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What is the main difference between the extreme points of a function of special types, such as: critical points, singular points, endpoints? Being more specific, I am talking about finding extreme values of derivatives, I understand the concept of critical points, which are where f'(x) = 0 and singular points, where f'(x) is not defined. But what about endpoint? Definition says that it is a point that do not belong to any open interval in D(f) which is kind of contradicting to the idea of finding endpoints

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Their definitions are different and not in a subtle way, so could you be more specific? –  Dylan Moreland Dec 5 '11 at 19:30
    
I have edited the post –  Mocialov Boris Dec 5 '11 at 19:44
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Hm. I don't see any contradiction: If my domain is $[a, b]$, then there is no open interval $I$ such that $a \in I \subset [a, b]$. So $a$ is an endpoint of the domain. What's bogus about this? –  Dylan Moreland Dec 5 '11 at 20:49
    
@DylanMoreland you say a is an endpoint of the domain, but why not b? Since the interval is a subset of [a, b] –  Mocialov Boris Dec 6 '11 at 13:04
    
$b$ is an endpoint as well, sure. –  Dylan Moreland Dec 6 '11 at 13:55
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1 Answer

You have learned in calculus that when a function is differentiable on an open interval $J$ and has a local extremum at a point $\xi\in J$ then necessarily $f'(\xi)=0$.

Now the usual situation in applications is the following: You are given an interval $I:=[a,b]\subset{\mathbb R}$ and a continuous function $f:\ I\to{\mathbb R}$, and you want to find the minimal or maximal value of $f$ on $I$. As $f$ is continuous and $I$ is compact, by general principles you know that these extremal values "are taken", i.e., there exist $\xi_1$ and $\xi_2\in I$ such that

$$f(\xi_1)\leq f(x)\leq f(\xi_2)\qquad (a\leq x\leq b)\ .$$

If $f$ is differentiable in the open interior $]a,b[$ of $I$, with the exception of finitely many points $s_i$, $\>a<s_1<s_2<\ldots< s_n<b$, then we can argue as follows: The point $\xi_1$ (and similarly $\xi_2$) is either one of the points $a$, $s_1$, $\ldots$, $s_n$, $b$; or it is a point in one of the open intervals $J_i$ between these special points. In the latter case $\xi_1$ is a a local minimum in the interior of an open interval, and we have $f'(\xi_1)=0$. It follows that $\xi_1$ is a solution of the ("algebraic") equation $f'(x)=0$. Hopefully there are only finitely many such solutions $x_1$, $\ldots$, $x_m$ lying in $]a,b[\ $.

We now produce a large candidate list

$$S:=\{a,s_1,\ldots, s_n,b,x_1,\ldots , x_m\}\ ,$$

and we know that $\xi_1$, $\xi_2\in S$. Which one(s) of the points in $S$ lead(s) to the minimum of $f$ and which one(s) to the maximum has to be found out by computing and comparing the values $f(x)$ for all $x\in S$.

One more note: Given such a situation one defines the set $\arg\min f\bigm|_{ [a,b]}$ (and similarly $\arg\max$) by

$$\arg\min f\bigm|_{ [a,b]}\ :=\ \bigl\{\xi\in[a,b]\ \bigm|\ f(\xi)\leq f(x)\quad\forall x\in [a,b]\bigr\}\ .$$

We have proven that $\arg\min f\bigm|_{ [a,b]}\subset S$, and similarly for $\arg\max$.

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