2 2 5 1 sets up ratio 1 limit evaluation 1 radius of

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2 2 5 : 1 : sets up ratio 1 : limit evaluation 1 : radius of convergence 1 : considers both endpoints 1 : analysis and interval of convergence (b) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 3 4 2 3 4 2 2 2 1 2 1 2 3 1 1 2 16 4 4 3 1 n n n n x x x x y n x x x x n = + + + = + + + " " " " ( ) ( ) 1 2 3 1 2 2 64 8 12 3 1 n n n x y x x x n ′ = + + + " " ( ) ( ) 2 3 4 1 2 64 8 12 3 1 n n n x xy x x x n ′ = + + + " " ( ) ( ) ( ) ( ) ( ) 2 3 4 2 2 2 4 8 16 1 2 4 1 2 4 1 2 n n n n xy y x x x x x x x x ′ − = + + + = + + + " " " " The series ( ) ( ) ( ) 2 2 0 1 2 4 1 2 2 n n n n x x x x = + + + = " " is a geometric series that converges to 1 1 2 x + for 1 . 2 x < Therefore 2 1 4 1 2 xy y x x ′ − = + for 1 . 2 x < 4 : 1 : series for 1 : series for 1 : series for 1 : analysis with geometric series y xy xy y ′ −
AP ® CALCULUS BC 2011 SCORING GUIDELINES Question 6 © 2011 The College Board. Visit the College Board on the Web: . Let ( ) ( ) 2 sin cos . f x x x = + The graph of ( ) ( ) 5 y f x = is shown above. (a) Write the first four nonzero terms of the Taylor series for sin x about 0, x = and write the first four nonzero terms of the Taylor series for ( ) 2 sin x about 0. x = (b) Write the first four nonzero terms of the Taylor series for cos x about 0. x = Use this series and the series for ( ) 2 sin , x found in part (a), to write the first four nonzero terms of the Taylor series for f about 0. x = (c) Find the value of ( ) ( ) 6 0 . f (d) Let ( ) 4 P x be the fourth-degree Taylor polynomial for f about 0. x = Using information from the graph of ( ) ( ) 5 y f x = shown above, show that ( ) ( ) 4 1 1 1 . 4 4 3000 P f < (a) 3 5 7 sin 3! 5! 7! x x x x x = + + " ( ) 6 10 1 2 4 2 sin 3! 5! 7! x x x x x = + + " 3 : ( ) 2 1 : series for sin 2 : series for sin x x (b) 2 4 6 cos 1 2! 4! 6! x x x x = + + " ( ) 2 4 6 121 1 2 4! 6! x x x f x = + + + " 3 : ( ) 1 : series for cos 2 : series for x f x (c) ( ) ( ) 6 0 6! f is the coefficient of 6 x in the Taylor series for f about 0. x = Therefore ( ) (6) 0 121. f = − 1 : answer (d) The graph of ( ) ( ) 5 y f x = indicates that ( ) 1 4 (5) 0 max 40. x f x ≤ ≤ < Therefore ( ) ( ) ( ) ( ) 1 4 (5) 5 4 5 0 max 1 40 1 1 . 5! 4 307 1 1 4 4 2 3000 120 4 x f x P f ≤ ≤ < = < 2 : { 1 : form of the error bound 1 : analysis
AP ® CALCULUS BC 2011 SCORING GUIDELINES (Form B) Question 6 © 2011 The College Board. Visit the College Board on the Web: . Let ( ) ( ) 3 ln 1 . f x x = + (a) The Maclaurin series for ( ) ln 1 x + is ( ) 2 3 4 1 1 . 2 3 4 n n x x x x x n + + + + + " " Use the series to write the first four nonzero terms and the general term of the Maclaurin series for f . (b) The radius of convergence of the Maclaurin series for f

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