设1>=x2>x1>=-1,则-1<=-x1<=1.所以:x2+(-x1)>0.由[f(x2)+f(-x1)]/[x2+(-x1)]>0.得f(x2)+f(-x1)>0.又由f(x)为奇函数. 则f(x2)-f(-x1)=f(x2)+f(-x1)>0所以f(x)为增函数由1小题增函数,可得-1<=x+1/2<1/x+1<=1. 展开
设1>=x2>x1>=-1,则-1<=-x1<=1.所以:x2+(-x1)>0.由[f(x2)+f(-x1)]/[x2+(-x1)]>0.得f(x2)+f(-x1)>0.又由f(x)为奇函数. 则f(x2)-f(-x1)=f(x2)+f(-x1)>0所以f(x)为增函数由1小题增函数,可得-1<=x+1/2<1/x+1<=1.解此不等式即有-3/2<=x<0 收起
