fr0568 #63

Forum for the GRE subject test in mathematics.
rhnsrbh
Posts: 5
Joined: Sat Mar 27, 2010 5:03 am

fr0568 #63

Postby rhnsrbh » Mon Mar 29, 2010 6:26 am

63 If f is the function defined by f(x)= x*exp(-x^2-x^(-2) if x!=0 and =0 if x=0, at how many values of x does the graph of f have a horizontal tangent line?

I've got f'(0) = 0 from the definition of a derivative at x = 0. I'm having trouble finding the other two roots of f'(x). I have f'(x) = Exp(-x^2-x^-2)*(1+2x^2-2x^-2), so if f'(x) = 0, (1+2x^2-2x^-2) = 0, but I can't find the roots to this. In particular, -1 and 1 don't seem to work. I don't think I'm calculating the derivative incorrectly. Any ideas?

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

Re: fr0568 #63

Postby enork » Thu Apr 01, 2010 12:55 pm

To find the roots of 1+2x^2-2x^{-2}=0 multiply through by x^2 and substitute y=x^2. Then you'll have a quadratic y+2y^2-2=0 that you can use the quadratic formula on. However, I'm not sure the signs are right on the derivative.




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



Who is online

Users browsing this forum: No registered users and 6 guests