Question

已知f(x)=x3-x2-8x,用函数单调性定义证明f(x)在[0,2]为单调减函数

Answer 1

设0≤x₁＜x₂≤2，所以f(x₁)-f(x₂)=x₁³-x₁²-8x₁-(x₂³-x₂²-8x₂)=x₁³-x₂³-(x₁²-x₂²)-8(x₁-x₂)
=(x₁-x₂)(x₁²+x₁x₂+x₂²-x₁-x₂-8)=(x₁-x₂)[x₁²-4+x₂²-4+x₁x₂-(x₁+x₂)]
因为0≤x₁＜x₂≤2，所以0＜x₁x₂＜4，即0＜√x₁x₂＜2，所以x₁²-4＜0，x₂²-4＜0，
又x₁x₂-(x₁+x₂)＜x₁x₂-2√x₁x₂=√x₁x₂(√x₁x₂-2)＜0，所以[x₁²-4+x₂²-4+x₁x₂-(x₁+x₂)]＜0
又x₁-x₂＜0，所以f(x₁)-f(x₂)＞0，所以f(x)在[0，2]上是单调减函数。

Answer 2

令0=<x1<x2<=2
则f(x1)-f(x2)=x1³-x2³-(x1²-x2²)-8(x1-x2)
=(x1-x2)(x1²+x1x2+x2²-(x1-x2)(x1+x2)-8(x1-x2)
=(x1-x2)(x1²+x1x2+x2²-x1-x2-8)
=0.5(x1-x2)(2x1²+2x1x2+2x2²-2x1-2x2-16)
=0.5(x1-x2)[(x1+x2)²+(x1-1)²+(x2-1)²-18]
因为x1-x2<0, (x1+x2)²<4²=16, (x1-1)²<=1, (x2-1)²<=1
所以(x1+x2)²+(x1-1)²+(x2-1)²-18<0
因此上式>0, 即f(x1)>f(x2)
所以函数在[0, 2]单调减。