Given $$f(x)=e^{ax}$$ where $0<a\ll e^{-1}$ and $x>0$
What are the fixed points of this function? Obviously, there are two points $x_{1}$ and $x_{2}$ satisfy the $f(x)=x$. One of them is close to 1 and Another one is far from 1. Is there any analytical solution to this or any approximation of the solution? Thanks!
In terms of the Lambert $W$ function, the fixed points are equal to $$ x = -\frac{W_k(-a)}{a} $$ where $W_k$ is the $k$th branch of the Lambert W function. The two real solutions that occur when $0 < a < e^{-1}$ come from the $k = 0$ and $k = -1$ branches.
To verify that the $x$ of the above form solve $x = e^{ax}$, it suffices to note that $$ x = e^{ax} \iff (-ax) = -ae^{ax} \iff (-ax)e^{(-ax)} = -a. $$