prove that if $f(x) \ge x^2$ and $f(x)$ is continuous then $f([0,\infty))$ has a minimum

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 $(0,\infty)$). – Robert Israel Dec 8 '11 at 8:43
Not true: the infimum of $f:x\mapsto x^2$ on $(0,+\infty)$ is $0$, which is not a minimum. – Did Dec 8 '11 at 8:43
@Robert: funny... – Did Dec 8 '11 at 8:44

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 :) – Jason Dec 8 '11 at 19:38
"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? – Jason Dec 8 '11 at 20:10
This is what is confusing me: imageshack.us/photo/my-images/593/confusiony.png – Jason Dec 8 '11 at 20:57
I Get it now (thanks to Dylan Moreland) I needed to choose M larger then $f(0)$ – Jason Dec 8 '11 at 21:22
@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.] – Dylan Moreland Dec 8 '11 at 23:40