Intuition and counterexamples for higher-order derivative test In the higher-order test we keep differentiating a function till we find the n'th derivative (n being even) to be greater than or less than zero thereby identifying it as a minimum or maximum.
My two questions refer to the one-dimensional case only:


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*What is the intuition behind this result? To be more exact: How could the first and second derivatives be zero yet we have an extreme value anyway? If you put a tangent line on a function and trace the function with it (from smaller to bigger x-values that is) intuition has it that it would e.g. first point down and then point up, being horizontal in between: Bingo, you found a minimum. How can you have a minimum without this happening (i.e. first and second derivatives being zero)?

*In all the text I have found so far they seem to be a little passive when it comes to the part when even this procedure won't work. There seem to be cases where all derivatives are zero yet we have an extreme value anyway. So in which cases will this happen? What will you have to do then? Could you give a (counter-)example?

 A: Consider a particle which is at position $x(t) = \frac{t^4}{4!}$ at time $t$. Its acceleration is $\ddot{x}(t) = \frac{t^2}{2!}$, which means that at $t=0$ it is stationary and it does not (at the moment) accelerate. Shouldn't it stay in place forever? No, the thing is that it will accelerate in a moment (precisely for any positive time), and the reason is, that jounce (i.e. the fourth derivative) is positive. 
This could be formulated as "promises", where velocity is a promise that the position will change. To go further consider the following table:
$$\begin{array}{|c|c|c|}\hline
\textbf{condition} & \textbf{name} & \textbf{description} \\\hline
|x| > 0 & \text{position} & \text{the position has changed} \\\hline
|\dot{x}| > 0 & \text{velocity} & \text{the position is changing} \\\hline
|\ddot{x}| > 0 & \text{acceleration} & \text{promise that } \\
&&\text{the position will be changing} \\\hline
|\dddot{x}| > 0 & \text{jerk} & \text{promise about promise that }\\
&&\text{the position will be changing} \\\hline
|\ddddot{x}| > 0 & \text{jounce} & \text{promise about promise about promise }\\
&&\text{that the postition will be changing}
\\\hline
\end{array}$$
Although we have no names for higher derivatives, it's easy to generalize the above. So, even if many derivatives are zero, we still has the appropriate promise, hence the function might have an extremum in that point.
However, this is not all! There are even more funny examples, like $$x \mapsto \begin{cases}e^{-\frac{1}{x^2}} & \text{ if} x \neq 0 \\ 0 & \text{ otherwise}\end{cases}$$ which also has minimum at zero, but all its derivatives there are zero (picture courtesy of Wolfram Alpha)!
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Of course, one also could consider $x \mapsto \mathrm{sgn}(x)\cdot e^{-\frac{1}{x^2}}$ (with the obvious smoothness fix at zero) which does not have minimum nor maximum there, i.e. all derivatives being zero tells us very little about how the function might look in the future.
You might argue that this is a weird example, but observe that in nature all functions/movements might be just like that. A car might be stationary, but then it starts moving. How did it happen? Is its movement infinitely differentiable? You might say that no, because combustion in engine and other things are not, but what about the tires and ground friction? I shall stop here, as this is not mathematics anymore.
I hope this helps $\ddot\smile$
