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?  
 A: $$f'(x) = (1-x)e^{-x}$$
$$0=(1-x)e^{-x}\implies x=1$$
Verify that $f(1)=e^{-1}$ is the maximum and therefore the supremum by the first derivative test.
There is no minimum. But observe as $x\to\infty$, we have $f\to0$. Furthermore, $f\ge0$ on the interval. Therefore the infimum is zero.
A: 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.)
A: For this problem, a good approach might be to: 
1). Split the function into two sections, $(0,1]$ and $(1,\infty)$
2). Show that $f(x)$ is increasing on $(0,1)$, find $\lim_{x\to 0}f(x)$, and  $f(1)$.   
3). Show that $f(x)$ is decreasing on $(1,\infty)$ find $\lim_{x\to 1}f(x)$, and  $\lim_{x\to \infty}f(x)$. (to find this limit, you might want to use L'Hopital's).  
4). For the increasing interval, the infimum is $\lim_{x\to 0}f(x)$ and supremum is $\lim_{x\to 1}f(x)$  
5). For the decreasing interval the infimum is $\lim_{x\to \infty}f(x)$ and supremum is $\lim_{x\to 1}f(x)$ 
5). Conclude that the supremum is the the supremum of the two subintervals and the infimum is the infimum of the two subintervals.  
A: Since we are only consider $x > 0$ we can see $f(x) > 0, \forall x \in (0, \infty)$. Moreover $\displaystyle \lim_{x \to \infty} \frac{x}{e^x} = 0$, hence the infimum is $0$.
Now consider $f'(x) = \frac{1}{e^x} - \frac{x}{e^x}$. Testing for maxima we obtain that at $x = 1$ the function $\frac{x}{e^x}$ achieves its global maxima of $\frac{1}{e}$ thus $\frac{1}{e}$ is the supremum.
