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Unformatted text preview: ASSIGNMENT 4  SOLUTIONS Problem 1. Differentiating with respect to x we get 2 3 x 1 / 3 + 2 3 y 1 / 3 y = 0 . It follows that y = 3 y/x . Problem 2. To find the points where the tangent to a curve is horizontal, we solve y = 0. Take y 2 = x 3 + x . Differentiating with respect to x we get 2 yy = 3 x 2 + 1, so: y = 3 x 2 + 1 2 y Then y = 0 precisely when 3 x 2 + 1 = 0, that is, never. In this case, the number of points is 0. Take y 2 = x 3 x . Differentiating with respect to x we get 2 yy = 3 x 2 1, so: y = 3 x 2 1 2 y Then y = 0 precisely when 3 x 2 1 = 0, that is, when x = 1 3 . Plugging x = 1 3 in the original relation y 2 = x 3 x , we get y 2 = 2 3 3 which has no solutions. Plugging x = 1 3 in the original relation y 2 = x 3 x , we get y 2 = 2 3 3 which has two solutions. In this case, the number of points is 2. Take y 2 = x 3 x + 1. Differentiating with respect to x we get 2 yy = 3 x 2 1, so: y = 3 x 2 1 2 y Then...
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This note was uploaded on 07/31/2011 for the course MATH 102 taught by Professor Maryamnamazi during the Spring '10 term at University of Victoria.
 Spring '10
 MARYAMNAMAZI
 Calculus, Logic

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