<s>[INST] Find the range of $f(x) =(x-1)^{1/2} + 2\cdot(3-x)^{1/2}$ How to take out the range of the following function : 
$$f(x) =(x-1)^{1/2} + 2\cdot(3-x)^{1/2}$$
I am new to functions hence couldn't come up with a solution.
 [/INST] Your function
$$f(x) =\sqrt{x-1} + 2\sqrt{3-x}$$
is defined for all $x \in [1, 3]$, otherwise the square root are not defined for real number. 
We can check what happens on the border of the segment $[1, 3]$:


*

*$f(1) = 2\sqrt{2}$;

*$f(3) = \sqrt{2}$.


Also, we can find the local maxima and minima of $x$ in the segment $[1, 3]$. In order to get them, let's evaluate the first derivative of $f(x)$:
$$\frac{\text{d} f(x)}{\text{d} x} = -\frac{1}{2\sqrt{x-1}} + \frac{1}{\sqrt{3-x}}$$
and find the values of $x$ such that $\frac{\text{d} f(x)}{\text{d} x} = 0$. Thus:
$$-\frac{1}{2\sqrt{x-1}} + \frac{1}{\sqrt{3-x}} = 0 \Rightarrow 2\sqrt{x-1}=\sqrt{3-x}.$$
Since both $x-1$ and $3-x$ are non-negative in the segment $[1, 3]$, we can elevate to the power of $2$ both members without particular problems:
$$4(x-1)=3-x \Rightarrow 4x+x = 3 + 4 \Rightarrow x = \frac{7}{5} \in [1, 3].$$
$x = \frac{7}{5}$ is a local maximum or minimum and $f\left(\frac{7}{5}\right) = \sqrt{10}.$
Summarizing:


*

*At the begining of the segment $[1, 3]$ the function assumes value $2\sqrt{2}$;

*at the end, the values is $2\sqrt{2}$;

*somewhere inside the segment (i.e. when $x = \frac{7}{5}$) the function has value $\sqrt{10}$.


It's clear now that the minimum value of $f(x)$ in $[1,3]$ is $\sqrt{2}$ while the maximum is $\sqrt{10}$.
We can conclude that the range of $f(x)$ is $[\sqrt{2}, \sqrt{10}]$. 
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