128a_su09_hw1_sol - 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 0.3066 >

# 128a_su09_hw1_sol - MATH 128A SUMMER 2009 HOMEWORK 1...

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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) 0 . 3066 > 0 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 0 ( x ) = ( - e x + 2) / 3, so the only critical point of f occurs at x = ln 2, which lies in the interval [0 , 1]. The maximum for | f ( x ) | must consequently be max {| f (0) | , | f (ln 2) | , | f (1) |} = max { 1 / 3 , (2 ln 2) / 3 , (4 - e ) / 3 } = (2 ln 2) / 3 . 10. The first two derivatives of f ( x ) = e x cos x are f 0 ( x ) = e x (cos x - sin x ) and f 00 ( x ) = e x ( - 2 sin 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 ( - 2 cos x - 2 sin 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 (cos 0+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 - 5 . (b) There are again multiple correct ways to find a bound on | R 2 ( x ) | for x [0 , 1]. Without any knowledge of ξ ( x
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