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# sB - MAT 137Y 2008-09 Winter Session Solutions to...

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MAT 137Y 2008-09 Winter Session, Solutions to Supplementary Problem Set #2 1 (a) By IVT, it is clear that the equation f ( x ) = x 3 - x + 1 has a root in the interval ( - 2 , - 1 ) , since f ( - 2 ) = - 5 < 0 and f ( - 1 ) = 1 > 0. We now show f ( x ) = 0 has only one unique solution. Differentiating f , we have f 0 ( x ) = 3 x 2 - 1. The function is increasing on ( - , - 1 3 ) , which is where one of the roots is located. There can’t be another root at that interval since the function is strictly increasing. The function is decreasing on ( - 1 3 , 1 3 ) , but f ( 1 3 ) = 9 - 2 3 9 > 0, so there can’t be any roots at that interval. Finally, f is increasing for x > 1 3 , so there obviously can’t be any roots there. f ( x ) has only one root. (b) We use Newton’s Method with the initial guess of x 0 = - 3 2 . Then, x n + 1 = x n - f ( x n ) f 0 ( x n ) = x n + 1 = x n - x 2 n - x n + 1 3 x 2 n - 1 . Applying a few iterations, we get x 1 = x 0 - x 2 0 - x 0 + 1 3 x 2 0 - 1 = x 1 = - 31 23 ≈ - 1 . 347826 x 2 = x 1 - x 2 1 - x 1 + 1 3 x 2 1 - 1 = x

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sB - MAT 137Y 2008-09 Winter Session Solutions to...

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