f(x)'=2(x-3)^(2/3)+(2x-1)*(2/3)*(x-3)^(-1/3)
=(10/3)*(x-3)^(-1/3)*(x-2)
求得间断点为x=3,驻点为x=2.
当x=3的时候,f(x)=0
当x>3或者x