## Calculus (3rd Edition)

The domain of $f(x)=\sqrt{x^2-x+5}$ is $R$ and the range of $f(x)$ is $[\frac{19}{2},\infty)$.
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 $[\frac{\sqrt{19}}{2},\infty)$.