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anonymous
 3 years ago
Has anyone proved that y=2y zero on that problem in the first lecture where he used symmetry to get that y=2y zero? Could you help because I keep getting y=2/x zero.
anonymous
 3 years ago
Has anyone proved that y=2y zero on that problem in the first lecture where he used symmetry to get that y=2y zero? Could you help because I keep getting y=2/x zero.

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Stacey
 3 years ago
Best ResponseYou've already chosen the best response.0I would need to see the actual problem.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yy zero=1/(x zero)^2 (xx zero) is the problem. The professor solved for x by plugging in y=0 and got x=2x zero which I understand. Then he said by symmetry you get y=2y zero and we could prove it by taking the same equation and plugging in x=0. Every time I do that, I get y= 2/x zero, not y=2y.

Stacey
 3 years ago
Best ResponseYou've already chosen the best response.0Based on what you are saying, I am assuming that previously in the lecture, it was stated that \[y _{0} = \frac{ 1 }{ x _{0} }\]If that is the case, then \[x _{0} = \frac{ 1 }{ y _{0} }\] and \[y=2y _{0}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i must be doing something wrong...thanks anyway

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The area of the triangle as he derives is 2XoYo. As y = f(x) = 1/X, f(Xo) = 1/Xo. Therefore Yo = 1/Xo. Thus, 2XoYo = 2Xo(1/Xo) = 2
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