Finding the right scalar for change of stability I've been given the following equation:
$$\tau\frac{dx}{dt}=-x+\frac{e^{\beta(x-\frac{1}{2})}}{1+e^{\beta(x-\frac{1}{2})}}$$
For some value $\tau$.
I have been trying to find its fixed points. I have been analyzing it graphically by varying the value of $\beta$ in MatLab and graphing. I have found that for $4<\beta\leq5$ I get 2 additional fixed points. 
My question is,how do I compute what value of $\beta$ that this fixed point occurs?

Let us assume that $\beta>0$. The equilibrium equation $dx/dt = 0$ rewrites as $f(x) = 0$, where
$$
f(x) = x e^{-\beta/2} + \left(x-1\right) e^{\beta (x-1)} \, .
$$
Note that $f$ is positive at least over $[1, +\infty[$ and negative at least over $]-\infty , 0]$. Therefore, equilibrium points are necessarily between $0$ and $1$. In particular, $f(1/2) = 0$, i.e. $x=1/2$ is an equilibrium point. Now, let us examine the variations of $f$, which derivative reads
$$
f'(x) = e^{-\beta/2} + \left( 1 + \beta (x-1) \right) e^{\beta (x-1)} \, .
$$
Note that $f'$ is positive at least over $[1-1/\beta, +\infty[$, and that $f'(1/2) = (2-\beta/2)e^{-\beta/2}$. The derivative $f'$ vanishes at the abscissas $x$ such that
$$
\left( 1 + \beta (x-1) \right) e^{1 + \beta (x-1)} = -e^{1-\beta/2} \, ,
$$
viz. $1 + \beta (x-1) = W(-e^{1-\beta/2})$, where $W$ denotes the Lambert W-function. This special function is not defined over $]-\infty ,-1/e[$, and negative double-valued over $]-1/e,0[$. Therefore, if $\beta\leq 4$, then $f$ is monotonously increasing and $x=1/2$ is the only equilibrium point. Otherwise, $\beta > 4$, and $f$ is not monotonous. More precisely, $f'$ changes sign twice: once at the abscissa
$$
x_{0} = 1 + \frac{W_0(-e^{1-\beta/2}) - 1}{\beta}
$$
in $]1-2/\beta , 1-1/\beta [$, and once at the abscissa
$$
x_{-1} = 1 + \frac{W_{-1}(-e^{1-\beta/2}) - 1}{\beta}
$$
in $] 1/2-\sqrt{\beta-4}/\beta , 1-2/\beta [$. There are two additional equilibrium points: one is in $]0, x_{-1}[$, and one is in $]x_0, 1[$.
