函数fx=(2a+3)lnx+ax^2+1讨论函数单调性.设a≤-2,证明:对任意x1x2∈(0,﹢∞),|fx1-fx2|≥4|x1-x2|
问题描述:
函数fx=(2a+3)lnx+ax^2+1讨论函数单调性.设a≤-2,证明:对任意x1x2∈(0,﹢∞),|fx1-fx2|≥4|x1-x2|
答
对fx求导,得fx‘=(2a+3)/x+2ax,a≤-2,fx`<0,fx单减,不妨设x1<x2,fx1>fx2,|fxi-fx2|=fx1-fx2,|x1-x2|=x2-x1,即证fx1-fx2≥4(x2-x1),fx1+4x1≥fx2+4x2,x1<x2,即证fx+4x单减,对新函数求导(2a+3)/x+(2a+4)x<0,得证