GRE 0568
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).
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).
Re: GRE 0568
can anybody post this test, GRE 0568 ?
Re: GRE 0568
mathrun
You can look for so many test samples on this forum, please search for them
You can look for so many test samples on this forum, please search for them
Re: GRE 0568
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,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).
f'(x) = e^(x^2x^2)+x(2x+2x^3)e^(x^2x^2) = (12x^2+2x^4)*e^(x^2x^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.
Re: GRE 0568
in fact it's
e(x^2x^2)(12x^2+2x^2) =0
multiply both sides by x^2 (since x is not zero) we get the equation:
2x^42x^2+1=0
take y=x^2 and solve the following equation for positive root:
2y^22y+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)
e(x^2x^2)(12x^2+2x^2) =0
multiply both sides by x^2 (since x is not zero) we get the equation:
2x^42x^2+1=0
take y=x^2 and solve the following equation for positive root:
2y^22y+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)
Re: GRE 0568
Ah, whoops. x^3 * x != x^4. Silly mistake...

 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?
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.
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.
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/ ... e_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.