Question

导数的题要过程不要讲解设常数a>=0,函数f(x)=x-(lnx)^2+2alnx-1(x>0)(1)令g(x)=xf'

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.