已知函数f(x)=(x-1)/(x^2+1),求证：-（根号2+1）/2〈=f(x)<=(根号2-1)/2
热心网友
由y=(x-1)/(x^2+1)，得yx^2-x+y+1=0方程关于x有解，所以△=1-4y(y+1)≥0y^2+y≤1/4[y+(1/2)]^2≤1/2|y+(1/2)|≤√2/2-√2/2≤y+(1/2)|≤√2/2-(√2+1)/2≤y≤(√2-1)/2既：-(√2+1)/2≤f(x)≤(√2-1)/2
热心网友
对
热心网友
已知函数f(x)=(x-1)/(x^2+1),求证：-（根号2+1）/2〈=f(x)<=(根号2-1)/2 解：令f(x)=(x-1)/(x^2+1)=y则yx^2-x+y+1=０x必定有实跟，故△＞＝０得：-（根号2+1）/2〈=f(x)<=(根号2-1)/2