(
and
)
- Prove that
for all complex numbers
- Prove that
HINT: We know from part (a) that
Using the Mean Value Theorem for the cosine function on the
interval
derive a contradiction from the assumption that
- Prove that
HINT: Recall that
for all complex
(Note that this does not yet tell us that
We do not yet know that
)
- Prove that
for all complex numbers
- Compute the derivatives of the tangent and hyperbolic tangent functions
and
Show in fact that
Here are two more elementary consequences of the Mean Value Theorem.
- Suppose
and
are two complex-valued functions of
a real (or complex) variable,and suppose that
for all
(or
)
Prove that there exists a constant
such that
for all
(or
)
- Suppose
for all
where
and
are complex constants with
Prove that there exists a constant
such that
What if
- (A generalization of part (a)) Suppose
and
are continuous real-valued functions on
the closed interval
and suppose there exists a partition
of
such that both
and
are differentiable on each subinterval
(That is, we do not assume that
and
are differentiable at the endpoints.)
Suppose that
for every
in each open subinterval
Prove that there exists a constant
such that
for all
HINT: Use part (a) to conclude that
where
is a step function, and then observe that
must be continuous
and hence a constant.
- Suppose
is a differentiable real-valued function on
and assume that
for all
Prove that
is 1-1 on
Let
be a function that is continuous
on its domain
and differentiable on
(We do not suppose that
is continuous on
)
- Prove that
is nondecreasing on
if and only if
for all
Show also that
is nonincreasing on
if and only if
for all
- Conclude that, if
takes on both positive and negative values on
then
is
not 1-1.
(See the proof of
[link] .)
- Show that, if
takes on both positive and negative
values on
then there must exist a point
for which
(If
were continuous, this would follow from the Intermediate Value Theorem.
But, we are not assuming here that
is continuous.)
- Prove the
Intermediate Value Theorem for Derivatives: Suppose
is continuous on the closed bounded interval
and differentiable on the
open interval
If
attains two distinct values
then
attains every value
between
and
HINT: Suppose
is a value between
and
Define a function
on
by
Now apply part (c) to
Here is another perfectly reasonable and expected theorem, but one whose
proof is tough.
Inverse function theorem
Suppose
is a function that is
continuous and 1-1 from
onto the interval
Assume that
is differentiable at a point
and that
Then
is differentiable at the point
and
The formula
is no surprise.
This follows directly from the Chain Rule. For, if
and
and
are both differentiable,
then
which gives the formula.
The difficulty with this theorem is in proving that the inverse function
of
is differentiable at
In fact, the first thing to check is that the point
belongs to the interior of the domain of
for that is essential if
is to be differentiable there,
and here is where the hypothesis that
is a real-valued function of a real variable is important.
According to
[link] , the 1-1 continuous function
maps
onto an interval
and
is in the open interval
i.e., is in the interior of the domain of
According to part (2) of
[link] ,
we can prove that
is differentiable at
by showing that
That is, we need to show that, given an
there exists a
such that if
then
First of all, because the function
is continuous at the point
there exists an
such that if
then
Next, because
is differentiable at
there exists
a
such that if
then
Now, by
[link] ,
is continuous at the point
and therefore there exists a
such that if
then
So, if
then
But then, by
[link] ,
from which it follows, using
[link] , that
as desired.
REMARK A result very like
[link] is actually true for complex-valued functions of a complex variable.
We will have to show that if
is in the interior of the domain
of
a one-to-one, continuously differentiable, complex-valued function
of a complex variable,
then
is in the interior of the domain
of
But, in the complex variable case, this requires a somewhat more difficult argument.
Once that fact is established, the proof that
is differentiable at
will be the same for complex-valued functions of complex variables as it is
here for real-valued functions of a real variable.Though the proof of
[link] is reasonably complicated for real-valued functions of a real variable,
the corresponding result for complex functions is much more deep, and that proofwill have to be postponed to a later chapter.
See
[link] .