MATH 411 Lecture Notes - Surface Integral, Jordan Measure, Bounded Function
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Math 411 fall 2011 boyle final exam solutions. Y(cid:16)f(cid:17)(cid:17)(cid:17)(cid:17)(0) and it is not the same as the product ( 3f / x3(0)) ( f / y(0)) Prove that there exists x in r2 such that f1(x) = f2(x). De ne g : [0, 1] r by g(t) = f1(tp) f2((1 t)q), where p = ( 1. Since g is continuous and assumes both positive and negative values, by the intermediate value theorem g asssumes the value zero. But g(t) = 0 means f1 = f2 at the point tp + ((1 t)q). Then the image of f is separated by the open sets on which f1 f2 > 0 vs. f1 f2 < 0. That means the image of f is not connected. But, r2 is connected, and the image of this connected set under the continuous function f is connected. A computation of the derivative matrix df shows. This matrix is invertible (nonzero determinant) and f is c 1.
