For what real value $k$ does the equation $e^{-kx}+e^{-x} = x$ have only one real root? I am working on this equation, and I think I need to consider Rolle's theorem. According to a teacher, I need to use IMV as well. Any ideas?
$e^{-kx}+e^{-x} = x$
The value of $k$ is that, wherein the exponential $f(x) = e^{-kx}+e^{-x}$ is tangent to $g(x) = x$.
 A: Take the derivatives and set equal...
$$-ke^{-kx}-e^{-x}=f'(x)=g'(x)=1$$
$$-ke^{-kx}=1+e^{-x}$$
Clearly $k\ge0$ is not possible, so we look at $k=-n<0$
$$ne^{nx}=1+e^{-x}$$
$$u=e^x\implies nu^{n+1}=1+u$$
When $n$ is even, then there are two solutions for $u$ and when $n$ is odd, there is only one solution.
(Prove with intermediate value theorem)
Then check if $f(x)=g(x)$ at those points.
A: If $k>0$, left side is monotonously decreasing to 0 and right side increasing, so it clearly always has 1 and only 1 real root. You can use Rolle's theorem for this case to prove it rigorously.
For $k<0$ left side is a convex function and it will cross $y=x$ either never (huge negative $k$), twice ($k$ just below zero), or (in a single $k$ inbetween when it has a double root). If it's a double root, it matches both in value and derivative. So the critical value of $k$ satisfies
$$e^{-kx}+e^{-x}=x$$
$$-ke^{-kx}-e^{-x}=1$$
which is two equations for variables $k$ and $x$.
Plotting a few cases for $k$, I fond the critical value to be somewhere around $-0.36$ which naturally gets me thinking of  $k=-e^{-1}$ but isn't quite correct. In fact, the pair of equations is transcendental and has no reason of having a closed form solution.

EDIT:
I don't think this can be expressed only with Lambert's function. Here's my try. Setting $n=-k$, above can be converted to
$$e^{nx}=\frac{1+x}{1+n}=u$$
$$e^{-x}=\frac{nx-1}{n+1}=v$$
where $x=u+v$. The last one can be converted to
$$ve^v=e^{-u}\rightarrow v=W(e^{-u})$$
I hoped that substitutio to $u$ and $v$ variables would fully separate the variables and enable me to solve for $u$ and $v$ separately, but the $nx=(u+v)(1+v)/u$ can't be nicely simplified to something that would lead to a solvable equation. There may be better substitutions but I don't think the system has enough symmetry for this to be successful.
This process, however, does lead to a nice iterative scheme which relates to the guessed approximations. The iteration begins with initial condition:
$$u_0=e, \quad v_0=0$$
These are consistent with the guess $n\approx e^{-1}$, $x\approx 1/n \approx e$, $nx\approx 1$.
The iterative step is:
$$u_{k+1}=e^{(1+v_k)(u_k+v_k)/u_k},\quad v_{k+1}=W(e^{-u_{k+1}})$$
This converges fairly quickly to $u=2.915444361$, $v=0.05146225679$, and therefore $x=2.96690661781$ and $n=0.36065248606699$.
