已知函数f(x)=(2-a)lnx+1/x+2ax,问当a

问题描述:

已知函数f(x)=(2-a)lnx+1/x+2ax,问当a

f'(x)=(2-a)/x-1/x^2 2a
=[2ax^2 (2-a)x-1]/x^2
=(2x-1)(ax 1)/x^2
=a(x-1/2)(x 1/a)/x^2
令f'(x)=0得x=1/2或x=-1/a
当0f(x)递增区间为(-1/a,1/2)
递减区间为(0,-1/a),(1/2, ∞)
当-1/a=1/2即a=-2时,f'(x)=-2(x-1/2)^2/x^2≤0
∴f(x)递减区间为(0, ∞)
当-1/a>1/2即-2f(x)递增区间为(1/2,-1/a)
递减区间为(0,1/2),(-1/a, ∞)

f'﹙x﹚=﹙2-a﹚/x-1/x²+2a
=﹙2x-1﹚﹙ax+1﹚/x²
=2a﹙x-1/2﹚﹙x+1/a﹚/x²
①当-2<a<0时,-1/a>1/2
函数在﹙0,1/2﹚上递减,在﹙1/2,-1/a﹚上递增,在﹙-1/a,+∞﹚上递减
②当a=-2时,导函数≤0
函数在﹙0,+∞﹚上递减
③当a<-2时,0<-1/a<1/2
函数在﹙0,-1/a﹚上递减,在﹙-1/a,1/2﹚ 上递增,在﹙1/2,+∞﹚上递减