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

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=<2a<1时0=<a<1/2则f(x)min=f(2a-1)=2a^2+1
若2a-1<0时a<1/2则f(x)min=f(0)=(1-2a)^2+a^2+1
=5a^2-4a+2
宗上所述得：
a>=1时g(x)=5a^2-8a+5
0=<a<1/2时g(x)=2a^2+1
a<0时g(x)=5a^2-4a+2

f'(x)=2x-2(2x-1)
=2-2x
x=1,f(1)=0,
x∈[0,1]递增 ,f'(x)>0
最小值g(a)=f(0)
=5a^2-4a+2

f(x)=x^2-2(2a-1)x+5a^2-4a+2
=x^2-2(2a-1)x+(4a^2-4a+1)+a^2+1
=[x-(2a-1)]^2+a^2+1
对称轴为：x=2a-1
1)当2a-1>=1 所以x=1的时候f(x)值最小
g(a)=5a^2-8a+5 (a>=1)
2)当2a-1<=0 所以x=0的时候f(x)值最小
g(a)=5a^2-4a+2 (a<=1/2)
3)当0<=2a-1<=1 g(a)=5a^2-4a+2 (1/2<=a<=1)