Theorem 4.4.1. The Extreme Value Theorem.
If \(f\) is continuous on a closed interval, Then \(f\) obtains an global max and a global min on that interval.
Section 4.4 Extreme Value Theorem
The function we worked with in the previous section had both an absolute max and an absolute min. So, is that true for every function? Can you picture a function that doesn’t have an absolute max or an absolute min? Consider this function defined on the closed interval \([0,2]\)
\begin{equation*}
f(x) = \left\{ \begin{array}{cc}
x \amp 0 \le x \le 1 \\
3-x \amp 1 \lt x \le 2
\end{array}
\right.
\end{equation*}
Does this function have an absolute max? Let’s plot it to see if we can understand the question better. 
Now that we see its graph, what do you think? Does it *obtain* an absolute max? You might say, "Well, yeah. Isn’t it 2? Right there at \(x=1\text{?}\)" But... no. At \(x=1\) the value of the function is \(1\text{.}\) It never actually obtains the value 2. In fact, no matter where you evaluate this function, there are always larger values that the function obtains. So, what went wrong? A little thought leads us to conclude that it is that annoying discontinuity at 1 that is the problem. So, maybe as long as a function is continuous, we can avoid situations where the function does not obtain its absolute max. So, suppose we have *better* behaved, continuous function similar to the one we’re studying. For example, suppose that we just use the nice continuous function \(f(x)=x\text{.}\) Now, that one is continous everywhere so it shouldn’t have a problem. Right? Not so fast. Suppose we define it on the partially open interval \([0,2)\) instead of the closed interval \([0,2]\text{.}\) Then we get this graph 
Once again, we see that the function does not obtain an absolute maximum. The problem this time is that the domain interval has that darned open end. At this point, you may be thinking that there might be tons of ways a function could have problems that prevent extremes from being obtained. Fortunately, that’s not really the case. The two problems identified above are the only real issues. In fact, we have ...
Notice that this theorem makes finding optimums pretty easy. Namely, if you have a continuous function on a closed interval, it guarantees that the function WILL obtain an absolute max and an absolute min on that interval. So, we can just do a first derivative analysis to narrow our options down to a few critical numbers in the domain (i.e. eliminate open intervals where optimums can’t happen), evaluate our function at those places and the largest result must be the absolute max and the smallest the absolute min.
