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Unformatted text preview: y =2 x/ 11 + 1. 4. (a) We need f = . From Problem 2, this gives us 2 x +2 y = 0 and 3 y 2 +8 y +2 x = 0. The rst equation give x =y , which turns the second equation into 3 y 2 +6 y = 0. The solutions are y = 0 and y =2. Since x =y , the critical points are (0 , 0) and (2 ,2). (b) We have f xx = 2, f xy = 2 and f yy = 6 y + 8. Thus f xx > 0. At (0 , 0), f yy = 8 and f xx f yy( f xy ) 2 > 0 so the point is a (local) minimum. At (2 ,2), f yy =4 and f xx f yy( f xy ) 2 < 0 so the point is a saddle....
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This note was uploaded on 06/17/2009 for the course MATH 20C taught by Professor Helton during the Fall '08 term at UCSD.
 Fall '08
 Helton
 Calculus, Chain Rule, The Chain Rule

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