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.

  1. Find the derivative, and then second derivative of f.

  2. 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).

  3. 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?


Concavity generally stays the same the same on intervals containing critical numbers. The exception is when the critical number is an inflection point instead of a maximum or minimum. Think about it: consider a typical maximum. The curve reaches a high point and then heads back down. Does it switch from concave down to concave up here? No. It is concave down on both sides. A typical minimum is the just the opposite. It will be concave up on both sides. I don't know where you got the idea that $f''$ is undefined at critical points. There is no particular reason for it to be.

For a function $f(x)$ whose 2nd derivative is continuous:

  • A critical point is a point $x$ where $f'(x) = 0$. If $f''(x) > 0$, then the point is a minimum. If $f''(x) < 0$, then the point is a maximum, and if $f''(0) = 0$, then it could be any of the three (maximum, minimum, or inflection point).
  • An inflection point is a point $x$ where $f''$ switches from positive to negative, or vice versa. Since $f''$ is continuous, this necessarily means that $f''(x) = 0$, but the inverse is not necessarily true: $f''(x)$ can be $0$ without it being an inflection point ($f(x) = x^4$ at $0$, for example).

As a specific example following your pattern, consider $f(x) = x^3 - 3x$. Then $f'(x) = 3x^2 - 3$ has roots at $-1$ and $1$. By your pattern, you could look at $f''(x) = 6x$ at say $-2, 1/2, 2$ and claim that $f$ is concave downward on $(-\infty, -1)$ and concave up on $(-1,1)$ and on $(1, \infty)$. But if you chose $-1/2$ instead of $1/2$, you would have it concave down on $(-1, 1)$.

The truth is that $f''(x) = 6x = 0$ at $x = 0$. Further $f''(x) < 0$ for $x \le 0$ and $f''(x) > 0$ for $x > 0$. So $f$ is concave down on $(-\infty, 0)$ and concave up on $(0, \infty)$.

  • $\begingroup$ I think I worded my question wrong. I should have just said Does the concavity have to change at f''(a) == 0? When I said critical points, I meant critical points of f', not critical points of f. Although even with my poor wording you still answered my question with f''(x) can be 0 without it being an inflection point. $\endgroup$ – DJMcMayhem Oct 16 '15 at 4:25
  • $\begingroup$ If would have helped if I had noticed you said critical points of $f'$ instead of $f$. I'm afraid my eyes are no longer up to spotting the difference between those easily. Glad I was able to answer the question despite the confusion. $\endgroup$ – Paul Sinclair Oct 16 '15 at 16:24


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.


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