Let $U\subset \mathbb{R}$ be open, let $f:U\to \mathbb{R}$ be locally Lipschitz, and let $x^*\in U$ satisfy $f(x^*)=0$. Suppose that $x^*$ is an isolated equilibrium point, in the sense that there exists $\eta>0$ such that $(x^*-\eta,x^*+\eta)\subset U$ and $f(x)\ne 0$ whenever $0<|x-x^*|<\eta$.

In this statement, a solution $x:J\to U$ of $\dot{x}=f(x)$ on an interval $J\subset \mathbb{R}$ means a continuous map that is differentiable at every interior point of $J$ and satisfies $\dot{x}(t)=f(x(t))$ at every interior point $t\in J$.

If $f(x)>0$ for every $x\in(x^*-\eta,x^*)$ and $f(x)<0$ for every $x\in(x^*,x^*+\eta)$, then every solution $x:J\to U$ of $\dot{x}=f(x)$ on an interval $J\subset \mathbb{R}$ with $0\in J$ and $x(0)\in(x^*-\eta,x^*+\eta)\setminus\{x^*\}$ stays on its initial side of $x^*$ for positive time while it remains in $(x^*-\eta,x^*+\eta)$, moves monotonically toward $x^*$ while it remains in that interval, and converges to $x^*$ as $t\to\infty$ whenever $[0,\infty)\subset J$ and $x([0,\infty))\subset(x^*-\eta,x^*+\eta)$.

If instead $f(x)<0$ for every $x\in(x^*-\eta,x^*)$ and $f(x)>0$ for every $x\in(x^*,x^*+\eta)$, then every solution $x:J\to U$ with $0\in J$ and $x(0)\in(x^*-\eta,x^*+\eta)\setminus\{x^*\}$ moves monotonically away from $x^*$ for positive time while it remains in $(x^*-\eta,x^*+\eta)$.