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.

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 $\min(f(1),f(3)).$
