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fr0568 #63

Posted: Mon Mar 29, 2010 6:26 am
by rhnsrbh
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?

Re: fr0568 #63

Posted: Thu Apr 01, 2010 12:55 pm
by enork
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.