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Unformatted text preview: Chapter 4: Hints and Solutions Exercise 4.1 All the functions are smooth. So try and find points where f vanishes and then figure out which level sets of f contain these points. (a) First note that f 1 ( c ) is the empty set if c < 0, so f 1 ( c ) is not an nsurface. Next, since f ( x 1 , . . . , x n +1 ) = 2( x 1 , . . . , x n +1 ), then, f ( x 1 , . . . , x n +1 ) = (0 , . . . , 0) if and only if ( x 1 , . . . , x n +1 ) = (0 , . . . , 0). Noting that the origin (0 , . . . , 0) is on the level set f 1 (0) (in fact, its equal to { (0 , . . . , 0) } ). So c = 0 does not give an nsurface either. Now suppose c > 0 then f 1 ( c ) negationslash = and if p f 1 ( c ) then p negationslash = and f ( p ) negationslash = . We conclude that f 1 ( c ) is an nsurface if and only if c > 0. (b) f ( x 1 , . . . , x n +1 ) = 2( x 1 , . . . , x n , x n +1 ) so f ( x 1 , . . . , x n +1 ) = (0 , . . . , 0) if and only if ( x 1 , . . . , x n , x n +1 ) = (0 , . . . , 0) if and only if ( x 1 , . . . , x n +1 ) = (0 , . . . , 0). Since f (0 , . . . , 0) = 0 then f 1 (0) is not an nsurface. For any other c , i.e., for c negationslash = 0, f 1 ( c ) is an nsurface in R n +1 . (c) x i f ( x 1 , . . . , x n +1 ) = x 1 x 2 x i 1 x i +1 x n +1 if and only if x j = 0 for some j negationslash = i . This shows that f ( x 1 , . . . , x n +1 ) = (0 , . . . , 0) if and only if x i = x j = 0 for for at least two indices i negationslash = j . The level set f 1 (1) consists of points for which x 1 x 2 x n +1 = 0 and so contains points at which f vanishes, which means it cannot be an nsurface. On the other hand, if c negationslash = 1, f 1 (...
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 Spring '09

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