9

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

user 1591719
  • 44,987

3 Answers3

10

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 $x>0$.

Sasha
  • 71,686
9

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.

Hagen von Eitzen
  • 382,479
2

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

Mikasa
  • 67,942
  • Nice! This seem rather elementary! Thanks (+1) – user 1591719 Feb 22 '13 at 15:19
  • 1
    @Chris'ssisterandpals: Yes, it is very elementary as I am. Thanks for sharing great questions, always. – Mikasa Feb 22 '13 at 15:20
  • 1
    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". – amWhy Feb 23 '13 at 03:19
  • @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. ;-) – Mikasa Feb 23 '13 at 03:26
  • But that's a good thing: simplification of a problem, rather than "overkill" when not necessary. Sometimes people offer answers that are beyond the level of the asker. And you are so humble (I admire this quality), but you underestimate your level of contribution and expertise. – amWhy Feb 23 '13 at 03:30
  • 1
    This does not work. $y<0$ implies (1-e^{-1/y})<0$, hence the second inequality must be reversed. – ziggurism Oct 22 '15 at 03:39
  • The second inequality is certainly not true as @ziggurism pointed out. If you select x=-1 and y=-2 you showed something like $-1.7183<-0.6487<-1.2974$. – obareey Dec 22 '23 at 12:01
  • 1
    @obareey The inequality required is ultimately an expression of the convexity of the exponential. So it would have been very surprising if Mikasa has been able to prove it using only monotonicity. I'm not sure how to formalize the intuition about which axiomatic facts of the exponential function are required for which results, but that's my intuition. – ziggurism Dec 23 '23 at 05:56