# What is the domain and range of f(x)=sqrt(x^2 +4)?

##### 1 Answer
Apr 17, 2015

$f \left(x\right) = \sqrt{{x}^{2} + 4}$ is defined for all Real values of $x$
The Domain is $x \epsilon \mathbb{R}$
(actually $f \left(x\right)$ is valid for $x \epsilon \mathbb{C}$ but I will assume we are not interested in Complex numbers).

If we restrict $x \epsilon \mathbb{R}$
then $f \left(x\right)$ has a minimum value when $x = 0$ of
$\sqrt{{0}^{2} + 4} = 2$
and the Range of $f \left(x\right)$ is $\left[2 , + \infty\right)$

(If we allow $x \epsilon \mathbb{C}$ the Range of $f \left(x\right)$ becomes all of $\mathbb{C}$)