题目
MAT136H5 S 2025 - All Sections 6.3 preparation check
多重下拉选择题
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
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思路分析
Let me break down each part and option carefully so you can see where each piece comes from and why it fits or doesn't fit.
a) What is the derivative of f(x) = (1+x)^{3/2}?
- The general power-rule application gives f'(x) = (3/2)(1+x)^{1/2}. This matches the first option: (3/2) (1+x)^{1/2}. It is the correct expression for f'(x).
b) What is f''(x)?
- Differentiating again, f''(x) = (3/2) * (1/2) * (1+x)^{-1/2} = (3/4)(1+x)^{-1/2}. The second option corre......Login to view full explanation登录即可查看完整答案
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