Solutions
for
PSet
5
1.
(8.14:10)
(a)
By
hypothesis,
we
know
±
f
(
x
)=
0
for
all
x
∈
B
(
a
).
By
deﬁnition
±
f
(
x
)=(
D
1
f
(
x
)
,...,D
n
f
(
x
)),
thus
D
1
f
(
x
···
=
D
n
f
(
x
) = 0.
By
the
1dimensional
theorem,
this
means
that
f
(
x
)
is
constant
on
every
line
x
+
t
e
1
,...
x
+
t
e
n
for
any
x
∈
B
(
a
).
Let
f
(
a
c
.
Given
y
∈
B
(
a
),
we
will
prove
that
f
(
y
c
.
Let
y
−
a
=
d
1
e
1
+
+
d
n
e
n
,
then:
c
=
f
(
a
f
(
a
+
d
1
e
1
=
f
(
a
+
d
1
e
1
+
+
d
n
e
n
f
(
y
)
.
(b)
The
condition
means,
that
a
is
a
local
maximum
of
the
function
f
.I
n
particular,
because
f
(
x
0
for
all
x
∈
B
(
a
),
0
is
a
maximum
for
all
the
functions
f
(
a
+
t
e
i
).
By
the
1dimensional
theorem
D
i
(
f
(
a
))
=
f
±
(
a
+
t
e
i
)
=
0,
and
thus
±
f
(
a
D
1
f
(
a
)
n
f
(
a
))
=
0
.
2.
(8.17:6)
f
(
x, y

xy

=

x


y

(a)
f
(
x, y
)
=
0
on
the
lines
(
x,
0)
and
(0
,y
).
Thus
∂f
=
=
0
at
the
origin.
∂x
∂y
(b)
For
f
to
have
a
tangent
plane
at
the
origin,
it
must
have
a
total
derivative
at
the
origin.
If
indeed
f
had
a
total
derivative
at
the
origin
then
we
expect
f
±
((
x, y
); (1
,
1)) to
be (
,
)
·
(1
,
1)
=
+
=0+0=0
.
But
in
reality,
f
±
((
x, y
=
x
); (1
,
1))
is
the
partial
derivative
along
the
line
x
=
y
.
On
this
line,
f
(
x, x
x
,thu
s
f
±
((