I am following a real analysis textbook. In this section it considers a function $f : X \rightarrow \mathbb{R}$, with $X \subset \mathbb{R}$, and $a \in \mathbb{R}$ a bilateral accumulation point. If $f$ is differentiable at $a$, with f'(a) > 0, it follows that there exists some $\delta > 0$ such that $a - \delta < x < a$ implies $f(x) < f(a)$ and $a < x < a + \delta$ implies $f(x) > f(a)$. This I understood. Then it goes on to say that we cannot use this to conclude that $f$ is increasing in some neighborhood of $a$ unless f' is continuous at $a$. My problem is seeing how f' being continuous at $a$ allows us to conclude that it is increasing in some neighborhood.
Positive derivative and growth
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real-analysis
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0I think it is the latter.. it still holds, doesn't it? – 2012-02-03
1 Answers
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Assume f' is continuous at $a$. Since f'(a) necessarily exists, f'(a)=\lim\limits_{\delta\to 0^+}\frac{f(x+\delta)-f(x)}{\delta} exists so must be positive (as for sufficiently small $\delta$, f(x)+\delta>f(x)). By continuity of f', we then have some neighborhood of $a$ on which f' is positive, which means $f$ is strictly increasing on this neighborhood.