Some of the topics discussed today in lecture came up on a question at Math.Stackexchange. I am reposting my answer here, slightly edited.
This is actually a somewhat subtle issue.
Suppose first that is an interval on , that , that , and that is an interior point of . We could just consider one-sided derivatives if is an end-point of , but that seems an unnecessary distraction.
That means that
so if is sufficiently small, we can assume both that , and that
Considering , this means that if is sufficiently close to , then . Considering , this means that if is sufficiently close to , then . If this is all we mean by “ is increasing at ” then indeed implies that. Similarly, would imply that is decreasing at .
The definition here would be that is increasing at iff there is an interval about such that if and , and if and . But, really, this is a silly notion: The function below given by
is increasing at .
However, what is not true is that if , then is increasing on an interval around (even if we assume that is differentiable everywhere). It is common to define the notions of increasing and decreasing so that they apply to functions defined on intervals, rather than to individual points of the domain: We say that is increasing on a set iff whenever are in then (with replaced by if we insist that “increasing” be interpreted in the strict sense.)
To see that indeed needs not be increasing on any interval containing , no matter how small, consider the example suggested in these slides by Louis A. Talman, on The Mother of All Calculus Quizzes: Let be the function given by and, if , then
This function is differentiable everywhere, with , and
for . Note that there are (both positive and negative) values of arbitrarily close to where , so “ is decreasing at “. This is because we can find arbitrarily small with , so and . (Note that the in the definition of could be replaced with any function with small enough derivative to ensure the same behavior.)
Finally, note that is not continuous at . This is an essential feature of the example. For suppose that exists on a neighborhood of and is continuous at . If , then for sufficiently close to we have as well. This means that is indeed increasing on a neighborhood of . For example, if and they are close enough to to ensure that at all , then use the mean value theorem to see that there is a with
that is, .