0
$\begingroup$

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 $\delta$?

$\endgroup$
1
$\begingroup$

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 $[a- \delta,a+ \delta]$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.