Question

已知函数f(x)=(x^2+b)分之ax在x=1处取得极值2.

已知函数f(x)=(x^2+b)分之ax在x=1处取得极值2.
(1)求函数f(x)的解析式；
（2）实数k满足什么条件时，函数f(x)在区间（2k，4k+1）上单调递增？

Answer 1

1)f'(x)=[a(x^2+b)-2ax^2]/[(x^2+b)^2]

由题意f(1)=2,f'(1)=0

有 a/(1+b)=2 (1)

(-a+ab)/[(1+b)^2]=0 (2)

由(1)(2)有a=4,b=1

f(x)=4x/(x^2+1)

（2）
f'(x)=4(1-x^2)/(x^2+1)^2,令f'(x)=0得x=1或-1
当f'(x)>0，即-1<x<1时，
f(x)单调递增，
若使区间（2k，4k+1）为函数f(x)的单调增区间
则必须有2k>-1,4k+1<1即-1/2<k<0;
所以k的条件是-1/2<k<0。

Answer 2

f(x) = ax/(x^2+b)
f'(x) = [ax(2x) - a(x^+b)]/ (x^+b)^2 =0
f'(1) = (2a-a(1+b)]/(1+b)^2 =0
=> 2a - a -b =0
a = b
f(1) = a/(1+b)
= a/(1+a) = 2
a = 2+2a
a = -2
b = -2
f(x) = -2x/ (x^2 - 2)
f'(x) >0
=> -4x^2 - 2(x^+2) > 0
=> 6x^2 - 4 < 0
x^2 - (2/3) <0
-√6/3 <x < √6/3
ie 2k> -√6/3 and 4k+1< √6/3
=> k > -√6/6 and k < (√6-3)/12
=> -√6/6 < k< (√6-3)/12