已知a>0,0已知a>0,0<=x<=1,求y=-x^+2ax的值域
已知a>0,0y=f(x)=-(x-a)^2+a^20y max=f(a)=a^2y min=f(1)=2a-1值域[a^2,2a-1]1/2<=a<1y max=f(a)=a^2y min=f(0)=0值域[0,a^2]a>=1y max=f(1)=2a-1y min=f(0)=0值域[2a-1,0]