函数f(x)=(x-5)x^(2/3)的拐点是多少,
问题描述：
函数f(x)=(x-5)x^(2/3)的拐点是多少,
答
f(x)=x^(5/3)-5x^(2/3)f'(x)=(5/3)x^(2/3)-5*(2/3)x^(-1/3)f''(x)=(5/3)(2/3)x^(-1/3)-5(2/3)(-1/3)x^(-4/3)=(10/9)x^(-1/3)+(10/9)x^(-4/3)=0令t=x^(-1/3),则有t+t^4=0,即t(1+t^3)=0∵t≠0（1除以三次根号下x 肯定...