Extreme Value Theorem For any function, given an interval with closed/open brackets, does a function have to have a min and max for the Extreme Value Theorem to apply? For example, does the extreme value theorem apply for the function f(x) = |x|?
In my opinion, I believe not because this function doesn't have a discrete maximum value. Is this thinking correct?
In addition, if at some point of any function is discontinuous within a interval, does it automatically mean that the application of the EVT is invalid? For example, take a simple floor function, since it's not continous, then it doesn't apply.
The EVT applies exactly when the hypothesis I mentioned in my comment exist, and there is no redundancy, meaning all of them must apply in order for the theorem to work. Importantly, it has nothing to do with differentiability.
$f(x)=\lvert x\rvert$  is indeed continuous so, pick a bounded, closed interval (say, $[a,b]$) then indeed, the EVT applies. Namely, the extreme values are $0$ if $ab&lt;0$ or $min(\lvert a\rvert,\lvert b\rvert)$ else, for the minimum; $max(\lvert a\rvert,\lvert b\rvert)$ for the maximum.
Again, for a discontinuous function such as the floor function the EVT does not apply, which doesn't mean that, by chance, the extreme values are not attained, which indeed they are, for any bounded interval.
A variant of this is Rolle's theorem (with is proved using the Extreme value theorem). If a real-valued function $f$ is continuous on a closed interval $[a, b]$, differentiable on the open interval $(a, b)$, and $f(a) = f(b)$, then there exists some $c$ in the open interval $(a, b)$ such that $f'(c)=0$. This means when you take two points of a function that satisfies this criteria (and produce equal $y$ values) there will be a relative min/max between those points.
However, if you take $f(x)=|x|$ (as an example $[a,b]=[-1,1]$) differentiability fails at an interior point of the interval (when $x=0$ the function isn't differentiable). Only when $(a,b)$ is differentiable and $f$ is continuous on $[a, b]$, we can deduce the curvature of the function (applies for Rolle's theorem).
So, Rolle's Theorem does not apply for $|x|$ because the function is not differentiable at $0$ (differentiation yields a step function). That is if you are trying to find a point when $f'(x)=0$. Hopefully, this is still useful for this problem or others down the line. However, the Extreme Value Theorem may yield a global minimum at $x=0$ because $f$ is continuous over all real numbers and our interval is bounded/closed.