Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

What is the proper way to finish the proof?

share|improve this question
    
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
3  
$f$ 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

3 Answers 3

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 $[a,b]$.

share|improve this answer
    
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 $(0,+\infty)$.

share|improve this answer

$\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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.