Answer
The domain of $ f(x)=\sqrt{x^2-x+5}$ is $ R $ and the range of $ f(x)$ is $ [\frac{19}{2},\infty)$.
Work Step by Step
Since $ x^2-x+5=(x-\frac{1}{2})^2+\frac{19}{4}$, then
$ x^2-x+5\gt 0$ for all $ x\in R $, and the domain of $ f(x)=\sqrt{x^2-x+5}$ is $ R $. The minimum value is at $ x=1/2$, where $y=\sqrt{\dfrac{19}{4}}$, hence the range of $ f(x)$ is $ [\frac{\sqrt{19}}{2},\infty)$.
