The range of the real valued function $f\left(x\right)=\sqrt{\frac{x^2+2x+8}{x^2+2x+4}}$

$f\left(x\right)=\sqrt{\frac{x^2+2x+8}{x^2+2x+4}}$

Let $y=\frac{x^2+2x+8}{x^2+2x+4}$

$\Rightarrow y{x}^2+2xy+4y=x^2+2x+8$

$\Rightarrow \left(y-1\right)x^2+2x\left(y-1\right)+\left(4y-8\right)=0$

$\forall x\in R,\Delta \ge 0$

${\left(2\left(y-1\right)\right)}^2-4\left(y-1\right)\left(4y-8\right)\ge 0$

$\left(y-1\right)\left(-3y+7\right)\ge 0$

$y=1,\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$

$y\in (1,\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$3$}\right.]$

$(\because At y=1 x^2+2x+4\ne x^2+2x+8)$

$\therefore y\in (1,\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$3$}\right.]$

$f\left(x\right)\in (1,\sqrt{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}]$