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for x or y in terms of the other variable. We can think of equation (4.9) defining either
x or y in terms of the other implicitly. The obvious question is under what conditions
we can in fact think of x as a function of y or, conversely, y a function of x. A good
example is
(
)
=+-
2
2
(4.10)
fxy
,
x
y
1
in which case equation (4.9) defines the unit circle. A neighborhood of the point
A
= (0,1) on the circle is clearly the graph of the function
()
=-
2
.
yx
1
x
(4.11)
The variable x is
not
a function of y in any such neighborhood because functions are
single-valued. On the other hand, a neighborhood of the point
B
= (1,0) on the circle
is clearly the graph of the function
2
.
()
=-
xy
1
y
(4.12)
Here, the variable y is
not
a function of x in any neighborhood. The point
C
=
(1/ ,1/ ) is much nicer because we can solve for either x
or
y. See Figure 4.11. The
difference between these points is that the tangent line at
A
and
B
is horizontal and
vertical, respectively. The tangent line at
C
is neither. In fact, near
C
each point cor-
responds to a
unique
x and y value and the functions y(x) and x(y) given by equa-
tions (4.11) and (4.12), respectively, which are the local solutions to (4.9), are inverses
of each other.
So what is a possible criterion that will guarantee that one can solve for one of
the variables in equation (4.9)? Well, it is at points like
C
that have nonvertical tan-
gents that we can guarantee both solutions. It is there that we can guarantee a unique
x and y value for nearby points. For the points
A
and
B
, which have horizontal or ver-
tical tangents (equivalently, derivatives of functions vanish or do not exist), we can at
most guarantee one solution. Our theorem will only give us sufficient but not neces-
sary conditions and not much can be said in general at points where appropriate deriv-
atives vanish. They would have to be analyzed in special ways. Note that horizontal
tangents are not necessarily bad because the curve
2
2
y
A(0,1)
(
√
1
)
1
,
C
√2
x
B(1,0)
x
2
+ y
2
-1 = 0
gr
aph of
Figure 4.11.
Solving for implicitly defined
functions.
y(x) = √1-x
2
x(y) = √1-y
2
or