The range of the function $f\left(x\right)=\sqrt{(x-1)(3-x)}$ is :

Let $y=\sqrt{(x-1)(3-x)}$

by squaring on both sides, we get

$\begin{array}{r}{y}^2=\left(3x-3-x^2+x\right)\\ \Rightarrow x^2-4x+y^2+3=0\end{array}$

Thus, $x=\frac{4\pm \sqrt{16-4y^2-12}}{2}$$=2\pm \sqrt{1-y^2}$

Now for x to be defined

$\begin{array}{l}1-{\mathrm{y}}^2\ge 0\\ \Rightarrow -1\le \mathrm{y}\le 1\end{array}$

Thus the range of the given function is $[-1,1]$