有关高数的问题
设函数f(x)连续,f‘(0)存在,并且对于任何的x,y属于R,有f(x+y)=[f(x)+f(y)]/[1-4f(x)f(y)],证明f(x)在R上可微.望给出详细的解...
设函数f(x)连续,f‘(0)存在,并且对于任何的x,y 属于R,有f(x+y)=[f(x)+f(y)]/[1-4f(x)f(y)],证明f(x)在R上可微. 望给出详细的解答过程
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证明:∵ 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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