GRE 0568

Forum for the GRE subject test in mathematics.
moo5003
Posts: 36
Joined: Mon Oct 06, 2008 7:33 pm

GRE 0568

Postby moo5003 » Fri Nov 07, 2008 10:45 pm

Hey, I was just reviewing all the tests for tmmrws GRE exam and I cant seem to find the correct answer for problem #63.

f(x) = xe^(-x^2 - x^-2) for x!= 0
and 0 for x=0

How many horizontal tangent lines are there for f(x)?

a) None
b) One
c) Two
d) Three
e) Four

Answer: D (I guessed c)

I found f' and I can only seem to get two values for when the derivative is 0. Can anyone tell me where the third is?

EDIT: I'm guessing its because as x approaches 0 the derivative approaches 0. Missed this :/, should have expected something like that since it isnt defined at 0 (at least for the derivative).

amateur
Posts: 42
Joined: Wed Sep 10, 2008 9:41 am

Postby amateur » Fri Nov 07, 2008 10:49 pm

The 3rd one is at x = 0. Try plotting the curve.
Or take the limit x -> 0 of f'(x).

moo5003
Posts: 36
Joined: Mon Oct 06, 2008 7:33 pm

Postby moo5003 » Fri Nov 07, 2008 10:53 pm

Sadly I got this question correct when I took the pre-test 3 weeks ago :/, so much for improvement ^_^

mathrun
Posts: 5
Joined: Fri May 08, 2009 10:33 pm

Re: GRE 0568

Postby mathrun » Wed Jun 17, 2009 11:59 pm

can anybody post this test, GRE 0568 ?

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm

Re: GRE 0568

Postby Nameless » Mon Jun 22, 2009 1:05 am

mathrun
You can look for so many test samples on this forum, please search for them

EHart
Posts: 6
Joined: Mon Sep 14, 2009 1:38 pm

Re: GRE 0568

Postby EHart » Tue Sep 22, 2009 5:35 pm

moo5003 wrote:Hey, I was just reviewing all the tests for tmmrws GRE exam and I cant seem to find the correct answer for problem #63.

f(x) = xe^(-x^2 - x^-2) for x!= 0
and 0 for x=0

How many horizontal tangent lines are there for f(x)?

a) None
b) One
c) Two
d) Three
e) Four

Answer: D (I guessed c)

I found f' and I can only seem to get two values for when the derivative is 0. Can anyone tell me where the third is?

EDIT: I'm guessing its because as x approaches 0 the derivative approaches 0. Missed this :/, should have expected something like that since it isnt defined at 0 (at least for the derivative).


I saw the obvious solution (x=0; it was obvious to me, anyway), but I'm struggling with the other two. Anybody care to help? Assuming I can actually derive correctly,

f'(x) = e^(-x^2-x^-2)+x(-2x+2x^-3)e^(-x^2-x^-2) = (1-2x^2+2x^-4)*e^(-x^2-x^-2)

I guess I just need to find how many real roots the coefficient polynomial has, then? Any chance someone could lend a hand with this? My root finding is lackluster.

aas56
Posts: 29
Joined: Sun Jun 28, 2009 3:42 pm

Re: GRE 0568

Postby aas56 » Wed Sep 23, 2009 9:57 am

in fact it's
e(-x^2-x^-2)(1-2x^2+2x^-2) =0
multiply both sides by x^2 (since x is not zero) we get the equation:
-2x^4-2x^2+1=0
take y=x^2 and solve the following equation for positive root:
--2y^2-2y+1=0
it turns out that one positive solution for y which is y= (1+root(15))/4 = x^2
and then x =+/- root((1+root(15))/4)

EHart
Posts: 6
Joined: Mon Sep 14, 2009 1:38 pm

Re: GRE 0568

Postby EHart » Wed Sep 23, 2009 1:40 pm

Ah, whoops. x^-3 * x != x^-4. Silly mistake...

chsguitarist
Posts: 4
Joined: Fri Sep 25, 2009 1:35 pm

Re: GRE 0568

Postby chsguitarist » Mon Oct 18, 2010 12:31 pm

I can't seem to figure out why the limit x-->0 of f'(x) is 0, or especially why it should be obvious or intuitive. Is there some easy way to sketch this or a theorem I am missing? I get this weird equation (1 - 2x^2 + 2x^(-2)) and when I try to evaluate it at 0, I get f'(x) goes to infinity. I have been trying all sorts of tricks to make it amenable to l'hopital, but nothing seems to work. Any suggestions?

enork
Posts: 33
Joined: Fri Sep 18, 2009 3:16 am

Re: GRE 0568

Postby enork » Mon Oct 18, 2010 9:48 pm

I think the derivative is f'(x) = (1 + x(-2x +2x^{-3}))e^{-x^2 - x^{-2}} but it doesn't really matter as long as you notice that the exponential e^{-x^2 - x^{-2}} survives, and it dominates the polynomial part as x goes to zero.

gregpy
Posts: 1
Joined: Thu Oct 21, 2010 9:45 pm

Re: GRE 0568

Postby gregpy » Thu Oct 21, 2010 9:50 pm

i'm taking the math subject test in a few weeks, and stumbled upon this website. i see people referring to various past math subject exams - gre0568, etc etc. how can i locate and download these exams? i tried googling past exams and got nowhere. how many past exams are out there in circulation? it seems the testing service is quite tight with them, eh? unlike the LSAT, SAT, etc., it seems. would appreciate any tips.

enork
Posts: 33
Joined: Fri Sep 18, 2009 3:16 am

Re: GRE 0568

Postby enork » Thu Oct 21, 2010 10:21 pm

The ETS posts the latest test to be released on their website. This one is at http://www.ets.org/Media/Tests/GRE/pdf/gre_0809_math_practice_book.pdf. Some searching came up with this thread http://www.mathematicsgre.com/viewtopic.php?f=1&t=11 which links at the end to three of the older tests that have been released. I think that may be all of them.




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