52.
}
d
d
y
x
}
55
}
1
2
1
si
c
n
os
t
t
}
At
t
52}
p
4
}
, we have
x
5
cos
1
2}
p
4
}
2
5 }
ˇ
2
2
w
}
,
y
p
4
}
1
sin
1
2}
p
4
}
2
p
4
}
2 }
ˇ
2
2
w
}
, and
}
d
d
y
x
}
5
ˇ
2
w
1
1.
The equation of the tangent line is
y
5
(
ˇ
2
w
1
1)
1
x
2 }
ˇ
2
2
w
}
2
2 }
p
4
}
2 }
ˇ
2
2
w
}
, or
y
5
(1
1
ˇ
2
w
)
x
2
ˇ
2
w
2
1
2 }
p
4
}
.
This is approximately
y
5
2.414
x
2
3.200.
53. (a)
[
2
1, 3] by
3
2
1,
}
5
3
}
4
(b)
Yes, because both of the onesided limits as
x
→
1 are
equal to
f
(1)
5
1.
(c)
No, because the lefthand derivative at
x
5
1 is
1
1 and
the righthand derivative at
x
5
1 is
2
1.
54. (a)
The function is continuous for all values of
m
, because
the righthand limit as
x
→
0 is equal to
f
(0)
5
0 for
any value of
m
.
(b)
The lefthand derivative at
x
5
0 is 2 cos (2
?
0)
5
2,
and the righthand derivative at
x
5
0 is
m
, so in order
for the function to be differentiable at
x
5
0,
m
must
be 2.
55. (a)
For all
x
±
0
(b)
At
x
5
0
(c)
Nowhere
56. (a)
For all
x
(b)
Nowhere
(c)
Nowhere
57.
Note that lim
x
→
0
2
f
(
x
)
5
lim
x
→
0
2
(2
x
2
3)
52
3 and
lim
x
→
0
1
f
(
x
)
5
lim
x
→
0
1
(
x
2
3)
3. Since these values agree
with
f
(0), the function is continuous at
x
5
0. On the other
hand,
f
9
(
x
)
5
5
, so the derivative is undefined at
x
5
0.
(a)
[
2
1, 0)
<
(0, 4]
(b)
At
x
5
0
(c)
Nowhere in its domain
58.
Note that the function is undefined at
x
5
0.
(a)
[
2
2, 0)
<
(0, 2]
(b)
Nowhere
(c)
Nowhere in its domain
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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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