Question

已知函数f(x)=(a+1/a)lnx+1/x-x(a>1) 1,试讨论f(x)在区间(0,1)的单调性

Answer 1

f(x1)-f(x2)=(a+1/a)lnx1+1/x1-x1-(a+1/a)lnx2-1/x2+x2

=(a+1/a)ln(x1/x2)+(x2-x1)/x1x2+(x2-x1)
=(a+1/a)ln(x1/x2)+(x2-x1)(1/x1x2+1)
因为0<x1<x2<1, 所以 (x2-x1)>0, (1/x1x2+1)>0, ln(x1/x2)<0

当a<0时，(a+1/a)<0, (a+1/a)ln(x1/x2)>0, f(x1)-f(x2)>0, 函数单调递增

Answer 2

f(x)=(a+1/a)lnx+1/x-x(a>1，0<x<1),
f'(x)=(a+1/a)/x-1/x^-1
=-[x^-(a+1/a)x+1]/x^
=-(x-a)(x-1/a)/x^,
1/a<x<1时f'(x)>0,f(x)↑；
0<x<1/a时f'(x)<0,f(x)↓。