I'm a little confused. After finishing the online multi-variable calculus course from the MIT OCW offerings (I wanted to brush up on the subject more concretely, after my Analysis II course), I looked at another brisk course on a single-variable calculus course.
The hope was to revisit calculus, after a couple years of rigorous analysis courses.
But, my question is: in the MIT OCW course, the prof. had mentioned a few times that:
If the first order derivatives of f were 0, then f is not actually constant, but rather it is just not changing, to first order. There are obviously higher order terms in its Taylor development.
But when I return to single-variable calculus, I have seen several times now that some theorems and proofs argue that if f' = 0, then f is constant. But isn't f just…not changing, to first order – and that it's not actually constant? It may or may not have higher order, non-vanishing terms in its Taylor development.
Thanks in advance,
Best Answer
The difference seems to be between $f'(x) = 0$ at a point $x \in Domain(f)$ vs. $f'(x) = 0$ for all $x \in Domain(f)$ or at least on an open interval in the domain.
In the first case, it doesn't mean $f$ is constant. E.g., $f(x) = x^2$ at $x = 0$. However in the second case it does.