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

Nov 4, 2015

The domain is easy, as the square makes everything under the root-sign non-negative, so there are no restrictions on $x$.
In other words domain $- \infty < x < + \infty$
Since ${x}^{2} \ge 0 \to {x}^{2} + 4 \ge 4 \to \sqrt{{x}^{2} + 4} \ge 2$
In other words range $2 \le f \left(x\right) < + \infty$