- 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.