Question

（要快）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

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

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

1.0≤2a-1≤1时,g(a)=a^2-1

2.2a-1≤0时,x取0,g(a)=(1-2a)^2+a^2-1

3.2a-1≥1时,x取1,g(a)=(2-2a)^2+a^2-1