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Unformatted text preview: MATH 128A, SUMMER 2009: HOMEWORK 1 SOLUTIONS 1.1 1(d). f ( x ) = x (ln x ) x is continuous when x > 1 (since ln( x ) > 0), hence on [4 , 5]. f (4) ≈ . 3066 > and f (5) ≈  5 . 799 < 0, so by the intermediate value theorem there is an x ∈ [4 , 5] with f ( x ) = 0. 4(a). First note that f ( x ) = ( e x + 2) / 3, so the only critical point of f occurs at x = ln2, which lies in the interval [0 , 1]. The maximum for  f ( x )  must consequently be max { f (0)  ,  f (ln2)  ,  f (1) } = max { 1 / 3 , (2ln2) / 3 , (4 e ) / 3 } = (2ln2) / 3 . 10. The first two derivatives of f ( x ) = e x cos x are f ( x ) = e x (cos x sin x ) and f 00 ( x ) = e x ( 2sin x ). Thus the second Taylor polynomial centered at π/ 6 is P 2 ( x ) = √ 3 2 e π/ 6 + √ 3 1 2 e π/ 6 ( x π/ 6) 1 2 e π/ 6 ( x π/ 6) 2 . Since f 000 ( x ) = e x ( 2cos x 2sin x ), the error R 2 ( x ) is given by 2 e ξ ( x ) (cos ξ ( x )+sin ξ ( x )) 3! ( x π/ 6) 3 = 1 3 e ξ ( x ) (cos ξ ( x ) + sin ξ ( x ))( x π 6 ) 3 , where ξ ( x ) is a number between π/ 6 and x . (a) P 2 (0 . 5) ≈ 1 . 44688. The error is 1 3 e ξ (cos ξ +sin ξ )( π 6 1 2 ) 3 , for some ξ ∈ [0 . 5 ,π/ 6]. This quantity’s maximum value is 1 . 01019 · 10 5 , obtained when ξ = π/ 6; one may also overestimate it to find an error bound: for example,  R 2 (0 . 5)  < 1 3 e π/ 6 (cos0+sin π 6 )( π 6 1 2 ) 3 < 1 3 (3) 1 (1+ 1 2 )( 4 6 1 2 ) 3 ≈ 6 . 94444 · 10 3 . The actual error is  R 2 (0 . 5)  ≈ 1 . 00265 · 10...
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This note was uploaded on 04/06/2010 for the course MATH various taught by Professor Tao/analysis during the Spring '10 term at UCLA.
 Spring '10
 tao/analysis
 Intermediate Value Theorem

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