Let $x^*$ be an equilibrium point of the differential equation $\dot x = f(x)$ (1) where $f \in C_{\mathbb{R}}^{1}$. Show that if $f'(x^*)<0$ then $x^*$ is asymptotically stable, and that if $f'(x^*)>0$ then $x^*$ is unstable.
So far I know that $y(t)=x(t) - x^*$ (the linear approximation of (1)) is almost equal to $ce^{f'(x^*)t}$, where $c \in \mathbb{R}$. When $f'(x^*)<0$, then $y(t) \rightarrow 0$ as $t \rightarrow \infty$ and so $x(t) \rightarrow x^*$ as $t \rightarrow \infty$. When $f'(x^*)>0$ then $y(t) \rightarrow \infty$ as $t \rightarrow \infty$ and so $x(t) \not\rightarrow x^*$.
If $f'(x^*)<0$, then $\forall \epsilon >0 \exists N >0: x>N \implies |x(t) - x^*| < \epsilon$
However I can't seem to find $\delta >0$ that satisfies the definition of Lyapunov stability ($\forall \epsilon >0 \exists \delta >0: x=\phi (t)$ is a solution of $\dot x = f(x)$ and $x(t_0)=x_0$ and $x_0 \in B_{\delta} (x^*)$, then $\phi (t) \in B_{\epsilon} (x^*) \forall t \geq t_0$) using what I already know from above.
Any hint would be thorougly appreciated!
Consider the equation \begin{equation}\tag{2} \dot y= g(y),\end{equation} $g(y)=f(y+x^*)$, $g(0)=0$, $g'(0)=f'(x^*)$. Note that $$ g'(0)\ne 0,g\in C^1_{\mathbb R}\;\Rightarrow\; \exists\epsilon>0 : \forall y\in(-\epsilon,0)\cup(0,\epsilon)\;\;\;g(y)\ne0. $$ The function $$ V(y)=g^2(y) $$ is a Lyapunov function of (2): $$ \left.\frac{dV}{dt}\right|_{(2)}=2g(y)\cdot g'(y)\cdot\dot y(t)= 2g^2(y)g'(y). $$