设f(x)=(x-1)(x-2)(x-3)(x-4)(x-5),则f’(x)=0在[2,5]内有实根个数为

来源:学生作业帮助网 编辑:六六作业网 时间:2024/05/19 04:03:57
设f(x)=(x-1)(x-2)(x-3)(x-4)(x-5),则f’(x)=0在[2,5]内有实根个数为设f(x)=(x-1)(x-2)(x-3)(x-4)(x-5),则f’(x)=0在[2,5]内

设f(x)=(x-1)(x-2)(x-3)(x-4)(x-5),则f’(x)=0在[2,5]内有实根个数为
设f(x)=(x-1)(x-2)(x-3)(x-4)(x-5),则f’(x)=0在[2,5]内有实根个数为

设f(x)=(x-1)(x-2)(x-3)(x-4)(x-5),则f’(x)=0在[2,5]内有实根个数为
3个.手机不便祥解