Given: f(-3)=1, f'(-3)=-2, f"(-3)=-1, further derivatives (start from third derivative) at x=-3 will be 0
Find f(x)
Thanks!!!!
F.6 Functions
2011-05-20 6:47 pm
回答 (2)
2011-05-20 7:33 pm
✔ 最佳答案
By Taylor's expansion:f(x) = f (-3) + f ' (-3) * [x-(-3)] / 1! + f '' (-3) * [x-(-3)]^2 / 2! + f ''' (-3) * [x-(-3)]^3 / 3! +...
= 1 + (-2)(x+3) + (-1)(x+3)^2 (1/2) +0+....
= (-1/2) (x^2+10x+19)
Hence f(x) = (-1/2) (x^2 +10x +19)
2011-05-21 11:50 pm
答案不唯一,例如還可再加p(x) exp[-1/(x-3)^2] (x=3時設為0)
收錄日期: 2021-04-23 20:09:50
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110520000051KK00213