Question

求函数f(x)=x-x三次方单调区间

详细过程，快

Answer 1

f'(x)=1-3x^2
领f'(x)=0,得,x1=根号1/3,x2=负的根号1/3;
可知,当x<=x2时,f'(x)<0;
x>=x1时,
f'(x)<0;
x2<x<x1时,f'(x)>0;
所以,x<=x2时,f(x)单调递减;
x>=x1时,
f(x)单调递减;
x2<x<x1时,f(x)单调递增;
就是这样