# Proving that $x(1-e^{-1/x})$ is strictly increasing

Prove that the function below is strictly increasing $$f(x)=x(1-e^{-1/x}), \quad x>0$$

Letting $y=-\frac1x$ we see that $$f(x) = \frac{e^y-1}{y}$$ is the slope of the secant line through $(0,e^0)$ and $(y,e^y)$. The claim then follows from the convexity of the exponential.

• Only nice answers here. Thanks! (+1) – user 1591719 Feb 22 '13 at 15:14
• This is nice. I can visualize the transformed function and the secant line in my head. +1. – user1551 Feb 22 '13 at 15:16
• Really nice! So simple! – nbubis Feb 22 '13 at 15:18
• Yeah, it's a very beautiful math piece! – user 1591719 Feb 22 '13 at 15:22

If $f(x)$ is strictly increasing for $x>0$, then $f\left(\frac{1}{x}\right)$ is strictly decreasing. $$\frac{\mathrm{d}}{\mathrm{d}x}\left( \frac{1-\exp(-x)}{x} \right) = \mathrm{e}^{-x} \frac{1+x-\mathrm{e}^{x}}{x} < 0$$ The last inequality is consequence of the well known $\mathrm{e}^x > 1+x$ valid for $x>0$.

• hehe, nice. Actually this answer is really awesome. (+1) – user 1591719 Feb 22 '13 at 15:05

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 $x>y\to f(x)>f(y)$.

• Nice! This seem rather elementary! Thanks (+1) – user 1591719 Feb 22 '13 at 15:19
• @Chris'ssisterandpals: Yes, it is very elementary as I am. Thanks for sharing great questions, always. – mrs Feb 22 '13 at 15:20
• Glad to see around so many approaching ways. :-) – user 1591719 Feb 22 '13 at 15:24
• +1 I really like how often you offer alternative perspectives and approaches. It helps us all (the asker and users alike) "think outside the box". – Namaste Feb 23 '13 at 3:19
• @amWhy: Usually, I think of the simple way for the problem and unexpectedly the way is true. I wish, I could think of the problem globally as others did. Anyway, thanks. ;-) – mrs Feb 23 '13 at 3:26