Scalar Autonomous ODE: Stability of critical points using the derivative Given a scalar autonomous differential equation of the form $\dot{x} = f(x)$. I want to investigate the stability of any critical point $x_0$. My textbook states that one can use the following criteria:

Suppose that
\begin{align}
f^{(k)}(x_0) = 0 \quad \text{for } k = 0, 1, \ldots, n-1 \quad \text{and} \quad f^{(n)}(x_0) \neq 0 \quad \text{for odd $n$}
\end{align}
then $f^{(n)}(x_0) < 0$ implies that $x_0$ is asymptotically stable and $f^{(n)}(x_0) > 0$ implies that $x_0$ is unstable.

What can one say about the stability of $x_0$ in the case that $n$ is even?
 A: For one-dimensional systems, there is a separate classification of equilibrium points, which includes stable, unstable, and semi-stable equilibrium points. Following this classification, the equilibrium points under consideration are semi-stable. According to the general classification, such equilibrium points are unstable. Let's prove it.
According to the Taylor's theorem with Peano's form of remainder
$$
f(x_0+\Delta x) = \underbrace{f(x_0) + f'(x_0)\Delta x +  \cdots + 
\frac{f^{(n-1)}(x_0)}{(n-1)!}(\Delta x)^{n-1}}_{=0} +
\frac{f^{(n)}(x_0)}{n!}(\Delta x)^n + o((\Delta x)^{n}),
$$
where
$$
\lim_{\Delta x\to 0} \frac{o((\Delta x)^{n})}{(\Delta x)^n}=0.
$$
Consider the limit
$$
\lim_{\Delta x\to 0}\frac{f(x_0+\Delta x)}{(\Delta x)^{n}}=
\lim_{\Delta x\to 0}\left(
\frac{f^{(n)}(x_0)}{n!} + \frac{o((\Delta x)^{n})}{(\Delta x)^n}
\right)
=\lim_{\Delta x\to 0}
\frac{f^{(n)}(x_0)}{n!} +
\lim_{\Delta x\to 0}
 \frac{o((\Delta x)^{n})}{(\Delta x)^n}
$$
$$
=\frac{f^{(n)}(x_0)}{n!} +0=\frac{f^{(n)}(x_0)}{n!}.
$$
The continuous function
$$
g(\Delta x)=\begin{cases}
\dfrac{f(x_0+\Delta x)}{(\Delta x)^{n}}, \Delta x\ne 0\\
\dfrac{f^{(n)}(x_0)}{n!}, \Delta x=0\\
\end{cases}
$$
preserves the sign of its limit, $\dfrac{f^{(n)}(x_0)}{n!}$, in some neighborhood
$\Delta x\in(-\varepsilon,\varepsilon)$.
For definiteness, let $f^{(n)}(x_0)>0$. If $f^{(n)}(x_0)<0$ , then the reasoning is similar.
For any $\Delta x\in (-\varepsilon,\varepsilon)\setminus \{0\} $ we have
$$
\dfrac{f(x_0+\Delta x)}{(\Delta x)^{n}}>0.
$$
Since $n$ is even, $(\Delta x)^{n}>0$ for $\Delta x\ne0$, therefore
$f(x_0+\Delta x)>0$ for $\Delta x>0$ and for $\Delta x<0$.
That is, the solution on both sides of $x_0$ moves upward.
This means that from the lower side, where $\Delta x<0$, the solution approaches $x_0$, and from the upper side, where $\Delta x> 0$, the solution moves away from
$x_0$, hence, the solution $x=x_0$ is semistable.
