設0< =a< 1,函數f(x)=(a-1)x^2-6ax+a+1橫為正，求f(x)定義域

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設0≤a< 1,函數f(x)=(a-1)x^2-6ax+a+1橫為正，求f(x)定義域 (a-1)x^2-6ax+a+1=a(x^2-6x+1)-(x^2-1)把a(x^2-6x+1)-(x^2-1)看成關于a一次函數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)定義域:(-1,1/3]