Question

求二次函数f(x)=x^2-2(2a-1)x+5a^2-4a+2在[0,1]上的最小值g(a)的解析式

Answer 1

解；f(x)=x^2-2(2a-1)x+5a^2-4a+2

=(x-2a+1)^2-4a^2+4a-1+5a^2-4a+2

=(x-2a+1)^2+a^2+1

若2a-1>=1时a>=1则f(x)min=f(1)=(2-2a)^2+a^2+1

=5a^2-8a+5

若0=