**Theorem. Let \(f\) be a real function, which is differentiable up to order \(n\) in \(c \in \Bbb R\), for some positive integer \(n\). Suppose that \(f^{(n)}(c) \neq 0\), and that all preceeding derivatives (if any) are null in \(c\). Then \(c\) is a local extremum if and only if \(n\) is even.**

*In the following, the notation \(f^{(k)}(x)\), for \(k>0\) is used to identify the \(k\)-th order derivative of \(f\) in \(x\), and, for \(k=0\), \(f^{(0)}(x) = f(x)\). We will also assume, WLOG, that \(c=0\), \(f(0)=0\), and \(f^{(n)}(0)>0\).*

Since we have information only about the \(n\)-th order derivative in \(c\) **and not in a neighborhood** of \(c\), we cannot make use of Taylor’s Theorem. That is the main reason why I thought to propose to the reader the proof of this assertion. However, if we use the definition of derivative given for example in Rudin’s *Principles of Mathematical Analysis* (def. 5.1, page 103), for \(k=1,2,\dots\), the existence of \(f^{(k)}(c)\) implies the existence of \(f^{(k-1)}(x)\) in a *whole interval containing \(c\)*. This will allow us to make approprate use of the Mean Value Theorem.

First observe that \[\lim_{x\to 0}\frac{f^{(n-1)}(x)}{x} = f^{(n)}(0) > 0,\]implying that there exists a neighborhood \(\mathcal N\) of \(0\) such that \[\frac{f^{(n-1)}(x)}{x} > 0,\tag{1}\label{eq3649:1}\]for all \(x \in \mathcal N\setminus \{0\}\). If \(n=1\), we are done, since by \eqref{eq3649:1} \(f^{(0)}(x)=f(x)\) does not have a local extremum in \(0\).

If \(n>1\), from \eqref{eq3649:1} we can conclude that \[f^{(n-2)}(x)>0\tag{2}\label{eq3649:2}\] for all \(x\in \mathcal N\setminus \{0\}\). We can prove this assertion by contradiction. Suppose in fact there is \(\overline x \in \mathcal N\setminus \{0\}\) such that \(f^{(n-2)}(\overline x)\leq 0\). Then, by the Mean Value Theorem we would find \(\xi\) such that \[f^{(n-1)}(\xi) = \frac{f^{(n-2)}(\overline x)}{\overline x}.\] If \(\overline x >0\), then \(0< \xi <\overline x\), and \(f^{(n-1)}(\xi) \leq 0\). If, instead \(\overline x < 0\), then \(\overline x < \xi < 0\) and \(f^{(n-1)}(\xi) \geq 0\). Both situations contradict \eqref{eq3649:1}.

If \(n=2\), our conclusion is reached, since from \eqref{eq3649:2} it follows that \(0\) is a local minimum (this depends on the choice we have made to let \(f^{(n)}(0) > 0\), of course).

If \(n>2\), note that \eqref{eq3649:2} implies that \(f^{(n-3)}(x)\) is strictly monotonic in \(\mathcal N\). Together with the condition \(f^{(n-3)}(0) = 0\), this yields \[\frac{f^{(n-3)}(x)}{x} > 0\]for all \(x \in \mathcal N \setminus \{0\}\).

Now it is straightforward to proceed further and show that \[f^{(n-2k)}(x)>0 \ \ \mbox{for} \ \ k=0,1\dots ,\left\lfloor \frac{n}2\right\rfloor-1\] for all \(x \in \mathcal N \setminus \{0\}\), and that \[\frac{f^{(n-2k-1)}(x)}{x} > 0\ \ \mbox{for} \ \ k=0,1,\dots ,\left\lfloor \frac{n}2\right\rfloor-1\] for all \(x \in \mathcal N \setminus \{0\}\). The last two assertions easily yield our thesis.

Here are some interesting examples. The first one is taken from Robert Burn’s *Numbers and Functions.* Take \[f(x) = \begin{cases}x + 2x^2 \cos\left(\frac1{x}\right) & (x\neq 0) \\ 0 & (x=0).\end{cases}\] This function has first derivative equal to \(1\) in \(0\) (use the definition of derivative!). However \(f(x)\) **is not increasing **in any neighborhood of \(0\). You can show in fact that \(f\left(\frac1{2k\pi}\right) > f\left(\frac1{(2k-1)\pi}\right)\) for any \(k =1,2,\dots\). Note that \(f'(x)\) is not continuous in \(0\).

The above example can be generalized by\[f(x) = \begin{cases}\frac{ax^n} {n!} + x^{2n}\cos\left(\frac1{x}\right) & (x\neq 0)\\ 0 & (x=0),\end{cases}\]where \(f^{(k)}(0) = 0\) for \(0<k<n\) and \(f^{(n)}(0) = a\), with \(f^{(n)}(x)\) discontinuous in \(0\).

Note that the Theorem requires that there exists **at least one non-null derivative**.

If all derivatives up to a certain order \(n-1\) exist and are zero in \(c\), and the \(n\)-th order derivative does not exist, no conclusion can be drawn. Take as an example \[f(x) = \begin{cases} x^{2(n-1)}\left[2 + \cos \left(\frac1{x}\right)\right] & (x\neq 0) \\ 0 & (x=0). \end{cases}\tag{3}\label{eq3649:3}\]The above function, for any \(n>1\), has a (global) minimum in \(0\), it has \(f^{(k)}(0) = 0\) for \(0<k<n\), and \(f^{(n)} (x)\) is not defined in \(0\) (see Figure below).

**Further exercise. **

- Show that, for \(x \neq 0\), the derivative of function \eqref{eq3649:3} can be upper bounded as follows.\[f'(x) \leq x^{2n-4}\sin\left(\frac1{x}\right) + 6(n-1)x^{2n-3}.\]
- Deduce that for \(x = \frac1{2k\pi -\frac{\pi}2}\), with \(k\) integer and \(k>\frac14 + \frac{3(n-1)}{\pi}\), \(f'(x)\) is
*negative*. - Conclude that \(f(x)\)
*is not monotonically increasing in any right neighborhood of \(0\).*

On the same track, for \(n>1\), the function\[f(x) = \begin{cases} x^{2n-1}\left[2 + \cos \left(\frac1{x}\right)\right] & (x\neq 0) \\ 0 & (x=0). \end{cases}\] (which does not have a local extremum in \(0\), being an odd function) has null (and continuous) derivatives in \(0\) up to order \(n-1\), whereas the \(n\)-th derivative in \(0\) does not exist. A picture of the situation in this case is shown below.

Similarly, if all derivatives are null in \(c\), then no conclusion can be drawn about \(c\) being a local extremum or not. Consider for example \[f(x) = \begin{cases} e^{-\frac1{x^2}} & (x\neq 0) \\ 0 & (x=0),\end{cases}\]and\[f(x) = \begin{cases} \frac{x}{|x|}e^{-\frac1{x^2}} & (x\neq 0) \\ 0 & (x=0).\end{cases}\]