求二次函数f﹙x﹚＝x²－2﹙2a－1﹚x＋5a²－4a＋2在[0,1]上的最小值g﹙a﹚的解析式.

讨论呗

f(x)在x=2a-1处取得最小值

若2a-1<=0,即a<=1/2,那么g(a)=f(0)=5a^2-4a+2

若0<2a-1<=1即1/2<a<=1,那么g(a)=f(2a-1)=(2a-1)^2-2(2a-1)^2+5a^2-4a+2=a^2+1

若2a-1>1即a>1,那么g(a)=f(1)=1-2(2a-1)+5a^2-4a+2=5a^2-8a+5

综上g(a)=5a^2-4a+2 (a<=1/2)

               a^2+1(1/2<a<=1)

               5a^2-8a+5(a>1)

写成分段函数哟