证明f(x)=x2-2x-2在(负无穷,1)上是单调减函数,变形到后面怎么化成商的形式

Question

证明f(x)=x2-2x-2在(负无穷,1)上是单调减函数,变形到后面怎么化成商的形式

Answer 1

f(x)=x^2-2x-2
令x1＜x2＜1
f(x2)-f(x1) = (x2^2-2x2-2) - (x1^2-2x1-2)
= x2^2-x1^2 - 2(x2-x1)
= (x2+x1)(x2-x1) - 2(x2-x1)
= (x2-x1)(x2+x1-2)
x2＞x1,∴x2-x1=0
x2＜1,x1＜x2＜1,∴x2+x1＜2,x2+x1-2＜0
∴ (x2-x1)(x2+x1-2)＜0
∴f(x2)＜f(x1),得证.