Complete study of x e^(−x)

Limits. $f(x)\to-\infty$ as $x\to-\infty$ (since $x\to-\infty$ and $e^{-x}\to+\infty$), and $f(x)=\frac x{e^x}\to0$ as $x\to+\infty$, so $y=0$ is a horizontal asymptote at $+\infty$.

Variations. $f'(x)=(1-x)e^{-x}$: increasing on $(-\infty,1]$, decreasing on $[1,\infty)$. Maximum $f(1)=\frac1e\approx0.368$.

Convexity. $f''(x)=(x-2)e^{-x}$: concave on $(-\infty,2]$, convex on $[2,\infty)$, inflection point at $\left(2,\frac2{e^2}\right)$.

Graph. It rises from $-\infty$ through the origin, peaks at $\left(1,\frac1e\right)$, then decays to $0$ from above (it is positive for $x>0$ and negative for $x<0$).

Number of solutions of $xe^{-x}=a$.
- $a>\frac1e$: none,
- $a=\frac1e$: one ($x=1$),
- $0<a<\frac1e$: two (one in $(0,1)$, one in $(1,\infty)$),
- $a\le0$: one. For $a=0$, $x=0$. For $a<0$: on $(-\infty,1]$ the function increases from $-\infty$ to $\frac1e$ and takes the value $a$ exactly once (at a negative $x$, since $f(0)=0>a$), while on $[1,\infty)$ it is positive, so it never equals $a$.

Remark. The solutions of $xe^{-x}=a$ are given by the Lambert $W$ function: the equation is $(-x)e^{-x}=-a$, so $x=-W(-a)$. The two real branches of $W$ correspond to the two monotonic pieces of $f$ on either side of $x=1$.