Scalar Autonomous ODE: Stability of critical points using the derivative

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.