求y=x+(1/x+1)的值域
1个回答

当 x < -1 时,x+1 < 0 ,

因此 y = x+1/(x+1) = (x+1)+1/(x+1) -1 = -[-(x+1)-1/(x+1)] -1 ≤ -2*√[(x+1)*1/(x+1)] -1 = -3 ;

当 x > -1 时,x+1 > 0 ,

因此 y = x+1/(x+1) = (x+1)+1/(x+1) -1 ≥ 2*√[(x+1)*1/(x+1)] -1 = 1 ,

所以函数值域为(-∞,-3] U [1,+∞).