Postby **freeabelian** » Sat Jun 25, 2011 9:46 pm

I'm also pretty sure the answer is (d), as this isnt' a proof based thing, I'll explain my reasoning, but it's not terribly rigorous.

as far as (I), note that the function is negative, and it's derivative is negative. That means that (at least locally, around zero) it's going to be decreasing. Thus one certainly won't hold.

for (II), we notice that f certainly won't have a zero until it starts increasing, that is, until the first derivative is positive, which will eventually happen, as the second derivative is unbounded and increasing. However, once the function starts increasing, it will never decrease, as the first derivative is always increasing past this point (since we moved to where the first derivative is positive, which forces the second to be so, as it's an increasing function, and the first started out negative). thus the function can never decrease, and can't ever go through zero again.

for (iii) note that the integral of something unbounded is necessarily unbounded (The easiest way to see this is with Darboux integration). The FTOC twice will give you that (iii) holds.

Two wasn't nearly as intuitively explained as i thought it was, but I hope it helps.