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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0@JonasMeyer Yes I did. Edited. – 2012-02-03
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0For the statement about $f'$ being continuous at $a$, are we assuming then that $f'$ is defined everywhere in some interval containing $a$? Or do we only assume that $f'$ exists at each point of $X$ in some neighborhood of $X$ (which might not contain intervals)? – 2012-02-03
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0I think it is the latter.. it still holds, doesn't it? – 2012-02-03
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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.