MATH237: EXAMAID SOS
NIALL W. MACGILLIVRAY (EDITOR); VINCENT CHAN (WRITER)
October 31st, 2010
1
.
ScalarFunctions
(
1
.
1

1
.
2
)
DEFINITION 1.1. Suppose
A
and
B
are sets. A function
f
is a rule that determines
how a subset of
A
is associated with a subset of
B
, such that each element in
A
for which
f
is defined is sent to exactly one element in
B
. This subset of A for which f is defined is
called the domain of
f
and denoted by D(
f
), while the subset of
B
which is attained by
f
is called the range of
f
and denoted by R(
f
). We typically denote
f
by
f
:
A
→
B
.
DEFINITION 1.2.
f
:
R
→
R
is a scalar function, i.e. when the domain is a subset of
R
and the range is a subset of
R
.
EXAMPLE 1.
Define
f
:
R
2
→
R
by
f
(
x, y
) =
x
2
+ 2
y
2

2
x
.
Find the domain and
range of f.
SOLUTION. Clearly D(
f
) =
R
2
. For the range, notice
x
2
+
y
2

2
x
≥
x
2

2
x
, which has a
minimum at
x
= 1 (using what we know from 1dimension). Then
f
(
x, y
)
≥
f
(

1
,
0) =

1.
For any value
z
∈
[0
,
∞
[, we can take
x
= 1 +
√
1 +
z
and
y
= 0. Indeed, this is a possible
value for
x
we can solve for using
z
=
x
2

2
x
and the quadratic equation. Thus, R(
f
) =
[

1
,
∞
[.
EXAMPLE 2. Define
f
:
R
2
→
R
by
f
(
x, y
) =

x

+

y


x

y

. Find the domain and range of
f
.
SOLUTION. We only need

x
 6
=

y

for
f
to be defined.
The domain is then
R
2
\
{
(
x, y
) :

x

=

y
}
. As for the range, notice we can obtain 1 and 1 by setting
y
= 0 and
x
= 0 respectively. Now, suppose for
z
∈
R
, we have
f
(
x, y
) =
z
. Then

x

+

y

=
z

x

z

y

,
so that:
(1

z
)

x

=

(1 +
z
)

y

We have already dealt with the case
z
=
±
1. Otherwise,

y

=

1

z
1+
z

x

. Since

x

,

y
 ≥
0,
we have

1

z
1+
z
≥
0. Then either both 1

z
and

1

z
are positive, or both are negative.
That is, we have
z <

1 or
z >
1.
1
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ATERLOO
SOS E
XAM
AID: MATH237
In either case, we can set
x
= 1
and
y
=
1
z
1+
z
j
x
j
to obtain
z
, so the range is (including the case
z
=
1
)
R
(
f
) = (
1
;
1]
[
[1
;
1
)
.
E
XERCISE
1.
Find the domain and range of the following functions.
(a)
f
(
x; y
) = ln(1
j
xy
j
)
.
(b)
g
(
x; y
) =
cos
x
sin
y
.
(c)
h
(
x; y
) =
p
1
x
2
y
2
It is difficult to sketch scalar functions from their definition alone. To help us, we look at level curves
and crosssections.
D
EFINITION
1.3.
For
f
:
R
2
!
R
, we define the
level curves
of
f
to be the curves given by
k
=
f
(
x; y
)
;
where
k
is a constant (in the range of
f
).
Sketching level curves is an exercise in solving for equations in two variables. The course notes cover
three excellent examples, which are fairly standard.
E
XAMPLE
3.
Define a function
f
by
f
(
x; y
) =
x
2
+ 2
y
2
2
x
. Sketch the level curves and use them to sketch
the surface
z
=
f
(
x; y
)
.
S
OLUTION
.
From example 1,
D
(
f
) =
R
2
and
R
(
f
) = [
1
;
1
)
. For values of
k
1
, we examine
x
2
+ 2
y
2
2
x
=
k
. Recall that the equation
(
x
h
)
2
a
2
+
(
y
k
)
2
b
2
= 1
determines an ellipse with major axis length
2
a
, minor axis length
2
b
, shifted
h
units right and
k
units up.
This formula is a more general form of what we have: since there are no
y
terms, we will have
k
= 0
and
since there is a
2
x
term, we will want
h
= 1
(to get
(
x
1)
2
=
x
2
2
x
+ 1
). Then if
k
6
=
1
,
x
2
+ 2
y
2
2
x
=
k
x
2
2
x
+ 1 + 2
y
2
=
k
+ 1
(
x
1)
2
k
+ 1
+
2
y
2
k
+ 1
= 1
(
x
1)
2
p
k
+ 1
2
+
y
2
q
k
+1
2
2
= 1
:
Thus,
x
2
+ 2
y
2
2
x
=
k
describes an ellipse with major axis length
2
p
k
+ 1
, minor axis length
2
q
k
+1
2
=
p
2(
k
+ 1)
, shifted 1 unit right. If
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 WILKIE
 Derivative, Scalar, Sets, SOS E XAM, WATERLOO SOS

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