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 :)

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).
