Solution of $f_n(x) = x + \dfrac{e^{-x}}{n}=0$ Suppose $\forall x \in \mathbb {R}\quad f_n(x) = x + \dfrac{e^{-x}}{n}$.
How can I prove that for $n\ge3$ $f_n(x)=0$ has exactly two solution $x_n$ and $y_n$ as $x_n\le -ln(n)$ and $\dfrac{−e}{n}\le y_n\le 0$
 A: We have $f'_n(x)=1- \frac{e^{-x}}{n}$, so $f'_n$ is increasing, vanishes on $t_n=-\log n$ and takes the sign $-$ then $f$ is dicreasing in the interval $(-\infty,t_n)$  and the sign + then $f$ is increasing in the interval $(t_n,+\infty)$.
Moreover, we check easily that 
$$\lim_{x\rightarrow+\infty}f_n(x)=+\infty\quad;\quad \lim_{x\rightarrow\infty}f_n(x)=+\infty$$
We have $f_n(t_n)=-\log n+1<-\log e+1=0$ because $n\geq 3$, hence there's exactly  two solution $x_n<t_n=-\log n$ and $y_n>t_n$.
We have $f_n(0)=\frac{1}{n}>0$ and $t_n<0$ so $y_n<0$. 
Since $e<3\leq n$, we have $\frac{e}{n}<1=\log e\leq \log n$ so $-\frac{e}{n}>t_n$ and $e^{e/n}>e$ then $f_n(-e/n)<0$, hence $\frac{-e}{n}<y_n$.
A: At first we denote that $f_n(0)>0$ holds for all $n$. As $-\frac{e}{n}$ is greater equal $-1$ and so $f\left(\frac{-e}{n}\right)$ is lower than $0$.
As limit of $f_n(x)$ for $x\to -\infty$ is $\infty$ we have two zeroes.
So we know that there are at least 2 zeroes, but this one is the more obvious one. 
Now we use the Mean value theorem. When $f$ would have $3$ zeroes its derivative must have 2 zeroes and its second derivative must have 1 zero (at least). But as the second derivativ is 
$$\frac{\exp(-x)}{n}$$
which doesn't have any zero, we know that there are exactly $2$ zeroes.
A: Let $u=e^{-x}$, then
$$
f=-\log(u)+\frac un\\
-\frac1ne^{-f}=-u/n\,e^{-u/n}\\
u=-n\,\mathrm{W}\left(-e^{-f-\log(n)}\right)
$$
where $\mathrm{W}$ is the Lambert-W function.
Therefore,
$$
x=-\log\left(-n\,\mathrm{W}\left(-e^{-f-\log(n)}\right)\right)
$$
Solving for $f=0$, we get
$$
x=-\log\left(-n\,\mathrm{W}\left(-1/n\right)\right)\tag{1}
$$
Since $x=-\dfrac{e^{-x}}{n}$, $(1)$ becomes
$$
x=\mathrm{W}\left(-1/n\right)\tag{2}
$$

Lambert-W has two branches, both negative, for arguments in $\left(-\frac1e,0\right)$; one in $(-1,0)$ and the other in $(-\infty,-1)$.
When $n\gt e$, $-\frac1n\in\left(-\frac1e,0\right)$. $(1)$ says
$$
\begin{align}
\mathrm{W}\in(-1,0)&\Rightarrow x>-\log(n)\tag{3}\\
\mathrm{W}\in(-\infty,-1)&\Rightarrow x<-\log(n)\tag{4}
\end{align}
$$
Thus, for $n>e$, there are two roots, one greater than $-\log(n)$ and one less than $-\log(n)$.

Let's study the branch of $\mathrm{W}$ taking values in $(-1,0)$, so that $x\gt-\log(n)$.
Since $\mathrm{W}'(0)=1$, the concavity of $\mathrm{W}$ implies that
$$
x=\mathrm{W}(-1/n)\le-\frac1n\tag{5}
$$
Since $\mathrm{W}(-1/e)=-1$ and $\mathrm{W}(0)=0$, the concavity of $\mathrm{W}$ implies that
$$
x=\mathrm{W}(-1/n)\ge-\frac en\tag{6}
$$
Therefore,
$$
-\frac en\le x\le-\frac1n\tag{7}
$$
A: It is more pleasant to work with positive numbers, so let $t=-x$. Consider the function $f(t)=\frac{e^t}{n}-t$. 
Note that $f'(t)=\frac{e^t}{n}-1$. Thus $f$ is decreasing until $t=\log n$, and then increasing. At $t=\log n$, $f(t)$  reaches is a minimum. That minimum value is $1-\log n$, which is negative if $n \ge 3$. 
By the Intermediate Value Theorem, there is a solution of the equation $f(t)=0$ between $0$ and $\log n$. There is only one solution in this interval, since $f(t)$ is steadily decreasing in the interval.
When $t$ is large, $f(t)$ is positive. So again by the Intermediate Value Theorem, $f(t)=0$ has a solution with $t\gt \log n$. By monotonicity, there is only one solution in the interval $(\log n,\infty)$. 
