求函数y=x/1一x的单调增区间,并用定义证明用分离常数法怎么做,

Answer 1

y=x/(1-x)=-[x/(x-1)]=-[(x-1+1)/(x-1)]=-[1+1/(x-1)]=-1-1/(x-1)
设任意的x1、x2≠1,且x1＞x2
则有f（x1）-f（x2）=-1-1/(x1-1)+1+1/(x2-1)=1/(x2-1)-1/(x1-1)=(x1-x2)/(x2-1)(x1-1)
可由x1＞x2 x1-x2＞0
则可知当x1＞x2＞1或1＞x1＞x2时,y=f(x)单调递增
即递增区间为（负无穷,1）和（1,正无穷）