How to find the range of the function: $f(x) = \sqrt{x-1}+2\sqrt{3-x}$ Problem : 
Find the range  of the function: $f(x) = \sqrt{x-1}+2\sqrt{3-x}$
Solution  : 
Domain of this function can be determined as : 
$x - 1 >0 ; 3-x >0 \Rightarrow x >0 ; x <3 ;$ 
$\therefore $ domain of $x \in [1,3]$ 
Now if I put the values of this domain in my function then it gives the following values : 
at 1 ;  the value of the function is $2\sqrt{2}$ 
at 2 : the value of the function is $ 1+2 = 3$ 
at 3 : the value of the function is $2$ 
Can we say that the maximum value of the function is  3 and minimum value of the function is 2;
Therefore the range of this function is [2,3] but this answer is wrong. please suggest..
Also suggest that how can we use differentiation method to find the range... thanks.

I try to give another solution without using derivative. Here, we only use the intermediate value theorem (https://en.wikipedia.org/wiki/Intermediate_value_theorem) for continuous functions and Cauchy Schwarz inequality (https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality).
Clearly, the domain of the function is [1;3]. For every $x\in[1,3]$, we have
\begin{align}
f(x)-f(3)&=\sqrt{x-1}-\sqrt{2}+2\sqrt{3-x}\\
&=\frac{x-3}{\sqrt{x-1}+\sqrt{2}}+2\sqrt{3-x}\\
&=\frac{\sqrt{3-x}(2\sqrt{x-1}+2\sqrt{2}-\sqrt{3-x})}{\sqrt{x-1}+\sqrt{2}}\\
&\geq 0.
\end{align}
By the Cauchy-Schwarz inequality we have
\begin{align}
f(x)-f(7/5)&\leq \sqrt{(1+4)(x-1+3-x)}-(\sqrt{2}/\sqrt{5}+2\sqrt{8}/\sqrt{5})=0
\end{align}
Hence, $f([1,3])\subset [f(3),f(5/7)]$. On the other hand, by the intermediate value theorem, we have
$$
f([1,3])\supset f([7/5,3])\supset [f(3),f(7/5)].
$$
The range of $f$ is $[f(3),f(5/7)]$.
