asst4-f10-solns

asst4-f10-solns - MATH 137 Assignment 4 - Solutions 1. (a)...

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MATH 137 Assignment 4 - Solutions 1. (a) If f ( x ) = x 3 + bx 2 + cx + d and x 6 = 0, then f ( x ) = x 3 (1 + b x + c x 2 + d x 3 ). Suppose that | x | is chosen large enough so that ± ± ± ± b x ± ± ± ± , ± ± ± c x 2 ± ± ± , ± ± ± ± d x 3 ± ± ± ± are all less than 1 4 . Then ± ± ± ± b x + c x 2 + d x 3 ± ± ± ± ± ± ± ± b x ± ± ± ± + ± ± ± c x 2 ± ± ± + ± ± ± ± d x 3 ± ± ± ± < 3 4 < 1. Let b x + c x 2 + d x 3 = A . Now - A ≤ | A | , so A ≥ -| A | and thus 1 + A 1 - | A | > 0. Therefore when | x | is large enough, the sign of f ( x ) is the same as the sign of x 3 which is the same as the sign of x . (b) From (a) there exist numbers u and v such that f ( u ) < 0 and f ( v ) > 0. A polynomial function is continuous, so by the Intermediate Value Theorem here is a number w such that f ( w ) = 0. (c) “Yes”. One could repeat the argument to show that if f ( x ) = x m + b m - 1 x m - 1 + ··· + b 0 then when | x | is large enough 1 + b m - 1 x + ··· + b 0 x m > 0 .
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asst4-f10-solns - MATH 137 Assignment 4 - Solutions 1. (a)...

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