Is it possible to identify whether an inflection point such as this example, contained in y = x^3 from the wikipedia: enter image description here

Is positive or negatively oriented (i.e. the gradient leading up to, and away, from it) without visualising it or taking a point from either side?

We went over determining between maximum, minimum and inflection points using second derivatives in AS mathematics today, and I felt that the method of taking a point on either side wasn't a very conclusive (Since, in theory, there could be a turning point very close to it, couldn't there? Which could affect this) or pure way of going about it, although I could well be wrong. I enquired about it to my lecturer but she stated that there's no method that's in the specification for our exam and that if there is one it's probably higher level.

I did look around for information but it seems my vocabulary is too basic to find any relevant questions. I did find this on the Wikipedia:

If x is an inflection point for f then the second derivative, f″(x), is equal to zero if it exists, but this condition does not provide a sufficient definition of a point of inflection. Source

Which basically sums up what I already knew about second derivatives. It goes on to explain about determining between undulation and inflection points, however I found little resources on-line to explain what an undulation point is, and therefore am not sure if this is relevant to my question.

Thanks very much, again, sorry if this question seems too basic to present here-- I'm just intensely curious about mathematical methods, and will be thinking about this for weeks otherwise.

Note that I am only interested in 2D Cartesian geometry.


1 Answer 1


Generally you find a list of "possible" inflection points by finding all solutions to $f''(x) = 0$. If $x_0$ is one of these possible inflection points then you can test that it's an inflection point, exactly as you were taught in class, by testing points $a$ and $b$ on either side of $x_0$. In order to prevent there being, as you say, a "turning point" in between your test point $a$ and your possible inflection point $x_0$ you just have to make sure that none of the other solutions to $f''(x) = 0$ lie between $a$ and $x_0$.

So by looking at the solutions to $f''(x) = 0$ you can decide how close your test points need to be to your possible inflection point in order to prevent the situation that you describe.

  • $\begingroup$ So is there no way of doing it purely using the coordinates of the inflection itself? $\endgroup$ Jan 23, 2014 at 21:24
  • $\begingroup$ What do you mean "purely using the coordinates of the inflection"? You mean without knowing what the function $f$ is? No, there's no way of doing that. $\endgroup$
    – Jim
    Jan 23, 2014 at 21:25
  • $\begingroup$ Ah, okay. Thanks for the information! $\endgroup$ Jan 23, 2014 at 21:27
  • $\begingroup$ I meant not subbing in alternative coordinates into the gradient function (Or any other way of using other coordinates to determine the gradient adjacent to inflection points), just in case that makes a difference, but I'm figuring it doesn't. $\endgroup$ Jan 23, 2014 at 21:29
  • $\begingroup$ If the third derivative at the possible inflection point is nonzero then the concavity changes there and so true inflection point. I think one can try higher derivatives in order until the first one which is nonzero, and if the degree where you have stopped is odd one has actual inflection. $\endgroup$
    – coffeemath
    Jan 23, 2014 at 21:49

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