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I showed that $\lim f(x) = 0$ at both the $0$ end and $+\infty$ end.

What is the proper way to finish the proof?

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Are you looking for the global minimum? – amWhy Jan 26 '13 at 14:47
Yes I am looking for the global minimum – homiee Jan 26 '13 at 14:48
Do you need to show that the global minimum is in the interval $(0, +\infty)$ (which it is), or do you need to pinpoint the value of $x$ where $f(x)$ is minimum? – amWhy Jan 26 '13 at 14:50
$f$ is negative on $(0,\infty)$. Choose $a>0$ and $b$ so that $f(x)>f(1)$ for all $x$ outside $(a,b)$. $f$ has a global minimum on $[a,b]$. Now argue this is the global minimum on $(0,\infty)$. – David Mitra Jan 26 '13 at 14:51
Yes I need to show that the global minimum is in the interval (0,+∞) – homiee Jan 26 '13 at 14:52
up vote 1 down vote accepted

As given in the comment by David Mitra. You find a value of $x$ for which $f(x) < 0$. Lets say that $y_0 = f(x_0)<0$. Then you know that $$ \lim_{x\to 0} f(x) = 0 \quad\text{and}\quad \lim_{x\to \infty} f(x) = 0. $$ So there must be an $a< 1$ such that for $x\in (0,a]$ you have $f(x) > y_0$. Also, there must be a $b>10$ such that for $x\in [b,\infty)$, you have $f(x) > y_0$.

You (probably) know that on a closed interval $f$ will attain a (global) maximum and a global minimum. So you know that on the closed interval $[a,b]$ $f$ attains a global minimum. The only way that this could not be a global minimum for $f$ on $(0,\infty)$ is if $f$ attains a smaller value outside of $[a,b]$.

But we have shown above that outside $[a,b]$, $f(x) > y_0 = f(x_0)$. So indeed there is a global minimum for $f$ and it is attained on $[a,b]$.

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It is better to say "on a compact (closed and bounded) interval" $f$ ... – vesszabo Jan 26 '13 at 15:20
I prefer this proof, because it uses only basic notions, nothing diff calculus. (It can be generalized for very general situations.) – vesszabo Jan 26 '13 at 15:24
For the argument to be correct, you need to make sure that $a<b$ and that $x_0$ belongs to $[a,b]$. – 1015 Jan 26 '13 at 15:26
@julien: Thanks. I edited the answer to make clear that $a < b$. – Thomas Jan 26 '13 at 15:34
Would the downvoter care to comment on his vote so that I might improve the answer? – Thomas Jan 26 '13 at 18:39

I am just expanding on David Mitra's comment.

Compute $f(1)=-0.58...$.

Since $\lim_{x\rightarrow 0^+} f(x)=\lim_{x\rightarrow +\infty}=0$, we can find $0<a<1<b$ such that $f(x)\geq -0.5$ for all $0<x<a$ and $x>b$.

Since $f$ is continuous on $[a,b]$, it reaches a minimum at $c$ on $[a,b]$. Since we arranged for $1$ to belong to $(a,b)$, we know that $f(c)\leq f(1)<-0.5$.

Now it is clear that $f(c)$ is also a minimum for $f$ on $(0,+\infty)$.

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$\displaystyle f(x)= \frac{x^2}{1-e^x} $ , differentiating :

$\displaystyle f'(x)= \frac{g(x)}{(1-e^x)^2 } $ where $g(x)=\left(e^x (x-2)+2\right) x$ and $x\neq 0$.

what we need to prove is that there exists $\displaystyle \xi \in (0;+\infty) $, such as :

$\displaystyle \begin{cases}x=\xi \Longrightarrow g(x)=0 \Longrightarrow f'(x)=0\\ \\ x>\xi \Longrightarrow g(x)>0 \Longrightarrow f'(x)>0\\ \\ 0<x<\xi \Longrightarrow g(x)<0 \Longrightarrow f'(x)<0 \end{cases}$

you can prove that a such $\xi $ exists and it's unique using IVT .

the the function $f$ has a minimum at $\xi$ .

note : $\xi$ cannot be found using algebra.

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