## GR 9768 Q55

Forum for the GRE subject test in mathematics.
ch628496
Posts: 5
Joined: Wed Jun 22, 2011 12:06 am

### GR 9768 Q55

suppose that f is a twice-differentiable function on the set of real numbers and that f(0), f'(0) and f"(0) are all negative.

suppose that f" has all three of the following properties.

I. It is increasing on the interval [0, +inf).
II. It has a unique zero in the interval [0, +inf).
III. It is unbounded on the interval [0, +inf).

Which of the same three properties does f necessarily have?

a. I only
b. II only
c. III only
d. II and III only
e. I , II and III

Posts: 4
Joined: Tue Oct 05, 2010 1:52 am

### Re: GR 9768 Q55

I think the answer may be (d). Here is why:

If $$f''(x)$$ is unbounded, then $$f'(x)=\int_0^x f''(s)ds +f'(0)$$. the integral blows up as f'' does, so $$f'(x)$$ is also unbounded.

Since there is a unique c s.t. $$f''(c)=0$$, then we know $$f'(x) -f'(0) = f''(e)(x) < 0$$ if $$e < c$$ which is true because of the mean value theorem. Then, f'(c) < 0, f'(x) is strictly increasing for x >=c, and f'(x) is unbounded. Therefore, f'(x) has a unique zero.

Therefore, only II,III hold for $$f'(x) on [0,\inf)$$. Then, using the exact same arguments as above we can show that only II,III hold for $$f(x)$$.

freeabelian
Posts: 9
Joined: Sat Jan 01, 2011 5:35 pm

### Re: GR 9768 Q55

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.

ch628496
Posts: 5
Joined: Wed Jun 22, 2011 12:06 am

### Re: GR 9768 Q55

Thank you all for your effort.

ch628496
Posts: 5
Joined: Wed Jun 22, 2011 12:06 am

### Re: GR 9768 Q55

Do any of you have any recommendation for the techniques of caculi?

freeabelian
Posts: 9
Joined: Sat Jan 01, 2011 5:35 pm

### Re: GR 9768 Q55

Take an analysis class. I'm totally serious.