What is the supremum and infimum of $f(x) = xe^{-x}$, where $x >0$? What is the supremum and infimum of $f(x) = x e^{-x}$, where $x > 0$? For the infimum, I do not know $\infty$. Zero equals what? How should I think to solve for the supremum? Could anyone help me?

The infimum is zero because $f(x)$ is positive but comes arbitrarily close to zero for $x\to\infty$ and also for $x\to0$.

The supremum is $f(1)=1/e$ because this is the maximum value of $f(x)$ for $x>0$. (We find the maximum by calculating $f'(x)$, looking where we have $f'(x)=0$, and checking that $f'(x)$ changes from positive to negative at that point.)