题目内容
已知函数
(1)证明方程f(x)=0在区间(0,2)内有实数解;
(2)使用二分法,取区间的中点三次,指出方程f(x)=0(x∈[0,2])的实数解x在哪个较小的区间内.
【答案】分析:(1)通过计算函数值,得f(0)•f(2)=-
<0,由零点存在性定理可得方程f(x)=0在区间(0,2)内有实数解;
(2)根据零点存在性定理,依次取x1=1,x2=
,x3=
,从而计算出f(
)•f(
)<0,得区间(
,
)即为符合题意的较小区间.
解答:解:(1)∵f(0)=1>0,f(2)=-
<0
∴f(0)•f(2)=-
<0,
由函数的零点存在性定理可得方程f(x)=0在区间(0,2)内有实数解;
(2)取x1=
(0+2)=1,得f(1)=
>0
由此可得f(1)•f(2)=-
<0,下一个有解区间为(1,2)
再取x2=
(1+2)=
,得f(
)=-
<0
∴f(1)•f(
)=-
<0,下一个有解区间为(1,
)
再取x3=
(1+
)=
,得f(
)=
>0
∴f(
)•f(
)<0,下一个有解区间为(
,
)
综上所述,得所求的实数解x在区间(
,
).
点评:本题给出三次多项式函数,求函数的零点所在的区间,着重考查了三次多项式函数的性质和零点存在性定理等知识,属于基础题.

(2)根据零点存在性定理,依次取x1=1,x2=






解答:解:(1)∵f(0)=1>0,f(2)=-

∴f(0)•f(2)=-

由函数的零点存在性定理可得方程f(x)=0在区间(0,2)内有实数解;
(2)取x1=


由此可得f(1)•f(2)=-

再取x2=




∴f(1)•f(



再取x3=





∴f(




综上所述,得所求的实数解x在区间(


点评:本题给出三次多项式函数,求函数的零点所在的区间,着重考查了三次多项式函数的性质和零点存在性定理等知识,属于基础题.

练习册系列答案
相关题目