title: 求函数定义域设0&lt;=a&lt;1,函数f(x)=(a
desc: 设0< =a< 1,f(x)=(a-1)x^2-6ax+a+1横为正，求f(x)定义域
answer: 设0≤a< 1,函数f(x)=(a-1)x^2-6ax+a+1横为正，求f(x)定义域 
  
  (a-1)x^2-6ax+a+1=a(x^2-6x+1)-(x^2-1)
  把a(x^2-6x+1)-(x^2-1)看成关于a一次函数g(a)=a(x^2-6x+1)-(x^2-1)
  
  g(0)=-(x^2-1)＞0且g(1)=-6x+2≥0
  即:-1＜x＜1且x≤1/3
  -1＜x≤1/3,f(x)定义域:(-1,1/3]
  
  
  
  
  
  
  
  