设函数f(x)满足f'(lnx)=1-x,且f(0)=0,求f(x)
人气:229 ℃ 时间:2020-06-12 06:11:17
解答
设Y=lnX,则X=e^y
所以f^(y)=1-e^y,即f(y)=y-e^y+a,a为常数
又因为f(0)=0,所以f(0)=0-1+a=0,所以a=1
所以f(x)=x-e^x=1,再加上一个x的范围负无穷到正无穷.
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