This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 301/ENAS 513: Homework 11 Solution Triet M. Le Page 107: 3(a): Let P be a polynomial of odd degree. One sees that lim x →∞ P ( x ) =∞ and lim x →∞ P ( x ) = ∞ . Let a, b ∈ R such that P ( a ) < 0 and P ( b ) > 0. Clearly, P is continuous on [ a, b ]. By the Intermediate Value Theorem, there exists c ∈ ( a, b ) such that P ( c ) = 0. Page 107: 3(b): Let P ( x ) = a n x n + ... + a , a n 6 = 0 and n is even. WLOG, WMA a n > 0 and n ≥ 2 (if n = 0, then P is a constant and of course P attains both its max and min). We have P ( x ) = na n x n 1 + ... + a 1 . Since na n > 0 and n 1 is odd, we have lim x →∞ P ( x ) =∞ and lim x →∞ P ( x ) = ∞ . Let a, b ∈ R such that P ( x ) < 0 for all x ≤ a and P ( y ) > 0 for all y ≥ b . This implies, P ( x ) ≥ P ( a ) for all x ≤ a and P ( y ) ≥ P ( b ) for all y ≥ b . Clearly, P is continuous on the compact set [ a, b ]. Therefore, it must attain its minimum at c ∈ [ a, b ], and hence it is also a global minimum. If we assume that a n < 0, then following the same proof we obtain a global maximum....
View
Full Document
 Fall '07
 TrietLe
 Math, Calculus, Intermediate Value Theorem, Continuous function, Compact space, M. Le

Click to edit the document details