题目内容
求下列函数的零点:
(1)y=x3-7x+6;(2)y=x+
-3.
(1)y=x3-7x+6;(2)y=x+

(1)函数y=x3-7x+6的零点为-3,1,2(2)函数y=x+
-3的零点为1,2

(1)∵x3-7x+6=(x3-x)-(6x-6)
=x(x2-1)-6(x-1)=x(x+1)(x-1)-6(x-1)
=(x-1)(x2+x-6)=(x-1)(x-2)(x+3)
解x3-7x+6=0,即(x-1)(x-2)(x+3)=0
可得x1=-3,x2=1,x3=2.
∴函数y=x3-7x+6的零点为-3,1,2.
(2)∵x+
解x+
即
=0,可得x=1或x=2.
∴函数y=x+
-3的零点为1,2.
=x(x2-1)-6(x-1)=x(x+1)(x-1)-6(x-1)
=(x-1)(x2+x-6)=(x-1)(x-2)(x+3)
解x3-7x+6=0,即(x-1)(x-2)(x+3)=0
可得x1=-3,x2=1,x3=2.
∴函数y=x3-7x+6的零点为-3,1,2.
(2)∵x+

解x+


∴函数y=x+


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