Theorem: CCD.EVT.0: If $f$ is a continuous function on the interval $[a,b]$  then there are numbers $c_*$ and $c^*$ in $[a,b]$  where for all $x \in [a,b]$,  $f(c_*) \le f(x) \le f(c^*)$.
Theorem: CCD.EVT.1: If $f$ is a continuous function on the interval $[a,b]$ then there are numbers $c_*$ and $c^*$ in $[a,b]$  where for $f( [a,b]) = [f(c_*) ,f(c^*)]$
The proof of CCD.EVT.0 relies on an important
Lemma: CCD.BVT:If $f$ is a continuous function on the interval $[a,b]$  then there are numbers $V_*$ and $V^*$ where for all $x \in [a,b]$,  $V_* < f(x) < V^*$.

A Bisection Method to prove CCD.EVT.0 can be created using the BVT. DO This! by finding $c_*$ and $c^*$ for any interval $[a,b]$: THIS NEEDS TO BE CREATED!

The procedure starts by letting $M_1 = \frac {f(a)+V^*}2$  and  $m_1 = \frac {f(a)+V_*}2$
If  for all $x \in [a,b]$ $le f(x) \le M_1$. then we can replace V_* with M_1 for the next step. If not  there is a number $a_1 \in [a,b]$ where $f(a) < f(a_1)<V^*$ and we can replace $a$ with $a_1$ for the next step...
Continue to arrive at $M^*$ with $M^*$ the least upper bound of $f([a,b])=\{f(x):x \in [a,b]$. It can then be shown that $M^* \in f([a,b])$ or there is a number  $c^*$ in $[a,b]$ where $f(c^*)= M^*$ and thus for all $x \in [a,b], f(x) \le f(c^*)$

A similar argument arrives at $M_*$ with $M_*$ the greatest lower bound of $f([a,b])=\{f(x):x \in [a,b]$. It can then be shown that $M_* \in f([a,b])$ or there is a number  $c_*$ in $[a,b]$ where $f(c_*)= M_*$ and thus for all $x \in [a,b], f(x) \ge f(c_*)$
Here is a GeoGebra visualization of the Extreme Value Theorem Illustrating how to find $c_*$ and $*c^*$ using mapping diagrams.