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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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    @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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    @DylanMoreland Yep I get it, thanks for the confirmation2011-12-09