Let $f $ be a continuous real-valued function on a closed interval $[a, b] \implies f$ is bounded.
Moreover, $\exists x_0, y_0 \in [a, b], \forall x \in [a, b], f(x_0) \leq f(x) \leq f(y_0)$.
If $f $ is a continuous real-valued function on an interval $I $, then $f $ has the intermediate value property on $I $:
$a, b \in I$, $a < b$. $\forall y $ lies between $f(a)$ and $f(b) $, $\exists x \in (a, b)$ such that $f(x) = y $.
If $f$ is a continuous real-valued function on an interval $I $, then the set $f(I) = \set{ f(x) \mid x \in I } $ is also an interval or a single point.
Let $f $ be a continuous strictly increasing function on some interval $I$.
Then $f(I) $ is an interval $J $ bt the previous corollary and $\inv{ f } $ represents a function with domain $J$. The function $\inv{ f } $ is a continuous strictly increasing function on $J $.
Let $g$ be a strictly increasing function on an interval $J$, $g(J)$ is an interval $I$ $\implies g$ is continuous on $J $.
$f $ is one-to-one and continuous function on an interval $I \implies f $ is strictly increasing or strictly decreasing.