tut2 - 1 Tutorial 2 Answers + Comments Q1 This question...

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1 Tutorial 2 Answers + Comments Q1 This question uses the Intermediate Value Theorem. In it's most general form, we have Théorème 0.1 Let I be an interval and let f : I 7→ R be continuous on I . If a,b I and if k R satis es f ( a ) < k < f ( b ) , then there exists a point c I between a and b such that f ( c ) = k. In this question, we are localising the roots of the polynomial p ( x ) = x 3 - 3 x 2 + c , where c > 2 . Note that in general, a polynomial of odd degree has always at least one root. This is guaranteed by the Intermediate Value Theorem since for a polynomial p of odd degree, the dominant term is of the form x 2 n +1 , and when x → -∞ p → -∞ , and when and when x + , p + . Taking k = 0 in the Intermediate Value Theorem (generalised), we have thus a point m R such that p ( m ) = 0 . Note that this theorem guarantees at least one real root, but NOT exactly one real root. Another way to show that there exists at least one real root is to recall that if the roots
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tut2 - 1 Tutorial 2 Answers + Comments Q1 This question...

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