求二次函数在[0,1]上的最小值g(a)的解析式. 答案:解:二次函数 其图象开口向上,对称轴为x=2a-1. 若2a-1<0,即a<时,二次函数f(x)在[0,1]上的最小值-4a+2; 若0≤2a-1≤1,即≤a≤1时,二次函数f(x)在[0,1]上的最小值+1; 若2a-1>1,即a>1时,二次函数f(x)在[0,1]上的最小值-8a+5. 综上所述,二次函数f(x)在[0,1]上的最小值为g(a)=24、----What time do you start work?
---- _____________.
A. It’s none of your
business. B. It varies.
C. From 8 a.m. to 4
p.m.
D. It doesn’t make any difference