Question

已知函数f(x)=ax/(x^2+b),在x=1处取得极值2.

Answer 1

(1):

f'(x)=a(b-x^2)/(x^2+b)^2,因为在x=1时取极值2,所以f'(1)=0;

由于分母不能为零,所以分子为零,即b-x^2=b-1=0,即b=1.

f(1)=a/(1+1)=2,所以a=4;

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