Is the function $y=xe^x$ invertible? I'm wondering if the equation $re^r=se^s$ has any answer. If there is any answer,and $r=-1+v,s=-1-v$ in which $v$ is a positive real number,what can we say about $v$?
Thank you in advance.
 A: The derivative is $e^x+xe^x$ which vanishes at $-1$. The second derivative is $2e^x+xe^x$, which is $2e^{-1}-e^{-1}=e^{-1}$ at $-1$. The limit at $-\infty$ is $0$ and the limit at $+\infty$ is $+\infty$, both of which are bigger than $-e^{-1}$. From this you can read off everything there is to say about the number of solutions to $xe^x=a$: the solution is unique for $a \geq 0$ and $a=-e^{-1}$, there are two solutions for $a \in (-e^{-1},0)$, and there are no solutions for $a<-e^{-1}$.
As an aside, the solutions for $a \geq 0$ create an inverse for (a restriction of) $xe^x$. This function (or rather its analytic continuation) is called the Lambert W function.
A: Let $r=-1+v$ and $s=-1-v$. You are looking for zeros of the function:
$$
f(v) = (-1+v)e^{-1+v} - (-1-v)e^{-1-v}.
$$
Observe that one solution is $v=0$ which corresponds to $r=s=-1$.
Compute the derivative
$$
f'(v) = ve^{-1-v} (-1+e^{2 v})
$$
and notice that it is never negative and it is zero only for $v=0$. Hence $f$ is strictly increasing. This means $f$ is injective and there cannot be other solutions. So you can say that $v=0$.
A: A straightforward solution is 
$$ r=s $$
for whatever parametrization. Other solution involves Lambert function $W$
$$ r + \log r =  s + \log s$$ 
