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)?
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.