- Prove that for
we have
- Prove that there is no constant
such that for all
we have
Why is this not a violation of the zero-derivative theorem (Theorem 5.2 in Apostol)?
- Proof. We can use the formulas for the derivatives of
and
(and the chain rule) to compute,
- Proof. First, let
. Then,
Then, since
, we have
Next, let
. Then,
Again, using that
, we have
Hence, there is no constant
such that
for all
This is not a violation of the zero-derivative theorem since the function
is constant on every open interval on which it is defined. Since it isn’t defined at
, any open subinterval must be a subinterval of only positive or only negative reals. The function is constant on any of these subintervals.
the function is not constant for negative x
ignore this