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.
 A: 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 $\min(f(1),f(3)).$
A: 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 $0$.
A: 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 $f(3)$).
A: 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 $[f(3),f(5/7)]$.
