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Unformatted text preview: Mathematics 105, Spring 2004 — Midterm Exam #1 Comments (1b) Give an example of a function f : R 3 → R 1 such that f is not differentiable at a = (0 , , 0), but all partial derivatives D i f ( a ) do exist. Comment : A common answer was f ( x, y, z ) = xyz/ p x 2 + y 2 + z 2 if ( x, y, z ) 6 = 0, and f (0 , , 0) = 0. This doesn’t work. Indeed, | f ( x, y, z ) | ≤ | ( x, y, z ) | 2 , and hence f (0 , , 0) equals the zero matrix (by problem (1c)). Some students wrote formulas like xyz/ p x 2 + y 2 + z 2 without addressing the issue of division by zero. It’s necessary to explicitly define f (0 , , 0). Several used f ( x, y, z ) = xy/ p x 2 + y 2 , and attempted to deal with the issue of division by zero by saying that the formula defines f when ( x, y ) 6 = (0 , 0), and defined f ( x, y, z ) = 0 whenever ( x, y ) 6 = (0 , 0). This doesn’t handle the problem of division by zero; the denominator p x 2 + y 2 vanishes at every point of the form (0 , , z ), not only at (0 , , 0). (3a) Let R = [ a, b ] × [ c, d ] be a closed, bounded rectangle in R 2 . Suppose that f is a continuously differentiable function defined in some open set containing...
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This note was uploaded on 05/07/2008 for the course MATH 105 taught by Professor Michaelchrist during the Spring '04 term at University of California, Berkeley.
- Spring '04