f(a+b)=f(a)+f(b)+2ab
1.求f (0) = ?
2.已知f ' (0) = 2
求 f ' (2) = ?
基本微分定義問題
2013-05-16 12:45 am
回答 (2)
2013-05-16 1:09 am
✔ 最佳答案
f(a+b)=f(a)+f(b)+2ab1.求f (0) = ?
f(0) = f(0+0) = f(0)+f(0)+2(0)(0) = 2f(0)
所以 f(0) = 0.
2.已知f ' (0) = 2
求 f ' (2) = ?
f(a+x) = f(a)+f(x)+2ax
所以, 若 f 可微分, 則 f'(a+x) = f'(x)+2a
取 a=2, 則 f'(+x) = f'(x)+4
將 x=0 代入, 得 f'(2) = f'(0)+4 = 2+4 = 6.
2013-05-16 10:17:03 補充：
[更正錯漏]
取 a=2, 則 f'(2+x) = f'(x)+4
2013-05-16 5:45 am
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130515000010KK02559