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

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

$=\sqrt{1+\frac{4}{x^2+2x+4}}=\sqrt{1+\frac{4}{{\left(x+1\right)}^2+3}}$

We know, $3\le {\left(x+1\right)}^2+3<\infty$

$\Rightarrow 0<\frac{4}{{\left(x+1\right)}^2+3}\le \frac{4}{3}\Rightarrow 1<1+\frac{4}{{\left(x+1\right)}^2+3}\le \frac{7}{3}$

$\Rightarrow 1<\sqrt{1+\frac{4}{{\left(x+1\right)}^2+3}}\le \sqrt{\frac{7}{3}}$

Hence, range of $f\left(x\right)$ will be $(1,\sqrt{\frac{7}{3}}]$