Find the point of inflection . f(x) = x^1/3
點解f^2(x) 是不存在，當x=0 .f(0)=0
可否給我祥細解釋f^2(x)何時存在和何時不存在和舉些例子。
Amath ：solve inflection point
2013-05-13 2:16 am
回答 (2)
2013-05-13 10:18 pm
✔ 最佳答案
f(x) = x^(1/3)f'(x) = (1/3)x^(-2/3)
f''(x) = (-2/9)x^(-5/3) = (-2/9)(1/x^3)^5
代x = 0時﹐因為x在分母位置﹐所以f''(x) = ∞ 或者不存在。而
當x < 0 時﹐f''(x) > 0
當x > 0 時﹐f''(x) < 0
因此(0,0)是拐點。
2013-05-13 12:59 pm
consider f ' ' (0)
= lim[h->0] [ f ' (0+h) - f ' (0) ] / h
for f (x) = x^ (1/3), f ' (x) = (1/3) x^ (-2/3)
since f ' (0) is undefined (not exist), f ' ' (0) also does not exist.
= lim[h->0] [ f ' (0+h) - f ' (0) ] / h
for f (x) = x^ (1/3), f ' (x) = (1/3) x^ (-2/3)
since f ' (0) is undefined (not exist), f ' ' (0) also does not exist.
收錄日期: 2021-04-27 17:45:27
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130512000051KK00222