讨论函数f(x)=ax/x^2-1(a≠0)在区间(-1,1)上的单调性

Question

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Answer 1

-1<x1<x2<1
f(x2)-f(x1)
=ax2/(x2^2-1)-ax1/(x1^2-1)
=a*(x2x1^2-x2-x1x2^2+x1)/[(x2+1)(x2-1)(x1+1)(x1-1)]
=a*[x1x2(x1-x2)+(x1-x2)]/[(x2+1)(x2-1)(x1+1)(x1-1)]
=a*(x1x2+1)(x1-x2)/[(x2+1)(x2-1)(x1+1)(x1-1)]

-1<x1<x2<1
看分母：x2+1>0 x1+1>0 x2-1<0 x1-1<0 所以分母大于0
分子：-1<x1x2<1 x1x2+1>0 x1-x2<0
1
a>0 分子<0
f(x2)<f(x1) 减函数
2
a<0 分子>0
f(x2)>f(x1)增函数！