Question

当x属于[0.1]时,求函数fx=x2+(2-6a)x+3a2的最小值

Answer 1

解：
f(x)=x²+(2-6a)x+3a²=[x+(1-3a)]²-(1-3a)²+3a²

=[x+(1-3a)]²-6a²+6a-1

∴在R上x=3a-1时f(x)有最小值

那么，当3a-1<0时，f(x)在[0,1]上单调递增，故f(x)在x=0时最小，f(x)min=f(0)=3a²
当0≤3a-1≤1时，f(x)在x=3a-1时有最小值，f(x)min=-6a²+6a-1

当3a-1>1时，f(x)在[0,1]上单调递减，故f(x)在x=1时最小，f(x)min=f(1)=3-6a+3a²

Answer 2

f(X)=[X+(1-3a)]^2+3a^2-(1-3a)^2
要使f(X)最小，让[X+（1-3a）]^2最小
令X+（1-3a）=0即X=3a-1
当a>2/3时，X取1
a<1/3时，X取0

Answer 3

f(x)=[x+(1-3a)]^2-(1-3a)^2+3a^2=[x+(1-3a)]^2-6a^2+6a-1∴在R上x=3a-1时f(x)有最小值那么，当3a-1<0时，f(x)在[0,1]上单调递增，故f(x)在x=0时最小，f(x)min=f(0)=3a^2当0≤3a-1≤1时，f(x)在x=3a-1时有最小值，f(x)min=-6a^2+6a-1当3a-1>1时，f(x)在[0,1]上单调递减，故f(x)在x=1时最小，f(x)min=f(1)=3-6a+3a^2