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?
 A: You don't have to argue indirectly (i.e., to asssume that we have a $\max$ somewhere), nor is it necessary to come up with derivatives.
The function $f(x):=x^2$ is strictly increasing on the interval $J:=\bigl[{1\over2},3\bigr)$. Given any $x\in J$, the point $x':={1\over2}(x+3)$ is still in $J$, and we have
$f(x')>f(x)$. It is therefore impossible that $f$ has a global $\max$ in $J$.
A: note that $$f'(x)=2x\geq 0$$ for $$x\geq0$$ thus the function is for $$x\geq0 $$ monotonously increasing in this interval. But $$\lim_{x \to 3}f(x)=f(3)=9$$
A: Note that $x^2$ is increasing on this interval; thus its supremum is at the right end of the interval which isn't attained from the domain.
A: Suppose $x\in [1/2,3)$ maximizes $f$ on $[1/2,3)$. But then, since $f$ is strictly increasing ($f'>0$), for any $y\in(x,3)$ we have $f(y)>f(x)$, a contradiction.
A: 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.
A: There are some good answers here already but since this is tagged real analysis, I'd like to formalize it a bit with an $\varepsilon$.  I think you're thinking the right thing but it is a little confusing near the end with the quantifiers.
Suppose that $f(x)$ does attain its maximum on $[1/2, 3)$.  Call this maximum $M$.  Note that $f(x)$ is strictly increasing on $[1/2,3)$.  This is because $f'(x) > 0$ for $x \in [1/2,3)$.  This means $f(x) < \lim_\limits{t \to 3^-} f(t) = 9$ for all $x \in [1/2, 3)$.  So it must be the case that $M < 9$.  Note also that $f(\sqrt{M}) = M$.  Because $f$ is strictly increasing, and because $M < 9$, this tells us that $\sqrt M < 3$.  Therefore there is positive distance between $\sqrt M$ and $3$.  Call this distance $\varepsilon$.  That is, $\varepsilon = 3 - \sqrt{M}$.
Because $[1/2, 3)$ is open on the right endpoint, we know that there is an open interval centered at $M$ that lies entirely inside of $[1/2,3)$.  Specifically, we can take the interval of radius $\varepsilon/2$.  So we have:
$$  \left(\sqrt M - \frac\varepsilon2, \sqrt M + \frac\varepsilon2\right) \subseteq \left[\frac12,3\right)
$$
This means we can choose $y \in [1/2, 3)$ such that $\sqrt{M} < y < \sqrt{M} + \varepsilon/2 < \sqrt{M} + \varepsilon = 3$.  Therefore, $f(\sqrt{M}) < f(y) < \lim_\limits{t \to 3^-} f(t)$, a contradiction.

Note that the use of limits above is technically required because $f(x)$ is not defined at $x=3$.  So although it's easily understood in this context what we mean when we say $f(3)$, it's not correct because $3$ is not in the domain of $f$.
