I'm very confused with all the different definitions and conventions regarding how we should treat endpoints.

One book says the following: let $f(x)$ be a function defined on a segment $[a,b]$. If the left-derivative of $b$ is greater than zero then $b$ is called endpoint maximum (if it's less than zero - endpoint minimum). The same goes for $a$, but there we shall check the right-derivative.

The second book doesn't even allow the possibility of endpoint extrema. It just says that if any endpoint is the highest/lowest of all the points in the segment then they are simply "highest/lowest" points.

The third book also doesn't allow the possibility of endpoint extrema because it's not really a differentiable point (the derivative must exist at both sides).

The fourth book says that any endpoint is automatically a local extremum (it may be the highest/lowest point and then it will be called a global extremum).

Wolfram Alpha says that contrary to what the first book has defined, endpoint is an extremum only if it is the highest/lowest point of all. Say, for example, we have the function $f(x)=(x-1)^2$ for $x \in [0,3]$. Then, according to Wolfram, there is only one minimum (both global and local) which is $x=1$. There's also a global maximum (not clear whether it is also a local one) at $x=3$. But $x=0$ is not an extremum - not even a local maximum of the function, despite the fact that it has a negative right-derivative.

So why $x=0$ is not an extremum here? Does it imply that endpoint extremum can exist only of it is the highest/lowest point of all the points in the segment?


Great question.

First note that to some degree this is about conventions and not mathematical correctness, so there is not necessarily one right answer to the query.

That said, I think some points should be clarified. (I will use your example throughout)

  1. With respect to the interval $[0,3]$ both endpoints must be local extrema. With respect to the entire domain the of the function ($\mathbb{R}$ in this case) both endpoints are not necessarily extrema, because they may or may not be the largest/smallest values of the function over the interval $[0,3]$. So it depends on what you are taking to be the "domain" of the function. In this sense the first and fourth books basically agree.

  2. I'm not entirely sure what books 2 and 3 are getting at by "now allowing endpoint extrema". Whether or not you say the function is differentiable at the endpoints, as mentioned above, depends on whether you are restricting the domain of the function to the interval or not. But either way, the endpoints can certainly have the highest or lowest values of the function over the interval.

  3. So in general: differentiate and set equal to $0$ to find "critical points". Then determine what local extrema lie inside the interval. Then check boundary points (in this case the two endpoints of the interval, in multi-variable calculus you will have to use Lagrange multipliers or a similar technique). Compare internal critical points and boundary points to find the absolute extrema (max and min) over the interval.

I think in general it is most consistent with mathematical convention to refer to the points of extrema as "extrema", whether or not they are at the endpoints.

  • $\begingroup$ Thanks for your answer. Regarding 2. - essentially they say that endpoints by definition can't be extrema because you can't evaluate the derivative there (the "usual" one, i.e. - from both sides). I want to make sure that I've understood you: if the endpoint isn't the highest/lowest one (like $x=0$ in my example), can I still call it a local extremum if it agrees with the definition of the first book (right/left derivatives are non-zero)? $\endgroup$ – conf Apr 20 '14 at 18:28
  • $\begingroup$ @conf You can certainly call it a local extremum if you want. Note though: I don't see why it matters what the one-sided derivative is. For example, all points $x\in\mathbb{R}$ are technically local extrema of the constant function $x=c$ with derivative $0$ everywhere. $\endgroup$ – user142299 Apr 20 '14 at 18:34
  • $\begingroup$ Well, I guess it matters for the same reason that calling points extrema where the derivative is zero (and changes sign) matters. Otherwise, why then bother comparing the local extrema in the interval with boundary points? The confusing thing is why if the endpoint is the highest/lowest of all the interior extrema we call it "absolute max/min" while if the endpoint isn't the highest/lowest we ignore it (yes, it won't be absolute but it will still be a local one). $\endgroup$ – conf Apr 20 '14 at 18:47
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    $\begingroup$ @conf I'd say you should call it local extremum - book 2 - is verry silly there. Otherwise, not every global extremum is a local extremum, which would be a bit weird. $\endgroup$ – fgp Apr 20 '14 at 19:06

This becomes much clearer once you are carefull to distinguish between local and global extrema.

A global minimum (respectively maximum) of $f$ is a point $x$ such that for all other $y$ in the domain of $f$, $f(x) \leq f(y)$ (respectively $f(x) \geq f(y)$ for a global maximum).

A local minimum (respectively maximum) of $f$ is a point $x$ such that for all other $y$ in a small (non-empty) interval $(x-\epsilon,x+\epsilon)$ around $x$, $f(x) \leq f(y)$ (respectively $f(x) \geq f(y)$ for a global maximum). We agree to simply ignore $y \in (x-\epsilon,x+\epsilon)$ for which $f$ is not defined here.

If $f$ is differentiable on $[a,b]$, you then have that

  1. A point $x \in [a,b]$ is a local extremum if $$ f'(x) = f''(x) = \ldots = f^{(n)}(x) = 0, \, f^{(n+1)}(x) \neq 0 \text{ for some odd $n > 0$,} $$ where $f^{(k)}$ denotes the $k$-th derivative of $f$ (or one-sided derivative if $x$ is an endpoint). If no non-zero derivative exist for some $x$, then this criterion is inconclusive. If $f^{(n+1)}(x) > 0$, then $x$ is a local minimum, otherwise $x$ is a local maximum.

  2. An enpoint $x \in \{a,b\}$ is additionally a local extremum if $$ f'(x) = f''(x) = \ldots = f^{(n)}(x) = 0, \, f^{(n+1)}(x) \neq 0 \text{ for some even $n \geq 0$,} $$ i.e. we can drop the requirement that $n$ is odd here because we don't need or want to exclude saddle points. The start point $x=a$ is a local minimum if $f^{(n+1)}(a) > 0$ and a local maximum if $f^{(n+1)}(a) < 0$, whereas for the end point $x=b$, the conditions are reversed.

In both cases, if $f$ isn't sufficiently smooth to find a non-zero derivative, i.e. if all existing derivatives are zero, then these conditions are inconclusive.

These conditions correctly identify the extrema of $x^k$ on $[0,1]$ and $[-1,0]$ at $x=0$. The first non-zero derivative of $x^k$ is the $k$-th derivative, which is $k!$. Thus, $n+1=k$, and we thus find that for even $k$ (i.e., odd $n$), $0$ is always local minimum, whereas for odd $k$, $0$ is a local minimum in $[0,1]$ but a local maxinum in $[-1,0]$.

Book 1's statement that a point is a (local, I assume) endpoint extrema if the one-sided derivative is non-zero is true, but incomplete with these definition. Take for example $f(x) = x^4$ on $[0,1]$. $x=0$ is obviously a local minimum - it's even global minimum of $x \to x^4$ on the whole real line!. Yet $f'(0) = 0$ is zero, same as $f''(0)$ and $f'''(0)$. But $f^{(4)}(0) = 4! > 0$, which indeed indicates that $0$ is local minimum.

Book 4's statement that every endpoint is automatically a local extremem is wrong with these definition. Take for example $f(x) = x^3\sin \frac{1}{x}$ for $x \in (0,1]$, $f(0) = 0$. You have $f'(x) = -x\cos \frac{1}{x} + 3x^2\sin \frac{1}{x}$, meaning $f'(0) = 0$. Yet every interval $[0,\epsilon)$ contains points where $f$ is positive as well as point where $f$ is negative, so $x=$ isn't a local extremum. Note that $f''(x)$ doesn't exists at $x=0$, so the conditions above are inconclusive about whether $x=0$ is a local extremem or not, which is consistent with the fact that it isn't.

For your function $f(x) = (x - 1)^2$, $x \in [0,3]$ you have $f'(x) = 2x - 2$, $f''(x) = 2$, so if we apply these definition we find

  • By (1), using $n=1$, that $f^{(n)}(1)=f'(1)=0$ and $f^{(n+1)}(1) = f''(1) = 2$, meaning $x$ is a local minimum.
  • By (2), using $n=0$, that $f^{(n+1)}(0) = f'(0) = -2 < 0$, meaning the start point $0$ is a local maximum.
  • By (2), using $n=0$, that $f^{(n+1)}(3) = f'(3) = 4 > 0$, meaning the end point $3$ is a local maximum.

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