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?
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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?
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?
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?
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
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
...
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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
=> k > -√6/6 and k < (√6-3)/12
=> -√6/6 < k< (√6-3)/12
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