當(dāng)x∈[0,1]時(shí),求函數(shù)f(x)=x^2+(2-6a)x+3a^2的最小值
f(x)=[x+(1-3a)]^2-(1-3a)^2+3a^2
=[x+(1-3a)]^2-1+6a-6a^2
該函數(shù)的對稱軸為直線 x=3a-1
1.
當(dāng) 3a-1=2/3時(shí),
函數(shù)f(x)在區(qū)間[0,1]上是遞減函數(shù),
所以 f(x)min=f(1)=3-6a+3a^2
3.
當(dāng) 0