Question: Find the 3rd Maclaurin Polynomial of . 𝑓 ( 𝑥 ) = ( 1 + 𝑥 ) 3 2 Solution steps: a) What is the derivative of   𝑓 ( 𝑥 ) = ( 1 + 𝑥 ) 3 2 ?    [ Select ] (3/2) (1+x) (3/2) (1+x)^(1/2) (2/5) (1+x)^(5/2) b) What is 𝑓 ″ ( 𝑥 ) ?    [ Select ] (3/2) (1+x)^(-1/2) 2 (1+x)^(1/2) (3/4) (1+x)^(-1/2) c) What is 𝑓 ‴ ( 𝑥 ) ?    [ Select ] (-3/8) (1+x)^(-2) (-3/8) (1+x)^(-3/2) (3/4) (1+x)^(-3/2) The definition of the 3rd Maclaurin polynomial is: 𝑝 3 ( 𝑥 ) = 𝑓 ( 0 ) + 𝑓 ′ ( 0 ) 𝑥 + 𝑓 ″ ( 0 ) 2 ! 𝑥 2 + 𝑓 ‴ ( 0 ) 3 ! 𝑥 3   Therefore we should evaluate the derivatives and the function at 𝑥 = 0 . d) What is 𝑓 ( 0 ) ?    [ Select ] 1/2 0 2/3 1 e) What is 𝑓 ′ ( 0 ) ?    [ Select ] 3/4 3/2 1 1/2 f) What is 𝑓 ″ ( 0 ) ?    [ Select ] 3/4 0 3/2 -3/8 g) What is 𝑓 ‴ ( 0 ) ?    [ Select ] -3/8 3/4 1 1/2 h) Putting everything together now, which of the below options is the correct 3rd Maclaurin Polynomial of 𝑓 ( 𝑥 ) = ( 1 + 𝑥 ) 3 2 ?    [ Select ] Option II Option V Option IV Option III Option I Option I.     𝑝 3 ( 𝑥 ) = 1 + 3 4 𝑥 + 3 8 𝑥 2 − 3 8 𝑥 3     Option II.     𝑝 3 ( 𝑥 ) = 1 + 3 2 𝑥 + 3 4 𝑥 2 − 3 8 𝑥 3     Option III.     𝑝 3 ( 𝑥 ) = 2 + 3 4 𝑥 − 3 8 𝑥 2 − 1 16 𝑥 3      Option IV.    𝑝 3 ( 𝑥 ) = 1 + 3 2 𝑥 + 3 8 𝑥 2 − 1 16 𝑥 3   Option V.     𝑝 3 ( 𝑥 ) = 1 + 3 2 𝑥 + 3 4 𝑥 2 + 3 42 𝑥 3     多重下拉选择题

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