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Unformatted text preview: Math 3110 Spring ’09 Homework 7: Selected Solutions [10.11]: Claim : If f ( x ) = cos πx then (a) f ((∞ , ∞ )) = [ 1 , 1], (b) f ((0 , 1)) = ( 1 , 1), (c) f (( 1 2 , 3 2 )) = [ 1 , 0) Proof Omitted; selfevident. [10.13]: Claim : (a) f ( x ) = x 2 1 x 2 4 is not bounded on its domain D ( f ) = (∞ , 2) ∪ ( 2 , 2) ∪ (2 , ∞ ). (b) f ( x ) = sin( x )tan 1 ( x ) is bounded on its domain D ( f ) = (∞ , ∞ ). (c) f ( x ) = ln(2 + cos x ) is bounded on its domain D ( f ) = (∞ , ∞ ). Proof (a) f ( x ) has vertical asymptotes at x = ± 2 and is unbounded in both the positive and negative directions along these asymptotes. (b) sin( R ) = [ 1 , 1] and tan 1 ( R ) = ‡ π 2 , π 2 · so f ( R ) ⊂ [ π 2 , π 2 ]. (c) 2 + cos( R ) = [1 , 3] so f ( R ) = [0 , ln3]. [10.19]: Claim : If f ( x ) and g ( x ) are defined for all x ∈ R , then (a) f ( x ) bounded ⇒ f ( g ( x )) bounded and (b) f ( x ) bounded ; g ( f ( x )) bounded....
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This note was uploaded on 11/09/2009 for the course MATH 3110 taught by Professor Ramakrishna during the Fall '08 term at Cornell University (Engineering School).
 Fall '08
 RAMAKRISHNA
 Math

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