Let $f(x)$ be a continuous function on $x\in[a, b]$. Let: $$ m(x) = \min_{t\in[a, x]}f(t)\\ M(x) = \max_{t\in[a,x]}f(t) $$ Prove both $m(x)$ and $M(x)$ are continuous.
This statement seems so obvious, but I was not able to write down a formal proof.
There is a couple of trivial facts I've figured out so far. First by definition of $\min x$ we know it is monotonically decreasing. Also $f(x)$ is bounded and by Bolzano-Weierstrass there exists $x_0\in[a,b]$ such that: $$ f(x_0) = \inf_{[a, b]}f $$
So we might only consider the interval $[a, x_0]$, because $m(x)$ is going to be constant for $x \ge x_0$. Similar facts are true for $M(x)$.
I've also been playing around with the following inequalities without any success: $$ \min_{x\in[a,b]} f(x) \le \min_{t\in[a, x]} f(t) \le f(x) \le \max_{t\in[a, x]} f(t) \le \max_{x\in[a,b]} f(x) $$
What would be a way to prove the statement?
