令g(x)=xe^x-lnx x∈(0,+∞)
g'(x)=e^x+xe^x-1/x
g''(x)=2e^x+xe^x+1/x²>0→g(x)是凹函数
y=kx+1 过定点(0,1) 由于g(x)是凹函数,故y=kx+1正好是g(x)的切线时,g(x)≥kx+1
当且当切点处,等号成立
设切点坐标为(x₁,kx₁+1)
g'(x₁)=e^x₁+x₁e^x₁-1/x₁=k
g(x₁)=x₁·e^x₁-lnx₁=kx₁+1
解得k=1,k<1时 kx+1-(x+1)=(k-1)x<0→ kx+1<x+1<g(x)
k∈[-∞,1]
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