求函数定义域设0<=a<1,函数f(x)=(a

求函数定义域设0<=a<1,函数f(x)=(a: 设0< =a< 1,f(x)=(a-1)x^2-6ax+a+1横为正，求f(x)定义域; 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]