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Unformatted text preview: 18.100B, Fall 2002, Homework 7 Was due by Noon, Tuesday November 5. This was a bit of a stinker. Rudin: (1) Chapter 5, Problem 12 Solution. In x > ,  x  3 = x 3 so is infinitely differentiable, being a poly 2 nomial, and has derivative 3 x . Similarly in x < ,  x  3 = − x 3 is again a polynomial and has derivative − 3 x 2 . The limit lim f (0) − f ( t ) = lim  t  3 /t = 0 = t → − t = t → so f is differentiable at and f ( x ) = 3 x  x  everywhere. As already noted this is differentiable in x = and has derivative 6  x  . The limit (1) lim f (0) − f ( t ) = lim 3  t  = 0 = t → − t = t → again exists, so f ( x ) = 6  x  exists everywhere. Finally the third derivative exists for x = and is f (3) ( x ) = 6 sgn x, sgn x = ± 1 as x > 0 or x < . The limit of f (0) − f ( t ) = f (0) − f ( t ) 6 sgn t − t − t does not exist as = t → 0, so f (3) (0) does not exist. (2) Chapter 5, Problem 14 Solution. By assumption, f ( x ) is convex and differentiable on ( a, b ) . Thus f ( tx + (1 − t ) y ) ≤ tf ( x ) + (1 − t ) f ( y ) ∀ t ∈ [0 , 1] , x ≤ y ∈ ( a, b ) . For any three points x < y < z ∈ ( a, b ) the difference quotient satisfies f ( x ) − f ( y ) f ( x ) − f ( z ) f ( y ) − f ( z ) ≤...
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This note was uploaded on 12/16/2011 for the course STAT 5446 taught by Professor Frade during the Fall '09 term at FSU.
 Fall '09
 Frade

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