Question

Range of the function f(x) =$\sqrt{x^2+x+1}$ is equal to

Answer 1

$\begin{array}{c}f\left(x\right)=\sqrt{x^2+x+1}\\ =\sqrt{{\left(x+\frac{1}{2}\right)}^2+\frac{3}{4}}\\ clearlyrangeoff\left(x\right)=[\frac{\sqrt{3}}{2},\infty )\\ minimunoccursatx=\frac{-1}{2}\end{array}$