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I have a homework question to prove that if $f(x) \ge x^2$ and $f(x)$ is continuous then $f([0,\infty))$ has a minimum .

This is fairly obvious why its true but I am having trouble writing it formally ( mainly the problem is selecting the min x)

Can someone help me please? Thanks :)

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    If you mean that $f$ has a minimum on $(0, \infty)$ it is not true: for example $f(x) = x^2$ (it does have a minimum at $0$, but that's not in $(0,\infty)$).2011-12-08
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    Not true: the infimum of $f:x\mapsto x^2$ on $(0,+\infty)$ is $0$, which is not a minimum.2011-12-08
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    @Robert: funny...2011-12-08

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You need to assume that $f$ is continuous; otherwise, there are counterexamples. You also need to specify the domain. Is it $(0,\infty)$? In this case, the statement isn't true.

If the domain is $[0,\infty)$:

Since $f(x)\ge x^2$, there is an $M>0$ such that $$\tag{1}f(x )\ge f( 0)\ \text{ for all }\ x\ge M.$$
Assuming $f$ is continuous, it does have a global minimum in the closed, bounded interval $[0,M]$.

By (1), this would also be the global minimum of $f$ in $[0,\infty)$.

Note that you just have to prove that there is a minimum of $f$, you don't have to explicitly find it (with the information given, this would be impossible to do).

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    Yes you are right those were my assumptions I was writing this on my phone and it was not easy not to make mistakes :) - Thanks for the answer :)2011-12-08
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    "By (1), this would also be the global minimum of f in [0,∞)." This line seems to get me confused tough I thought I understood it originally :S - can you explain please?2011-12-08
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    This is what is confusing me: http://imageshack.us/photo/my-images/593/confusiony.png/2011-12-08
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    I Get it now (thanks to Dylan Moreland) I needed to choose M larger then $f(0)$2011-12-08
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    @Jason Hm, is that what you meant to say? I hope I was pointing out that David's excellent answer selects $M$ such that what you've drawn can't happen. Do you understand why $M$ exists? [Give me a holler in chat if this doesn't make sense.]2011-12-08
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    @DylanMoreland Yep I get it, thanks for the confirmation2011-12-09