Question

当0≤x≤1时,求函数y=x²+（2-6a)x+3a²的最小值

Answer 1

函数f(x)=x^2+(2-6a)x+3a^2

开口向上,对称轴x=-(2-6a)/2=3a-1

若3a-1≤0,即:a≤1/3

函数f(x)=x^2+(2-6a)x+3a^2在[0,1]上单调递增

当x=0,最小值f(0)=0^2+(2-6a)*0+3a^2=3a^2

若0