# solHMK10 - Section 5.3 #1) By the Max-min theorem, f...

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Unformatted text preview: Section 5.3 #1) By the Max-min theorem, f attains a minimum value α = f ( c ) for some c ∈ I . Since f is positive- valued, α > 0. Since α is minimum, f ( x ) ≥ α for all x ∈ I . #4) Let f ( x ) = a n x n + ∑ n- 1 k =0 a k x k , where a n 6 = 0 and n odd. Note that lim x →∞ f ( x ) / ( a n x n ) = lim x →-∞ f ( x ) / ( a n x n ) = 1 . Thus we can definitely find A < 0 and B > 0 such that f ( A ) / ( a n A n ) > 0 and f ( B ) / ( a n B n ) > 0. Note that a n A n and a n B n have different sign, so f ( A ) and f ( B ) have different sign. Therefore, by the Location of Roots Theorem, f has a root in [ A,B ]. #6) Let g : [0 , 1 / 2] → R be defined by g ( x ) = f ( x )- f ( x + 1 / 2). g ( c ) = 0 if and only if f ( c ) = f ( c + 1 / 2), so we must find a zero of g . Note that g (0) = f (0)- f (1 / 2) and g (1 / 2) = f (1 / 2)- f (1), and since f (1) = f (0), we have that g (1 / 2) =- g (0). If g (0) = 0, we set c = 0 and we are done. If g (0) 6 = 0, then g (0) and g (1...
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## This note was uploaded on 09/30/2009 for the course MATH 444 taught by Professor Junge during the Spring '08 term at University of Illinois at Urbana–Champaign.

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