1.f'(x)=2a+b/x^2+1/xf(x)在x=-1 ,x=1/2处取得极值所以f'(-1）＝2a+b-1=0f'(1/2)=2a+4b+2=0解得a=1,b=-1 f(x)=2x+1/x+lnx2.f'(x)=2-1/x^2+1/x=(1+1/x)(2-1/x)=(x+1)(2x-1)/x^2当x