Question

已知函数f(x)=-x³+3x。求证:函数f(x)在区间(-1,1)上是增函数.

Answer 1

证明：设-1<x1<x2<1,则
f(x1)-f(x2)
=-x1³+3x1-(-x2³+3x2)
=x2³-x1³+3(x1-x2)
=(x2-x1)(x2²+x2x1+x1²)-3(x2-x1)
=(x2-x1)(x2²+x2x1+x1²-3)
∵-1<x1<x2<1
∴x2-x1>0,x2²+x2x1+x1²<3,x2²+x2x1+x1²-3<0
∴f(x1)-f(x2)<0
即f(x1)<f(x2)
∴函数f(x)=-x³+3x在区间(-1,1)上是增函数

Answer 2

利用函数单调性的定义证明。取值-作差-变形-定号-下结论
证明：设-1<x1<x2<1,则
f(x1)-f(x2)
=-x1³+3x1-(-x2³+3x2)
=x2³-x1³+3(x1-x2)
=(x2-x1)(x2²+x2x1+x1²)-3(x2-x1)
=(x2-x1)(x2²+x2x1+x1²-3)
∵-1<x1<x2<1
∴x2-x1>0,x2²+x2x1+x1²<3,x2²+x2x1+x1²-3<0
∴f(x1)-f(x2)<0
即f(x1)<f(x2)
∴函数f(x)=-x³+3x在区间(-1,1)上是增函数