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Pre-Calc Homework Solutions 396

# Pre-Calc Homework Solutions 396 - 396 Chapter 9 Review 1 xx...

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39. f (3) 5 } 1 x } ) x 5 3 5 } 1 3 } f 9 (3) 52 x 2 2 ) x 5 3 52} 1 9 } f 0 (3) 5 2 x 2 3 ) x 5 3 5 } 2 2 7 } , so } f 0 2 ( ! 3) } 5 } 2 1 7 } f - (3) 6 x 2 4 ) x 5 3 2 2 7 } , so } f - 3 ( ! 3) }52} 8 1 1 } } f ( n n ) ( ! 3) } 5 } ( 2 3 n 1 +1 ) n } } 1 x } 5 } 1 3 } 2 } 1 9 } ( x 2 3) 1 } 2 1 7 } ( x 2 3) 2 2 } 8 1 1 } ( x 2 3) 3 1 1 ( 2 1) n } ( x 3 2 n + 3 1 ) n } 40. f ( p ) 5 sin x ) x 5 p 5 0 f 9 ( p ) 5 cos x ) x 5 p 1 f 0 ( p ) sin x ) x 5 p 5 0, so } f 0 2 ( p ! ) } 5 0 f - ( p ) cos x ) x 5 p 5 1, so } f - 3 ( ! p ) } 5 } 1 6 } f ( k ) ( p ) 5 5 sin x ( x 2 p ) 1 } 3 1 ! } ( x 2 p ) 3 2 } 5 1 ! } ( x 2 p ) 5 1 } 7 1 ! } ( x 2 p ) 7 2 1 ( 2 1) n 1 1 } (2 n 1 1 1)! } ( x 2 p ) 2 n 1 1 1 41. Diverges, because it is 2 5 times the harmonic series: n 5 1 } 2 n 5 }52 5 n 5 1 } 1 n }52‘ 42. Converges conditionally. If u n 5 } ˇ 1 n w } , then { u n } is a decreasing sequence of positive terms with lim n u n 5 0, so n 5 1 } ( 2 ˇ 1 n w ) n } converges by the Alternating Series Test. The convergence is conditional because n 5 1 } ˇ 1 n w } is a divergent p -series 1 p 5 } 1 2 } 2 . 43. Converges absolutely by the Direct Comparison Test, since 0 # } ln n 3 n } , } n 1 2 } for n \$ 1 and n 5 1 } n 1 2 } converges as a p -series with p 5 2.
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