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Topic: Re: y=x^2...
Replies: 0

 Lou Talman Posts: 876 Registered: 12/3/04
Re: y=x^2...
Posted: Apr 28, 1998 12:08 PM

> y=x^2 is increasing on (0,1) but not on [0,1] since it is not
> increasing at x=0

The notion of a function that is "increasing at x = 0" is an
extraordinarily difficult one. Any definition of this notion must
necessarily be a local one. Most books that attempt it make the
definition of increasing at a point simply the positivity of the
derivative at the point in question. There are important deficiencies
in that definition, as I will show shortly.

The standard definition of monotonicity is global--not local: f is
increasing on the interval I if f[a] < f[b] whenever a and b are points
of I for which a < b. We can establish a very nice theorem connecting
the positivity of the derivative of f on I with its monotonicity there;
we usually do so in our elementary calculus courses, and that theorem
(and its cousins) are on the AP syllabus. The reason that this theorem
is of mathematical interest is that it connects an essentially *local*
hypothesis (f'[x] > 0 at each point of I) with a *global* conclusion
(f is increasing on the interval I). This connection can be mediated
by any of several devices; most people choose the Mean Value Theorem,
but there are others.

One reason for not introducing the notion of functions increasing at
a point is the loss of this mathematically interesting step from the
local to the global. If we simply *define* the increasing property
by the positivity of the derivative, mathematically interesting
connections vanish (by fiat).

But there are deeper reasons. Whatever the local notion of increasing
may be, we should certainly like it to have some connection with the
standard global notion. The obvious, and very desirable, connection
would be this: If f is increasing at x = 0, then f is increasing on
some open interval centered at x = 0.

It is an unfortunate fact that there is no such connection.

Consider the function f given by

f[x] = x^2 Sin[1/x] + x/2, when x is not 0, and

f[0] = 0.

It is easily checked that f'[0] = 1/2, so that f is "increasing at
x = 0". But when x is different from 0 we have

f'[x] = 2 x Sin[1/x] - Cos[1/x] + 1/2,

and so f'[x] < 0 whenever x = 1/(2 k Pi) for any integer k, while
f'[x] > 0 whenever x = 1/((2 k + 1) Pi) for any integer k. Thus,
f is neither increasing nor decreasing on any open interval centered
at 0, for, by continuity, every such interval contains sub-intervals
on which f'[x] < 0 and other sub-intervals on which f'[x] > 0.

It is for this reason that careful textbook authors avoid the idea
of "increasing at a point".

--Lou Talman