If $a<b$  then $f\left(x\right)=\sqrt{\frac{x-b}{a-x}}$ is continuous on

$f$ is defined

If $\frac{x-b}{a-x}\ge 0$ and $a-x\ne 0$

$\Rightarrow \left(x-b\right)\left(a-x\right)\ge 0$  and $a-x\ne 0$

$\Rightarrow \left(x-b\right)\left(x-a\right)\le 0$ and $x\ne a$

$\Rightarrow x\in \left[a,b\right]$ and $x\ne a$

$\Rightarrow x\in \left(a,b\right]$