**Proof plan.** The sign of $f'(x^*)$ determines, by [continuity](/page/Continuity), the sign structure of $f$ in a neighbourhood of $x^*$. In one dimension, this sign structure completely constrains the direction of motion, and uniqueness of solutions prevents trajectories from crossing the equilibrium. **Step 1 (Sign structure near $x^*$).** Since $f \in C^1(I)$ and $f'(x^*) < 0$ (respectively $> 0$), there exists $\delta > 0$ such that $(x^* - \delta, x^* + \delta) \subseteq I$ and $f'(x) < 0$ (respectively $> 0$) for all $x \in (x^* - \delta, x^* + \delta)$. Combined with $f(x^*) = 0$, the [mean value theorem](/theorems/186) gives: \begin{align*} f(x) = f(x) - f(x^*) = f'(\xi)(x - x^*) \end{align*} for some $\xi$ between $x$ and $x^*$. When $f'(x^*) < 0$, $f'(\xi) < 0$ throughout the $\delta$-neighbourhood, so $f(x)$ and $x - x^*$ have opposite signs: \begin{align*} f(x) > 0 \quad &\text{for } x \in (x^* - \delta, x^*), \\ f(x) < 0 \quad &\text{for } x \in (x^*, x^* + \delta). \end{align*} **Step 2 (Case $f'(x^*) < 0$: asymptotic stability).** Take $x_0 \in (x^*, x^* + \delta)$. Then $\dot{x}(0) = f(x_0) < 0$, so $x(t)$ initially decreases. The trajectory cannot cross $x^*$: since the constant [function](/page/Function) $x(t) \equiv x^*$ is a solution passing through $x^*$ at every time, uniqueness of solutions to the initial value problem prohibits any other trajectory from reaching $x^*$ in finite time. Hence $x(t)$ remains in $(x^*, x^* + \delta)$ for all $t > 0$, where $f < 0$ continues to hold. The function $t \mapsto x(t)$ is therefore monotonically decreasing and bounded below by $x^*$, so $\lim_{t \to \infty} x(t) = \ell$ exists with $\ell \geq x^*$. Since $\dot{x}(t) = f(x(t)) \to f(\ell)$ and the [derivative](/page/Derivative) of a convergent function must tend to zero, $f(\ell) = 0$. But $x^*$ is the only zero of $f$ in the interval (since $f$ is strictly decreasing there), so $\ell = x^*$. An analogous argument applies for $x_0 \in (x^* - \delta, x^*)$: now $f(x_0) > 0$, so $x(t)$ increases monotonically toward $x^*$. For Lyapunov stability: given $\varepsilon > 0$, set $\delta' = \min(\delta, \varepsilon)$. Every trajectory starting in $(x^* - \delta', x^* + \delta')$ moves monotonically toward $x^*$ and hence satisfies $|x(t) - x^*| < \varepsilon$ for all $t \geq 0$. **Step 3 (Case $f'(x^*) > 0$: instability).** The same continuity argument now yields $f(x) > 0$ for $x \in (x^*, x^* + \delta)$ and $f(x) < 0$ for $x \in (x^* - \delta, x^*)$. Trajectories starting above $x^*$ move upward; those starting below move downward. For any $\varepsilon > 0$, the initial condition $x_0 = x^* + \min(\delta, \varepsilon)/2$ produces a trajectory that is monotonically increasing and must eventually leave the $\varepsilon$-neighbourhood (otherwise, convergence to a [limit](/page/Limit) $\ell > x^*$ with $f(\ell) = 0$ would contradict the positivity of $f$ on $(x^*, x^* + \delta)$).