已知函数f(x)可导,且对任何实数x,y满足:f(x+y)=e^xf(y)+e^yf(x)和f'(0)=e 证明:f'(x)=f(x)+e^(x+1)
e^x*f(x)
人气:226 ℃ 时间:2020-02-02 20:32:43
解答
put x=y =0f(0) = f(0) + f(0)=>f(0) = 0f'(x)= lim(y->0)[f(x+y) - f(x)]/y= lim(y->0)[e^xf(y)+e^yf(x) - f(x)]/y= e^x lim(y->0)[f(0+y)-f(0)]/y + f(x) lim(y->0)( e^y - 1)/y= e^xf'(0) + f(x)= e^(x+1) + f(x)
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