Question

已知函數(shù)f(x)=(x-a)2(x-b)(a,b∈R,a<b).

(1)當(dāng)a=1,b=2時(shí),求曲線y=f(x)在點(diǎn)(2,f(2))處的切線方程;

(2)設(shè)x1,x2是f(x)的兩個(gè)極值點(diǎn),x3是f(x)的一個(gè)零點(diǎn),且x3≠x1,x3≠x2.證明:存在實(shí)數(shù)x4,使得x1,x2,x3,x4按某種順序排列后構(gòu)成等差數(shù)列,并求x4.

Answer 1

 (1)解:當(dāng)a=1,b=2時(shí),f(x)=(x-1)2(x-2),

f′(x)= (x-1)(3x-5),

故f′(2)=1.

又f(2)=0,

所以f(x)在點(diǎn)(2,0)處的切線方程為y=x-2.

(2)證明:由題意得f′(x)=3(x-a)(x-),

由于a<b且a,b∈R,故a<,

所以f(x)的兩個(gè)極值點(diǎn)為x=a,x=.

不妨設(shè)x1=a,x2=,

因?yàn)閤3≠x1,x3≠x2,

且x3是f(x)的零點(diǎn),

故x3=b.

又因?yàn)?img src='http://thumb.zyjl.cn/pic1/files/down/test/2014/09/09/13/2014090913333459467021.files/image003.gif'>-a=2(b-),

x4=(a+)=,

此時(shí)a, ,,b依次成等差數(shù)列,

所以存在實(shí)數(shù)x4滿足題意,且x4=.