解一下这道函数题
高中知识范围,麻烦写清楚过程,谢谢了
已知定义在[-1,1]的增函数f(x),解不等式:
f(x+1/2)<f[1/(x-1)]
参考答案:-1<=x+1/2<=1 -3/2<=x<=1/2
-1<=1/(x-1)<=1 x<=0 or x>=2
x+1/2<1/(x-1) x<-1 or 3/2>x>1
取交集 -3/2<=x<-1
高中知识范围,麻烦写清楚过程,谢谢了
已知定义在[-1,1]的增函数f(x),解不等式:
f(x+1/2)<f[1/(x-1)]
参考答案:-1<=x+1/2<=1 -3/2<=x<=1/2
-1<=1/(x-1)<=1 x<=0 or x>=2
x+1/2<1/(x-1) x<-1 or 3/2>x>1
取交集 -3/2<=x<-1