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My teacher told me (and Wikipedia backs this up) that the second derivative of a stationary point doesn't tell us anything about whether the point is a maximum, minimum or an inflection point. But I don't understand how it can be anything other than an inflection point.

Please can you:

a) explain why it can be a maximum/minimum with a second derivative of $0$

b) give an example where this happens (where there is a maximum/minimum and it has a second derivative of $0$)

Thank you.

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  • $\begingroup$ you mean (as per the title) " that the second derivative of a stationary point BEING ZERO doesn't tell (...)"? $\endgroup$ – Mefitico Mar 15 '18 at 21:25
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Consider polynomials $x^4$ for minimum and $-x^4$ for maximum.

The intuition is that "interesting things" might happen beyond the second derivative. To give you a more curious example: $$f(x) = \begin{cases}e^{-1/x^2} & \text{ for }x\neq 0\\0&\text{ for } x=0\end{cases}$$

is a smooth function which has all its derivatives at zero equal to zero, and yet it has a minimum at that point.

I hope this helps $\ddot\smile$

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  • $\begingroup$ Yes that's very useful thank you $\endgroup$ – Dan Mar 15 '18 at 21:29
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Pick $x^3$ , then both $f'(0)=0$ and $f''(0)=0$, and we have an inflection point.

Pick $x^4$ , then both $f'(0)=0$ and $f''(0)=0$, and we have an minimum point.

Pick $-x^4$ , then both $f'(0)=0$ and $f''(0)=0$, and we have an maximum point.

This is why the information is inconclusive, all three cases may happen. With higher order derivatives, for a single variable function, you can tell what is the case though.

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An example: $x^4$.

A intuition for why is that stationary point implies "stop" in the increasing/decreasing of the function. So in other words it is the point our derivative is equal to $0$, if the second derivative is positive the rate of change is increasing hence it is minimum, if negative the rate of change is negative hence maximum, but if a point before we have negative (still on the second derivative) at the point we have inflection point and after that point we have positive? In this case it is not different from the first case.

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Your teacher is correct. However, most of the times, sign of second derivative would give correct assignment if its magnitude is greater than $0$. The trick we used in high school was whenever $f''(x)=0$, check the sign change of the graph when from $f'(x-\delta x)$ to $f'(x+\delta x)$ at the critical point. For example, see above three examples given by dtldarek and Mefitico:

(a) $f(x)=x^4$, thus $f'(x)=4x^3$; $f'(x)=0$ when $x=0$ (and $f''(x)=0$ as well). But, $f'(x-\delta x)=(-\delta x)^3=(-)$ and $f'(x+\delta x)=(+\delta x)^3=(+)$, when $x$ increases through critical point ($x=0$). Therefore, $x=0$ point is a minimum (tangent to curve just before critical point is negative but just after is positive).

(b) $f(x)=-x^4$, thus $f'(x)=-4x^3$; both $f'(x)=0$ and $f''(x)=0$ when $x=0$. But, $f'(x-\delta x)=-(-\delta x)^3=-(-)=(+)$ and $f'(x+\delta x)=-(+\delta x)^3=-(+)=(-)$, when $x$ increases through critical point ($x=0$). Therefore, $x=0$ point is a maximum (tangent to curve just before critical point is positive but just after is negative).

(c) $f(x)=x^3$, thus $f'(x)=3x^2$; both $f'(x)=0$ and $f''(x)=0$ when $x=0$. But, $f'(x-\delta x)=(-\delta x)^2=(+)$ and $f'(x+\delta x)=(+\delta x)^2=(+)$, when $x$ increases through critical point ($x=0$). Therefore, $x=0$ point is neither a maximum nor a minimum (tangent to curve just before and after critical point is positive; no change), meaning it is an inflection point.

Also, there are some curves where tangent to the curve just before and after a critical point would be negative as well. Above example (c) happens to be positive.

I hope this would help you to understand.

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