Showing there's no maximum in the given interval Let $f$ be function as $f(x) = x^2, x \in [\frac{1}{2}, 3).$ If we want to show there's no maximum in the given interval, is this the right way to do it:
Assume there's maximum $y \in (8, 9)$ for some $x \in [\frac{1}{2}, 3)$. Then show that there's some $y' \in (8, 9)$ for all $y \in (8, 9)$ such that $y' > y$ for some $x' \in [\frac{1}{2}, 3)$ so that we get a contradiction. Does that make sense?

I'm a little confused by the ordering of your quantifiers. Just to make sure I understand, here's what I would do. 
Assume $f$ attains a maximum $M$ on $[\frac{1}{2},3)$. Since $f$ is strictly increasing on $[0,\infty)$, it must be that $M<f(3)=9$. Let $y_{0}=\frac{M+9}{2}$. Then, $M<y_{0}<9$ so that $\sqrt{M}<\sqrt{y_{0}}<3$. Hence, $\sqrt{y_{0}}\in[\frac{1}{2},3)$ and $f(\sqrt{y_{0}})=y_{0}>M$, a contradiction.
