Question

已知f(x)=x+a/x^2+bx+1是奇函数，且x∈【-1,1】，试判断它的单调性，并给予证明

要的是计算过程....文字解释的死一边去....

Answer 1

f(x)=(x+a)/(x^2+bx+1)
f(0)=(0+a)/(0+0+1)=a=0
f(-x)=-f(x)
x/(x^2+bx+1)=-(-x)/[(-x)^2+b(-x)+1]=x/(x^2-bx+1)
x/(x^2+bx+1)=x/(x^2-bx+1)
x^2+bx+1=x^2-bx+1
bx=-bx
2bx=0
x∈【-1,1】，b=0
f(x)=x/(x^2+1)
任取1>x2>x2>-1
x2-x1>0 -1< x2x1<1

f(x2)-f(x1)=x2/(x2^2+1)-x1/(x1^2+1)
=[x2(x1^2+1)-x1(x2^2+1)]/(x2^2+1)(x1^2+1)
=(x2x1^2+x2-x1x2^2-x1)/(x2^2+1)(x1^2+1)
=(x2-x1)(1-x1x2)/(x2^2+1)(x1^2+1)>0
所以是增函数

Answer 2

f(x)=x+a/x^2+bx+1是奇函数f(0)=0 a=0 , f(x)=-f(-x)
b=0
f(x)=x