiply the corresponding derivative formulas by
dx
.
(a)
Since
}
d
d
x
}
(
c
)
5
0,
d
(
c
)
5
0.
(b)
Since
}
d
d
x
}
(
cu
)
5
c
}
d
d
u
x
}
,
d
(
cu
)
5
c du
.
(c)
Since
}
d
d
x
}
(
u + v
)
5 }
d
d
u
x
} 1 }
d
d
v
x
}
,
d
(
u
1
v
)
5
du
1
dv
(d)
Since
}
d
d
x
}
(
u
?
v
)
5
u
}
d
d
v
x
} 1
v
}
d
d
u
x
}
,
d
(
u
?
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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

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