How to find the range of the function: $f(x) = \sqrt{x-1}+2\sqrt{3-x}$

We get $$f^\prime (x)=\frac{1}{2\sqrt{x-1}}-\frac{2}{2\sqrt{3-x}}=\frac{\sqrt{3-x}-2\sqrt{x-1}}{2\sqrt{(x-1)(3-x)}}.$$

So, $$f^\prime(x)\ge0\iff \sqrt{3-x}\ge2\sqrt{x-1}\iff 3-x\ge4(x-1)\iff x\le\frac{7}{5}.$$ Now we know that $f(x)$ is increasing in $1\le x\lt 7/5$ and that $f(x)$ is decreasing in $7/5\lt x\le 3$.

So, we know that the max is $f(7/5)$, and that the min is $\min(f(1),f(3)).$

This function is only defined on $[1,3]$. We can differentiate it, and search for points where the derivative is $0$. Since it is a continuous (on $[1,3]$) and differentiable (on $(1,3)$) function, extreme values must happen either at $1$, $3$ or at such points with derivative $0$.

Since the function is continuous, the image of the closed interval $[1, 3]$, i.e. $f([1, 3])$ is also a closed interval.

Since the function is also differentiable, we can find the maximum and minimum values (which the function attains) by comparing all the critical points (set $f'(x) = 0$) and the end points ($f(1)$ and $f(3)$).