是“f(x)=(ax+b)/(x²+1)”吧?
1、
∵f(x)是奇函數(shù)
∴f(0)=b/1=b=0
∴f(1/2)=(a/2)/(5/4)=2/5
∴a=1
∴f(x)=x/(x²+1)
2、
證明：
設(shè)：-1＜x1＜x2＜1
f(x1)-f(x2)
=x1/(x1²+1) - x2/(x2²+1)
=[x1(x2²+1) - x2(x1²+1)] / [(x1²+1)(x2²+1)]
=(x1x2²+x1-x2x1²-x2) / [(x1²+1)(x2²+1)]
=[x1x2(x2-x1)-(x2-x1)] / [(x1²+1)(x2²+1)]
=[(x1x2-1)(x2-x1)] / [(x1²+1)(x2²+1)]
∵-1＜x1＜x2＜1,∴x1x2＜1
∴x1x2-1＜0,x2-x1＞0,x1²+1＞0,x2²+1＞0
∴f(x1)-f(x2)=[(x1x2-1)(x2-x1)] / [(x1²+1)(x2²+1)] ＜0
∴f(x1)＜f(x2)
∴f(x)在(-1,1)上單調(diào)遞增.
3、個人感覺第三題要加一個“t-1,t∈(-1,1)”
∵t-1,t∈(-1,1),
∴t∈(0,1)
∵f(x)是奇函數(shù)
∴-f(t)=f(-t)
∴f(t-1)+f(t)