in Van Brunt's "The Calculus of Variations", the author says the following (I'm paraphrasing) -
"Suppose that $f$ is smooth in the interval $(x - \epsilon, x + \epsilon)$, where $\epsilon > 0$, and that $f$ has a local maximum at $x$. Let $\hat{x} - x = \epsilon \eta$. Taylor's theorem indicates that $f$ can be represented by
$$ f(\hat{x}) = f(x) + \epsilon\eta f'(x) + \frac{\epsilon^2}{2}\eta^2 f''(x) + O(\epsilon^3).$$
If $f'(x) \neq 0$ and $\epsilon$ is small, the sign of $f(\hat{x}) - f(x)$ cannot change in $(x - \epsilon, x + \epsilon)$, so that $\eta f'(x)$ must have the same sign for all $\eta$."
The last bit seems very spurious to me. Since $f$ achieves a local maximum at $x$, it's clear that the sign of $f(\hat{x}) - f(x)$ cannot change in $(x - \epsilon, x + \epsilon)$ - I have no problem with this. However, it's not clear to me why $\eta f'(x)$ cannot change sign. It seems possible, at least in principle, that a change in the sign of $\eta f'(x)$ could be 'undone' by a change in the sign of $f''(x)$, assuming $|f''(x)|$ sufficiently large. So, does it follow $necessarily$ from his reasoning that $\eta f'(x)$ cannot change sign or is the author simply making a hand-wavey argument that is not strictly correct?