判断函数f（x）=x/（x^2-1在区间（-1,1）上的单调性,并给出证明

Question

判断函数f（x）=x/（x^2-1在区间（-1,1）上的单调性,并给出证明

Answer 1

判断：
f(x) = x/(x^2-1)
f'(x) = {(x^2-1)*1 - x*2x}/(x^2-1)^2 = -(x^2+1)/(x^2-1)^2 ＜0,函数在定义域上单调减
分母不为零,x≠±1
在区间（-1,1）单调减
或者：f(x) = x/(x^2-1)= 1/(x-1/x)
在定义域上x单调增,1/x单调减,x-1/x单调增,1/(x-1/x)单调减,
函数在定义域上单调减
证明：
令-1＜x1＜x2＜1
f(x2) - f(x1) = x2/(x2^2-1) - x1/(x1^2-1)
= { x2(x1^2-1) - x1(x2^2-1) } / { (x2^2-1) (x1^2-1) }
= (x1^2x2-x2-x1x2^2+x1) / { (x2^2-1) (x1^2-1) }
= (x1^2x2-x1x2^2+x1-x2) / { (x2^2-1) (x1^2-1) }
= {x1x2(x1-x2)+(x1-x2)} / { (x2^2-1) (x1^2-1) }
= (x1-x2)(x1x2x1+1) / { (x2^2-1) (x1^2-1) }
∵-1＜x1＜x2＜1
∴(x1-x2)＜0；
-1＜x1x2＜1,x1x2x1+1＞0；
x2^2-1＜0,x1^2-1＜0
∴(x1-x2)(x1x2x1+1) / { (x2^2-1) (x1^2-1) }＜0
∴f(x2) ＜ f(x1) ,得证