作业帮函数零点个数判断
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/11/05 18:42:57
函数零点个数判断
先看在（0,4）区间使得f(x)=0时x=1,x=3满足要求；即f(1)=f(3)=0
f(4)=f(2+2)=f(3-2)=f(1)=0;f(2)=f(0+2)=f(3-0)=f(3)=0;f(5)=f(-2+7)=f(2+2)=f(4)=0
f(6)=f(-1+7)=f(2+1)=f(3)=0;f(7)=f(2)=0,f(8)=f(1)=0;f(9)=f(0)≠0;f(10)=f(-1)=f(6)=0.
f(11)=f(-2)=f(7)=0;f(12)=f(-3)=f(8)=0;f(13)=f(-4)=f(9)≠0;f(14)=f(-5)=f(10)=0;f(15)=f(-6)=f(11)=0;f(16)=f(-7)=f(12)=0;f(17)=f(-8)=f(13)≠0.同理可得f(9)=f(13)=f(17)=f(23)=f(27)=f(31)=f(35)=f(39)=f(43)=f(47)≠0即1-49之间有10个不为零点则49-10=39选C
f(4)=f(2+2)=f(3-2)=f(1)=0;f(2)=f(0+2)=f(3-0)=f(3)=0;f(5)=f(-2+7)=f(2+2)=f(4)=0
f(6)=f(-1+7)=f(2+1)=f(3)=0;f(7)=f(2)=0,f(8)=f(1)=0;f(9)=f(0)≠0;f(10)=f(-1)=f(6)=0.
f(11)=f(-2)=f(7)=0;f(12)=f(-3)=f(8)=0;f(13)=f(-4)=f(9)≠0;f(14)=f(-5)=f(10)=0;f(15)=f(-6)=f(11)=0;f(16)=f(-7)=f(12)=0;f(17)=f(-8)=f(13)≠0.同理可得f(9)=f(13)=f(17)=f(23)=f(27)=f(31)=f(35)=f(39)=f(43)=f(47)≠0即1-49之间有10个不为零点则49-10=39选C