Does third derivative verify if we have a point of inflection?

[dumbQuestion]

Hi.


I know that where the first derivative equals 0 or doesn't exist, we might have a local min or max. We can plug these critical numbers into the second derivative and depending if we get positive or negative answer, that tells us if the point is in fact a local min or max


Well, where the second derivative equals 0 or doesn't exist we might have a point of inflection. Can we plug these critical numbers into the third derivative and depending if we get positive or negative, verify such a result?


Thanks

 

[HallsofIvy]

A "point of inflection" is, by definition, a point at which the concavity, which is given by the change in the sign of the second derivative. Certainly, as long as the second derivative does change sign passing x= a, it must be 0 at that point. But it is possible that the the second derivative goes down to 0 but instead of becoming negative, goes back up or vice versa. We can, however, say that if the second derivative does go from being negative to positive it is an increasing function and so its derivative, the the third derivative of the function, must be positive and vice versa for a function whose second derivative is decreasing, from positive to negative.

You should be able to see that this argument for a "third derivative test" for a point of inflection is exactly the same as the argument for the "second derivative test" for a max or min- except that we don't normally worry about whether the third derivative is positive or negative.

 

[haruspex]

I know that where the first derivative equals 0 or doesn't exist, we might have a local min or max. We can plug these critical numbers into the second derivative and depending if we get positive or negative answer, that tells us if the point is in fact a local min or max

Well, where the second derivative equals 0 or doesn't exist we might have a point of inflection. Can we plug these critical numbers into the third derivative and depending if we get positive or negative, verify such a result?

In the case of the second derivative being 0, yes. In general, you can just keep taking derivatives (so long as they all exist) until you get one that's nonzero. There are four cases:
- even derivative, +ve: minimum
- even derivative, -ve: maximum
- odd derivative, +ve: 'upward' inflexion (i.e. locally non-decreasing)
- odd derivative, -ve: 'downward' inflexion
You can see this by considering a Taylor expansion.
But I'm not comfortable with your references to derivatives not existing. All bets are off in that case, no?