求y=(x^2+3x+3)/(x+1) 且(x大于-1)的值域.
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/06 20:58:21
![求y=(x^2+3x+3)/(x+1) 且(x大于-1)的值域.](/uploads/image/z/14572137-57-7.jpg?t=%E6%B1%82y%3D%EF%BC%88x%5E2%2B3x%2B3%EF%BC%89%2F%EF%BC%88x%2B1%EF%BC%89+%E4%B8%94%EF%BC%88x%E5%A4%A7%E4%BA%8E-1%EF%BC%89%E7%9A%84%E5%80%BC%E5%9F%9F.)
求y=(x^2+3x+3)/(x+1) 且(x大于-1)的值域.
求y=(x^2+3x+3)/(x+1) 且(x大于-1)的值域.
求y=(x^2+3x+3)/(x+1) 且(x大于-1)的值域.
y=(x²+3x+3)/(x+1) (x+1)>0
=[(x²+2x+1)+(x+1)+1]/(x+1)
=[(x+1)²+(x+1)+1]/(x+1)
=x+1+1/(x+1)+1
>=2√(x+1)[1/(x+1)]+1
=3
所以y>=3
值域[3,+∞)