# Find whether the given function is strictly increasing/decreasing in interval

So i have this question. $F(x) = x(1-e^{-\frac{1}{x}})$ and I need to find whether the function is strictly increasing/decreasing in $(0, \infty)$. I calculated first derivative, could not figure out the sign as such so calculated 2nd derivative to find out the nature of first derivative. Turns out first derivative is strictly decreasing in $(0, \infty)$ but then i was looking at the graph of first derivative I wondered if the question can be proceeded with limits. Lim of $f'(x)$ as $x\to\infty$ is zero. So can I proceed this way?

$F'(x) = -\frac{e^{-\frac{1}{x}}}{x} - e^{-\frac{1}{x}} +1$

• You can try graphing it to get an idea. – I.Padilla Jul 13 '17 at 2:27
• Yes, i did. But i need a proper subjective proof according to curriculum – user354545 Jul 13 '17 at 2:28
• my intuition says i should be able to use limits – user354545 Jul 13 '17 at 2:29
• You said that you looked at the first derivative. What did you get for the first derivitave? That is: $F'(x)=?$ – paw88789 Jul 13 '17 at 2:55
• Write $u=\exp(-1/x)$. You now want to show $u\mapsto \frac{1-u}{-\log u}$ is increasing on $[0,1]$. After differentiating, you need to show $\log u + 1/u \ge 1$ for $u\in(0,1]$. Which looks to me to be easier that doing all the calculus with respect to the variable $x$. – kimchi lover Jul 13 '17 at 3:10

There is a useful little trick for studying $F(x)$ over $\mathbb{R}^+$, that is just to study $$g(x) = F\left(\frac{1}{x}\right) = \frac{1-e^{-x}}{x} = \int_{0}^{1}e^{-xt}dt.$$ With this double-hat-trick, everything becomes trivial: since $z\to e^{-z}$ is positive, decreasing and convex on $\mathbb{R}^+$, $g(x)$ is positive, decreasing and convex over $\mathbb{R}^+$, and $$F(x) = g\left(\frac{1}{x}\right)$$ is positive, increasing and concave over $\mathbb{R}^+$.