Question

数学题。。做一下第三问

已知函数f(x)=(lnx+k)/e^x(k为常数,e=2.71828...是自然对数的底数)曲线y=f(x)在（1，f(1))处的切线与x轴平
行（1）求k（2)f(x)单调区间（3）g(x)=(x^2+x)f'(x),对任意x>0,g(x)<1+e^(-2)

Answer 1

g(x)=(1-xlnx-x)(x 1)/e^x
g(1)=0,h(x)=1-xlnx-x. h'(x)=-lnx-2.
x>1时，h'(x)<0恒有 h(x)<0. g(x)<0恒成立，所以讨论0<x<1. h'(x)=0. x=e^-2. 0<x<e^-2, h'(x)>0. e^-2<x<1,h'(x)<0, h(x)max=h(e^-2)=1 e^-2
易知(x 1)/e^x<1恒成立。故g(x)<1 e^-2

Answer 2

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Answer 3

Answer 4

Answer 5