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

Apr 4, 2018

Domain : $\mathbb{R}$
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 $\mathbb{R}$.
Let's find the range Now !
We know that x² is a positive or null value, so the minimum is for f(0).
$f \left(0\right) = \sqrt{1 + 0} = 1$
So the minimum is 1. And because x² is divergent, there's no limits.
So the range is : [1;+oo[