基本微分定義問題

f(a+b)=f(a)+f(b)+2ab

1.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

唉呦....

好厲害.....