已知x>2,则函数f(x)=2x^2/x-2的最小值

f(x)=2x^2/(x-2)
f'(x) = {(x-2)*4x-2x^2*1}/(x-2)^2 = 2x(x-4)/(x-2)^2
当2＜x＜4时,f'(x)＜0,单调减；
当x≥4时,f'(x)≥0,单调增
x=4时有最小值：f(4)=2*4^2/(4-2)=16