IkUNgk-09

# IkUNgk-09 - 22 SPRING 2012 Consider f R2 R defined by f(x y...

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22 SPRING 2012 9. The Implicit Function Theorem. Consider f : R 2 R defned by f ( x, y )= x 2 + y 2 1, which is C 1 with f ° ( x, y )= ° 2 x 2 y ± For all ( x, y ). The level set { ( x, y ) | f ( x, y )=0 } is a nice curve in R 2 , namely the unit circle x 2 + y 2 =1 . Moreover ,iFwelooknearthepo int(0 , 1), this equation defnes y as a Function oF x , namely y = 1 x 2 . Similarly, near (0 , 1), we have y = 1 x 2 . However, it is evidently not possible to solve For y as a Function oF x in any neighborhood oF (1 , 0). What goes right at (0 , 1) is that D 2 f (0 , 1) ° =0 ,wh ichbecause D 2 f is continuous means that f (0 ,y ) is either strictly increasing or strictly decreasing For y in a small neighborhood oF 1; thus there can be only one y in that neighborhood, namely y =1 ,Forwh ich f (0 ,y )=0 . As with the Inverse ±unction Theorem, the converse does not hold: For example, iF f ( x, y )= x y 3 , then the equation f ( x, y )=0c anb es o lv edF o r y as a Function oF x near (0

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IkUNgk-09 - 22 SPRING 2012 Consider f R2 R defined by f(x y...

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