f(x+y)=(f(x)+f(y))/(1-4f(x)f(y)),f可导 ,f'(0)=1/2 ,求f(x)
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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.
∴ 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.
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