Question

已知f(x)=log3(x^2-4mx+4m^2+m+1/m-1)求函数f(x)的最小值

Answer 1

x^2-4mx+4m^2+m+1/m-1 = (x - 2m)^2 + m + 1/(m-1) >= m + 1/(m-1) 当 x = 2m时,取得最小值,m + 1/(m-1) = (m^2 - m + 1)/(m-1) 所以 f(x)在x=2m处取得最小值,log(3)(m^2 - m + 1)/(m-1)