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hw8sol - Honors Introduction to Analysis I Homework VIII...

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Honors Introduction to Analysis I Homework VIII Solution April 4, 2009 Problem 1 Define x + = ( x x 0 0 x < 0 . 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 0 ( x ) = 1 if x 0 and f 0 ( x ) = 0 for x < 0, giving a discontinuity at 0. If k > 1 then we have that f ( x ) = ( x k x 0 0 x < 0 . 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 0 + f ( x ) - f (0) x - 0 = lim x 0 + x k - 0 x = lim x 0 + x k - 1 = 0 lim x 0 - f ( x ) - f (0) x - 0 = lim x 0 + 0 - 0 x = 0 The two limits are the same, so f is differentiable at 0, with f 0 (0) = 0. So we can define f 0 ( x ) = ( kx k - 1 x 0 0 x < 0 . We need to check this function is continuous at 0. lim x 0 + f 0 ( x ) = lim x 0 + kx k - 1 = 0 = lim x 0 - 0 = lim x 0 - f 0 ( 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 non-zero exactly on the interval ( a
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