21. continued
Check
x
5
3:
Since
g
(3)
5
(4
2
3)
2
5
1 and
lim
x
→
3
2
g
(
x
)
5
lim
x
→
3
2
(2
x
1
1)
5
2(3)
1
1
5
7, the function is
not continuous (and hence not differentiable) at
x
5
3.
The function is differentiable for all reals except
x
5
3.
22.
Note that
C
(
x
)
5
x
)
x
)
5
5
, so it is
differentiable for all
x
except possibly at
x
5
0.
Check
x
5
0:
lim
h
→
0
}
C
(0
1
h
h
)
2
C
(0)
}
5
lim
h
→
0
}
h
)
h
)
h
2
0
}
5
lim
h
→
0
)
h
)
5
0
The function is differentiable for all reals.
23. (a)
x
5
0 is not in their domains, or, they are both
discontinuous at
x
5
0.
(b)
For
}
1
x
}
: NDER
1
}
1
x
}
, 0
2
5
1,000,000
For
}
x
1
2
}
: NDER
1
}
x
1
2
}
, 0
2
5
0
(c)
It returns an incorrect response because even though
these functions are not defined at
x
5
0, they are
defined at
x
5 6
0.001. The responses differ from each
other because
}
x
1
2
}
is even (which automatically makes
NDER
1
}
x
1
2
}
, 0
2
5
0) and
}
1
x
}
is odd.
24.
[
2
5, 5] by [
2
10, 10]
}
d
d
y
x
}
5
x
3
25.
[
2
2
p
, 2
p
] by [
2
1.5, 1.5]
}
d
d
y
x
}
5
sin
x
26.
[
2
6, 6] by [
2
4, 4]
}
d
d
y
x
}
5
abs (
x
) or
)
x
)
27.
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 Fall '08
 GERMAN
 lim, Continuous function, H0

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