函数f(x)=ax3+bx2+cx+d的部分函数值如下表,则函数y=lgf(x)的定义域为_____

函数f(x)=ax3+bx2+cx+d的部分函数值如下表,则函数y=lgf(x)的定义域为______________.

x

-

-2

-1

0

1

2

3

4

F(x)

-80

-24

0

4

0

0

16

60

答案

答案:

(-1,1)∪(2,+∞)  【解析】本题考查函数与不等式的综合应用;据题意可知x=1,-1,2是函数的三个零点,故f(x)=a(x-1)(x+1)(x-2),又由f(0)=4可得a=2,故

f(x)=2(x-1)(x+1)(x-2),由函数y=lgf(x)的定义域满足f(x)=2(x-1)(x+1)(x-2)>0,利用穿根引线法解不等式即得其解集为(-1,1)∪(2,+∞),即为所求函数的定义域.

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