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On This Page, We consider the function $$f(x)=\frac{1}{1+x^2}$$

on the interval $[0,\infty)$

The first point argued is that $[0,\infty)$ is not bounded. The next point is minima does not occur.

Is it because only $\mathbb{R}$ rather than $\mathbb{R^*}$ is considered? Why is not $0$ a minimum.

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Because the function is nowhere equal to $0$. –  André Nicolas Feb 2 '13 at 16:39

2 Answers 2

up vote 0 down vote accepted

$0$ is an infimum but not a minimum.

The set of values of this function is $(0,1]$. The number $0$ is not a member of this set, so it cannot be the smallest member of this set. It is, however the largest number that member of this set can never be less than. That is what an infimum is.

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Ah! good point. $\inf\neq\min$ –  user45099 Feb 2 '13 at 16:43

The function is decreasing on $[0,\infty)$ because $$ f'(x) = -\frac{2x}{(1+x^2)^2} $$ The minimum is not attained for any finite $x$; however, the maximum is attained at $x=0$.

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