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Let $I$ be an interval and suppose $f : I \to \mathbb{R}$ is differentiable on $I$. Let $a \in I$, $f'(a) > 0$, and $f'(x)$ is continuous at $a$.

Prove that there exists some $\delta > 0$ such that $f$ is strictly increasing on $[a−\delta, a+\delta]$.

How would I go about choosing $\delta$?

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1 Answer 1

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Let $\epsilon = \frac{f'(a)}{2}$. Since $f'$ is continuous at $a$, there is $\delta >0$ such that

$x \in [a- \delta,a+ \delta]$ implies that $|f'(x)-f'(a)|< \epsilon = \frac{f'(a)}{2}$.

This gives $f'(x)> \frac{f'(a)}{2}>0$ if $x \in [a- \delta,a+ \delta]$. Therfore $f$ is strictly increasing on $[a- \delta,a+ \delta]$.

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