TITLE: Proving that $x(1-e^{-1/x})$ is strictly increasing
QUESTION [9 upvotes]: Prove that the function below is strictly increasing
$$f(x)=x(1-e^{-1/x}), \quad x>0$$

REPLY [3 votes]: If $x<0$ and $x>y$ then $-x,-y$ are positive and $-1/x>-1/y$ and so $\exp(-1/x)>\exp(-1/y)$ and then $1-\exp(-1/x)<1-\exp(-1/y)$ and so $$-x(1-e^{-1/x})<-x(1-e^{-1/y})<-y(1-e^{-1/y})$$ this shows that when $x<0$, then $x>y\to f(x)>f(y)$.