10.
[
2
2.7, 6.7] by [
2
6, 6]
A graph of
f
(
x
) is shown. The range of
f
(
x
) is
(
2‘
, 1)
<
[2,
‘
). The values of
c
for which
f
(
x
)
5
c
has
no solution are the values that are excluded from the range.
Therefore,
c
can be any value in [1, 2).
Section 2.3 Exercises
1.
The function
y
5
}
(
x
1
1
2)
2
}
is continuous because it is a
quotient of polynomials, which are continuous. Its only
point of discontinuity occurs where it is undefined. There is
an infinite discontinuity at
x
52
2.
2.
The function
y
5
}
x
2
2
x
1
4
x
1
1
3
}
is continuous because it is a
quotient of polynomials, which are continuous. Its only
points of discontinuity occur where it is undefined, that is,
where the denominator
x
2
2
4
x
1
3
5
(
x
2
1)(
x
2
3) is
zero. There are infinite discontinuities at
x
5
1 and at
x
5
3.
3.
The function
y
5 }
x
2
1
1
1
}
is continuous because it is a
quotient of polynomials, which are continuous.
Furthermore, the domain is all real numbers because the
denominator,
x
2
1
1, is never zero. Since the function is
continuous and has domain (
2‘
,
‘
), there are no points of
discontinuity.
4.
The function
y
5
)
x
2
1
)
is a composition (
f
+
g
)(
x
) of the
continuous functions
f
(
x
)
5
)
x
)
and
g
(
x
)
5
x
2
1, so it is
continuous. Since the function is continuous and has
domain (
2‘
,
‘
), there are no points of discontinuity.
5.
The function
y
5
ˇ
2
w
x
w
1
w
3
w
is a composition (
f
+
g
)(
x
) of
the continuous functions
f
(
x
)
5
ˇ
x
w
and
g
(
x
)
5
2
x
1
3, so
it is continuous. Its points of discontinuity are the points
not in the domain, i.e., all
x
,2}
3
2
}
.
6.
The function
y
5
ˇ
3
2
w
x
w
2
w
1
w
is a composition (
f
+
g
)(
x
) of
the continuous functions
f
(
x
)
5
ˇ
3
x
w
and
g
(
x
)
5
2
x
2
1, so
it is continuous. Since the function is continuous and has
domain (
2‘
,
‘
), there are no points of discontinuity.
7.
The function
y
5 }
)
x
x
)
}
is equivalent to
y
5
h
It has a jump discontinuity at
x
5
0.
8.
The function
y
5
cot
x
is equivalent to
y
5 }
c
s
o
in
s
x
x
}
, a quotient
of continuous functions, so it is continuous. Its only points
of discontinuity occur where it is undefined. It has infinite
discontinuities at
x
5
k
p
for all integers
k
.
9.
The function
y
5
e
1/
x
is a composition (
f
+
g
)(
x
) of the
continuous functions
f
(
x
)
5
e
x
and
g
(
x
)
5 }
1
x
}
, so it is
continuous. Its only point of discontinuity occurs at
x
5
0,
where it is undefined. Since lim
x
→
0
1
e
1/
x
5‘
, this may be
considered an infinite discontinuity.
10.
The function
y
5
ln (
x
1
1) is a composition (
f
+
g
)(
x
) of
the continuous functions
f
(
x
)
5
ln
x
and
g
(
x
)
5
x
1
1, so it
is continuous. Its points of discontinuity are the points not
in the domain, i.e.,
x
,2
1.