Honors Introduction to Analysis I
Homework VIII
Solution
April 4, 2009
Problem 1
Define
x
+
=
(
x
x
≥
0
0
x <
0
. Prove that
f
(
x
) =
x
k
+
is continuously differentiable if
k
is an integer greater
than one.
Solution.
For
k
= 1 one can see that
f
0
(
x
) = 1 if
x
≥
0 and
f
0
(
x
) = 0 for
x <
0, giving a discontinuity at 0.
If
k >
1 then we have that
f
(
x
) =
(
x
k
x
≥
0
0
x <
0
.
f
is continuously differentiable at any
x <
0, since it is constant.
It is also continuously differentiable at any
x >
0 because it is a polynomial. We are left to see if
f
is differentiable
at 0 and if the derivative is continuous at this point.
lim
x
→
0
+
f
(
x
)

f
(0)
x

0
=
lim
x
→
0
+
x
k

0
x
=
lim
x
→
0
+
x
k

1
= 0
lim
x
→
0

f
(
x
)

f
(0)
x

0
=
lim
x
→
0
+
0

0
x
= 0
The two limits are the same, so
f
is differentiable at 0, with
f
0
(0) = 0. So we can define
f
0
(
x
) =
(
kx
k

1
x
≥
0
0
x <
0
.
We need to check this function is continuous at 0.
lim
x
→
0
+
f
0
(
x
) =
lim
x
→
0
+
kx
k

1
= 0 =
lim
x
→
0

0 =
lim
x
→
0

f
0
(
x
)
Hence
f
is continuously differentiable everywhere.
Problem 2
Given a closed set
A
, construct a continuously differentiable function that has
A
as its set of zeroes.
Solution.
A
is closed, so
A
c
is open, hence we can write
A
c
=
S
k
(
a
k
, b
k
)  a disjoint union of open intervals. On
each interval (
a
k
, b
k
) define
f
k
= (
x

a
k
)
2
+
(
b
k

x
)
2
+
. Using the previous problem (and the hint),
f
k
is a continuously
differentiable function, that is nonzero exactly on the interval (
a
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 '08
 PROTSAK
 Derivative, Continuous function, Rational function

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