128a_su09_hw1_sol

128a_su09_hw1_sol - MATH 128A, SUMMER 2009: HOMEWORK 1...

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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 / 6 is P 2 ( x ) = 3 2 e / 6 + 3- 1 2 e / 6 ( x- / 6)- 1 2 e / 6 ( x- / 6) 2 . Since f 000 ( x ) = e x (- 2cos x- 2sin x ), the error R 2 ( x ) is given by- 2 e ( x ) (cos ( x )+sin ( x )) 3! ( x- / 6) 3 =- 1 3 e ( x ) (cos ( x ) + sin ( x ))( x- 6 ) 3 , where ( x ) is a number between / 6 and x . (a) P 2 (0 . 5) 1 . 44688. The error is 1 3 e (cos +sin )( 6- 1 2 ) 3 , for some [0 . 5 ,/ 6]. This quantitys maximum value is 1 . 01019 10- 5 , obtained when = / 6; one may also overestimate it to find an error bound: for example, | R 2 (0 . 5) | < 1 3 e / 6 (cos0+sin 6 )( 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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128a_su09_hw1_sol - MATH 128A, SUMMER 2009: HOMEWORK 1...

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