# Chapter 1 - Precalculus Review - Chapter Review Exercises - Page 36: 14

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 $[\frac{\sqrt{19}}{2},\infty)$.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.