What is the domain and range of #y= sqrt (x^2 + 1)#?

1 Answer
Apr 4, 2018

Answer:

Domain : #RR#
Range:#[1;+oo[#

Explanation:

Let's first search the domain. What we know about square root is that inside have to be a positive number.
So : #x²+1>=0#
#x²>=-1#
We also know that #x²>=0#, so #x# can take every values in #RR#.
Let's find the range Now !
We know that x² is a positive or null value, so the minimum is for f(0).
#f(0)=sqrt(1+0)=1#
So the minimum is 1. And because x² is divergent, there's no limits.
So the range is : #[1;+oo[#