Question

一道函数数学题

用定义证明y=2x²-4x+3在区间（负无穷，1】上是减函数.

Answer 1

设x1<x2<=1;
y1-y2=2x1^2-2x2^2-4x1+4x2
=2(x1+x2)(x1-x2)-4(x1-x2)
=2(x1-x2)(x1+x2-2)
因为x1-x2<0,且x1+x2<2,即x1+x2-2<0;
所以y1-y2>0;
此题得证。

Answer 2

设，x1<x2<=1
f(x1)-f(x2) = 2(x1^2 -x2^2)-4(x1-x2) = 4(x1-x2)((x1+x2)/2) >0
所以f(x1)>f(x2)
即，f(x)是减函数~