Question

高中数学导数证明

Answer 1

f'(x)=(1-lnx)/x^2
当0<x<e时，f'(x)>0，f(x)递增；当x>e时，f'(x)<0，f(x)递减
所以2f(x)<=2f(e)=2e^(-1)
令g(x)=(x+1)e^(x-2)
g'(x)=e^(x-2)+(x+1)(x-2)e^(x-2)=(x^2-x-1)e^(x-2)
当0<x<(1+√5)/2时，g'(x)<0，g(x)递减；当x>(1+√5)/2时，g'(x)>0，g(x)递增
所以g(x)>=g[(1+√5)/2]=[(3+√5)/2]*e^[(√5-3)/2]
因为(3+√5)/2>2，(√5-3)/2>-1，所以[(3+√5)/2]*e^[(√5-3)/2]>2e^(-1)
即2f(x)<=2e^(-1)<[(3+√5)/2]*e^[(√5-3)/2]<=g(x)
2f(x)<(x+1)e^(x-2)