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I need to show that, for the function $f(x) = x^2 (e^{1/x}-1) - x^2/(x+1)$, there is some $x_0 > 0$ such that $f(x)$ is decreasing on $(0,x_0)$ and increasing on $(x_0,\infty)$.

By setting $f'(x) = 0$ I can see that $f(x)$ is decreasing on $(0,1/2)$, and taking the series expansion of $f(x)$ at $\infty$ tells me that $f(x) \to 3/2$ as $x \to \infty$, but this isn't enough. The problem boils down to the question in the title.

I just can't seem to make it give up the goods.

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1 Answer 1

up vote 3 down vote accepted

I could simplify the problem quite a bit but left a simpler problem for you to solve. We have

$$f(x) = \left(e^{\frac{1}{x}}-1\right) x^2-\frac{x^2}{x+1}$$

using some basic algebra you can factor its derivate to:

$$f'(x) = \frac{2 e^{\frac{1}{x}} x^3-2 x^3+3 e^{\frac{1}{x}} x^2-5 x^2-4 x-e^{\frac{1}{x}}}{(x+1)^2}$$

So what is left to show is that

$$h(x)= e^{\frac{1}{x}} x^3-2 x^3+3 e^{\frac{1}{x}} x^2-5 x^2-4 x-e^{\frac{1}{x}}$$

has exactly one real root in $(0, \infty)$, but this is not hard because if we look at the graph of h we see that

This is a plot of h

And therefore assume that $$h'(x)=(x+1) \left(e^{\frac{1}{x}} \left(\frac{1}{x^2}+6 x-\frac{1}{x}-2\right)-6 x-4\right)>0$$

Where you can also use the notation

$$h'(t)=h'(1/x)=\frac{(t+1) \left(-4 t+e^t ((t-2) t (t+1)+6)-6\right)}{t^2}$$

So it remains again to show that

$$ (t+1) \left(t^3-t^2-2 t+6\right) e^t\geq 2 \left(2 t^2+5 t+3\right) $$

Then we are done if we apply the mean value theorem. It should be doable to show this.

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Setting $t= 1/x$ and working with $t$ should simplify things a bit... –  Aryabhata Jul 22 '11 at 20:29
    
@Aryabhata: Thanks! I used your suggestion to reduce it even further. –  Listing Jul 22 '11 at 20:41
    
Thanks Listing, following in this manner did eventually yield the solution. Also, @Aryabhata: yes, I had the same idea in the shower :) –  Antonio Vargas Jul 23 '11 at 0:42
    
No problem, good to see I could help. –  Listing Jul 23 '11 at 9:19

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