求值域 ① y=(x²—1)/(x²+1)② y=(x+3)/(x—1)① y=(x²—1)/(x²+1)② y=(x+3)/(x—1)

来源:学生作业帮助网 编辑:作业帮 时间:2024/04/28 23:10:18

求值域 ① y=(x²—1)/(x²+1)② y=(x+3)/(x—1)① y=(x²—1)/(x²+1)② y=(x+3)/(x—1)
求值域 ① y=(x²—1)/(x²+1)② y=(x+3)/(x—1)
① y=(x²—1)/(x²+1)
② y=(x+3)/(x—1)

求值域 ① y=(x²—1)/(x²+1)② y=(x+3)/(x—1)① y=(x²—1)/(x²+1)② y=(x+3)/(x—1)
答:
(1)
y=(x^2-1)/(x^2+1)
=(x^2+1-2)/(x^2+1)
=1-2/(x^2+1)
因为:x^2+1>=1
所以:0

答:(1)y=(x^2-1)/(x^2+1)=(x^2+1-2)/(x^2+1)=1-2/(x^2+1)因为:x^2+1>=1所以:0<2/(x^2+1)<=2所以:1-2<=y<1-0所以:-1<=y<1所以:值域为[-1,1)(2)y=(x+3)/(x-1)=(x-1+4)/(x-1)=1+4/(...

全部展开

答:(1)y=(x^2-1)/(x^2+1)=(x^2+1-2)/(x^2+1)=1-2/(x^2+1)因为:x^2+1>=1所以:0<2/(x^2+1)<=2所以:1-2<=y<1-0所以:-1<=y<1所以:值域为[-1,1)(2)y=(x+3)/(x-1)=(x-1+4)/(x-1)=1+4/(x-1)所以:y≠1所以:值域为(-∞,1)∪(1,+∞)

收起