So this is my function, and I'm trying to find the root where f(x)=0:
$c_1-\frac{2}{c_2}(x+2)e^{-x/2}=0$
where $0< c_1\le1$ and $c_2\ge2$
This is what I got thus far:
$c_1c_2-2(x+2)e^{-x/2}=0$
$c_1c_2=2(x+2)e^{-x/2}$
$\ln(c_1c_2)=\ln(2(x+2)e^{-x/2})$
$\ln(c_1c_2)=\ln(2(x+2))+\ln(e^{-x/2})$
$\ln(c_1c_2)=\ln(2x+4)-x/2$
So I do not know where to go from here... I've read that you can use the lambert function but it requires the form $xe^x=c$, but I do not know how to transform the equation. I've thought about using Newton's method but it requires a 'close' enough initial guess; Usually I will use the bisection method as an initial guess but it requires a change of signs onto the interval, which you can see from the graph below, when $c_1=0.05$ and $c_2$ increases it becomes negative over its entire domain, so the definition of root, f(x)=0, becomes impossible to find
Making $y = x+2$ we have
$$ c_1-\frac{4}{c_2}\frac{y}{2}e^{-\frac{(y-2)}{2}}=c_1-\frac{4}{c_2}\frac{y}{2}e^{-\frac{y}{2}}e = 0 $$
and then
$$ c_1 + \frac{4e}{c_2}\left(-\frac y2\right)e^{-\frac y2}=0\Rightarrow x = -2\left(1+W\left(-\frac{c_1c_2}{4e}\right)\right) $$
Note the pair $\left(-\frac y2\right)e^{-\frac y2}$