In school I'm learning about concavity and finding points of inflection using the second-derivative test. A typical question will look like
Determine the open intervals on which the graph is concave upward or concave downward.
Here is how I would solve a problem like that.
Find the derivative, and then second derivative of f.
Find the critical numbers of f'(x) by setting f''(x) = 0 and f''(x) is undefined, then simplify. Let's say I get the critical numbers a and b (a < b).
Take f'' of a number in the open interval $(-\infty, a)$ If it is negative, f is concave downward on $(-\infty, a)$ if it is positive, it is concave upward. Repeat this process for a number in $(a, b)$ and in $(b, \infty)$
In general, I will get a result like
f is concave upward on $(-\infty, a) \cup (b, \infty)$
and concave downward on $(a, b)$
But now I'm wondering, is it possible for the concavity to stay the same even on an interval containing a critical number of f prime? For example, if f'(x) is increasing when x < a, f'(x) = 0 @ x = a, and f'(x) is still increasing when a < x < b? Does that mean that f is concave upward on $(-\infty, b)$ Can this even happen?