# How do you find the domain and range of f(x)= sqrt( x^2+4)?

May 16, 2018

$x$ is squared, so any negative x-values will result in a positive value for ${x}^{2}$. The domain is not restricted for x = a positive number or x = 0 since the expression under the radical will always be positive.
For the range, ${x}^{2}$ takes its minimum at $x = 0$, so plugging in $x = 0$ gives $y = \sqrt{{x}^{2} + 4} = \sqrt{0 + 4} = 2$ therefore making the range [2,∞).