Question

导数题，不懂，要过程

Answer 1

f'(x)=1-2(lnx)/x+2a/x
(1)g(x)=x-2lnx+2a,g'(x)=1-2/x,g"(x)=2/x^2>0,所以g(x)是凹函数.因此g(x)在x=2有最小值2-2ln2+2a>0
(2)f"(x)=-2/x^2+2(lnx)/x^2-2a/x^2=-2(1-lnx+2a)/x^2=0,x=e^(2a+1),当x≥e^(2a+1),f"(x)≥0,在0＜e^(2a+1)≤e^(2a+1),f"(x)≤0,f'(x)在e^(2a+1)达到最小值,f'(e^(2a+1))=1-2(2a+1)/e^(2a+1)+2a/e^(2a+1)=(e^(2a+1)-2a-2)/e^(2a+1)>0,所以f'(x)>0,即f(x)在(0,+∞)严格单调增
f(1)=1-1=0,因为f(x)严格单调增,所以x>1时f(x)>0.

Answer 2

解1=lim[f(x0-t)-f(x0+t)]/t
=lim[f(x0+t-t)-f(x0+2t)]/t
=lim[f(x0)-f(x0+2t)]/t
=-lim[f(x0+2t)-f(x0)]/t
=-2lim[f(x0+2t)-f(x0)]/2t
则lim[f(x0+2t)-f(x0)]/2t=-1/2
则f'(x0)=-1/2