y为2为周期可导函数,极限x趋于0时(f(1+x)+4f(1-x))/x=3,y=f(x)在(3,f(x))的切线是什么？

Question

如题所述

Answer 1

应用洛必达法则求出导数。

追问

那f(3)怎么消掉，答案为x+y=3

追答

由已知知f(1)=0，f(3)=0，∴x+y-3=0，

Answer 2

f(0+0) = 2f(0)/[1-4f^2(0)]
∴ f(0) - 4f^3(0) = 2f(0)
即有：f(0)=0
f'(x)
= lim(y->0) [f(x+y)-f(x)]/y
= lim(y->0) { [f(x)+f(y)]/[1-4f(x)f(y)] - f(x) }/y
= lim(y->0) f(y)[1+4f^2(x)]/y[1-4f(x)f(y)]
= lim(y->0) [f(y)-f(0)]/y * [1+4f^2(x)]/[1-4f(x)f(y)]
= f'(0) * [1+4f^2(x)]
∵对任意 x∈R , f'(x) 存在,∴f（x）在R上可微.