## GRE 0568

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

### GRE 0568

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
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
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

can anybody post this test, GRE 0568 ?

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

### Re: GRE 0568

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

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

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

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

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

### Re: GRE 0568

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

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

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

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.

Return to “Mathematics GRE Forum: The GRE Subject Test in Mathematics”

### Who is online

Users browsing this forum: No registered users and 10 guests