已知x>2,求函数y=(2x^2-8x+16)/(x^2-2x+4)的值域.用均值不等式解答,

问题描述:

已知x>2,求函数y=(2x^2-8x+16)/(x^2-2x+4)的值域.
用均值不等式解答,

y=(2x^2-8x+16)/(x^2-2x+4) =2+(8-4x)/(x^2-2x+4) =2+4/[(x^2-2x+4)/(2-x)] =2-4/{[x(x-2)+4]/(x-2)} =2-4/[x+4/(x-2)] =2-4/[x-2+2+4/(x-2)] =2-4/[x-2+4/(x-2)+2] >=2-4/{[(2√(x-2)*4/(x-2)]+2} =4/3故...