Question

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

Answer 1

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[`