Question

来道数学题.设函数f（x）=2x^2-4ax+2a^2+a的定义域为[x0,+∞）,值域为[yo,+∞）.若对任意实数a

Answer 1

f(x)=2x^2-4ax+2a^2+a=2[(x-a)^2]+a

1)x0a时,y0=f(x0)=2(x0)^2-4a(x0)+2a^2+a>=3

转换为a的函数,g(a)=2a^2+a-4a(x0)+2(x0)^2-3