-3
-2
-1
1
2
3
-2
-1
1
2
Section 13.2
C13S02.001:
Because
f
(
x, y
) = 4
−
3
x
−
2
y
is defined for all
x
and
y
, the domain of
f
is the entire
two-dimensional plane.
C13S02.002:
Because
x
2
+ 2
y
2
0 for all
x
and
y
, the domain of
f
(
x, y
) =
x
2
+ 2
y
2
is the entire
two-dimensional plane.
C13S02.003:
If either
x
or
y
is nonzero, then
x
2
+
y
2
>
0, and so
f
(
x, y
) is defined—but not if
x
=
y
= 0.
Hence the domain of
f
consists of all points (
x, y
) in the plane other than the origin.
C13S02.004:
If
x
=
y
then the denominator in
f
(
x, y
) is nonzero, and thus
f
(
x, y
) is defined—but not if
x
=
y
. So the domain of
f
consists of all those points (
x, y
) in the plane for which
y
=
x
.
C13S02.005:
The real number
z
has a unique cube root
z
1
/
3
regardless of the value of
z
. Hence the domain
of
f
(
x, y
) = (
y
−
x
2
)
1
/
3
consists of all points in the
xy
-plane.
C13S02.006:
The real number
z
has a unique cube root
z
1
/
3
regardless of the value of
z
. But
√
2
x
is real
if and only if
x
0. Therefore the domain of
f
(
x, y
) = (2
x
)
1
/
2
+ (3
y
)
1
/
3
consists of all those points (
x, y
)
for which
x
0.
C13S02.007:
Because arcsin
z
is a real number if and only if
−
1
z
1, the domain of the given function
f
(
x, y
) = sin
−
1
(
x
2
+
y
2
) consists of those points (
x, y
) in the
xy
-plane for which
x
2
+
y
2
1; that is, the
set of all points on and within the unit circle.
C13S02.008:
Because arctan
z
is defined for every real number
z
, the only obstruction to the computation
of
f
(
x
) = arctan(
y/x
) is the possibility that
x
= 0. This obstruction is insurmountable, and therefore the
domain of
f
consists of all those points (
x, y
) in the
xy
-plane for which
x
= 0; that is, all points other than
those on the
y
-axis.
C13S02.009:
For every real number
z
, exp(
z
) is defined and unique. Therefore the domain of the given
function
f
(
x, y
) = exp(
−
x
2
−
y
2
) consists of all points (
x, y
) in the entire
xy
-plane.
C13S02.010:
Because ln
z
is a unique real number if and only if
z >
0, the domain of
f
(
x, y
) = ln(
x
2
−
y
2
−
1)
consists of those points (
x, y
) for which
x
2
−
y
2
−
1
>
0; that is, for which
y
2
< x
2
−
1. This is the region
bounded by the hyperbola with equation
x
2
−
y
2
= 1, shown shaded in the following figure; the bounding
hyperbola itself is
not
part of the domain of
f
.
C13S02.0011:
Because ln
z
is a unique real number if and only if
z >
0, then domain of
f
(
x, y
) = ln(
y
−
x
)
consists of those points (
x, y
) for which
y > x
. This is the region
above
the graph of the straight line with
equation
y
=
x
(the line itself is
not
part of the domain of
f
).
1

This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
C13S02.012:
Because
√
z
is a unique real number if and only if
z
0, the domain of the given function
f
(
x, y
) =
4
−
x
2
−
y
2
consists of those points (
x, y
) for which
x
2
+
y
2
4. That is, the domain consists
of all those points (
x, y
) on and within the circle with center (0
,
0) and radius 2.
C13S02.013:
If
x
and
y
are real numbers, then so are
xy
, sin
xy
, and 1 + sin
xy
.
Hence the only
obstruction to computation of
f
(
x, y
) =
1 + sin
xy
xy
is the possibility of division by zero. So the domain of
f
consists of those points (
x, y
) for which
xy
= 0;
that is, all points in the
xy
-plane other than those on the coordinate axes.

This is the end of the preview.
Sign up
to
access the rest of the document.