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Unformatted text preview: Honors Introduction to Analysis I Homework VIII Solution April 4, 2009 Problem 1 Define x + = ( x x x < . Prove that f ( x ) = x k + is continuously differentiable if k is an integer greater than one. Solution. For k = 1 one can see that f ( x ) = 1 if x 0 and f ( x ) = 0 for x < 0, giving a discontinuity at 0. If k > 1 then we have that f ( x ) = ( x k x x < . f is continuously differentiable at any x < 0, since it is constant. It is also continuously differentiable at any x > 0 because it is a polynomial. We are left to see if f is differentiable at 0 and if the derivative is continuous at this point. lim x + f ( x ) f (0) x = lim x + x k x = lim x + x k 1 = 0 lim x  f ( x ) f (0) x = lim x + x = 0 The two limits are the same, so f is differentiable at 0, with f (0) = 0. So we can define f ( x ) = ( kx k 1 x x < . We need to check this function is continuous at 0. lim x + f ( x ) = lim x + kx k 1 = 0 = lim x  0 = lim x  f ( x ) Hence f is continuously differentiable everywhere. Problem 2 Given a closed set A , construct a continuously differentiable function that has A as its set of zeroes. Solution. A is closed, so A c is open, hence we can write A c = S k ( a k ,b k )  a disjoint union of open intervals. On each interval ( a k ,b k ) define f k = ( x a k ) 2 + ( b k x ) 2 + . Using the previous problem (and the hint), f k is a continuously differentiable function, that is nonzero exactly on the interval (...
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