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.
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Find the derivative, and then second derivative of f.
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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).
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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?
Best Answer
No.
Consider $f(x) = x^4$.
In order for there to be a change in concavity there needs to be a change of sign for the second derivative around our potential inflection point. However, $f''(x)=12x^2$ is positive to both sides of $x=0$ so it's not an inflection point.
You can visually tell this isn't an inflection point because the graph looks like a steeper parabola.