About the solution to "Finding the range of $y= \sqrt x + \sqrt{3-x}"$ I was reading the solution of "Find the range of $y = \sqrt{x} + \sqrt{3 -x}$" and I had some points of confusion about the solution posted in the OP.
I wrote here my interpretation of the solution. Could you please let me know if my understanding contains any redundant, incomplete, missing, or erroneous steps?
My understanding of the solution is as follows:
The domain is clearly $[0, 3]$
$\{(x, y):y=\sqrt x + \sqrt {(3-x)} \}=\{(x, y):y^2-3=2\sqrt{x(3-x)}$ and $y \ge 0\}$
Since the RHS in $y^2-3=2\sqrt{x(3-x)}$ is non-negative $\forall x$ in the domain, it is implied that $y^2-3 \ge 0$ $\forall y$ in the solution set. So we can now say that
$\{(x, y):y=\sqrt x + \sqrt {(3-x)} \}=\{(x, y):y^2-3=2\sqrt{x(3-x)}$ and $y \ge \sqrt3\}$ 
(We do not have to worry about $y \le -\sqrt 3$ because $y \ge 0 \forall x$ in the domain). Now $\sqrt 3$ is just a "boundary" for $y$; we need to check that the $y=\sqrt 3$ is actually a part of the range. Because $2\sqrt{x(3-x)}$ attains the value of $0$ at $x=0$ and $x=3$, $y$ actually attains the value of $\sqrt 3$.Therefore we can establish a minimum of $\sqrt 3$ for the range.
Now, we know that 
$\{(x, y):y^2-3=2\sqrt{x(3-x)}$ and $y \ge \sqrt3\}=\{(x, y):4(3x-x^2)=(y^2-3)^2$ and $y \ge \sqrt3\} = \{(x, y): 4x^2-12x+(y^2-3)^2=0$ and $y \ge \sqrt 3 \}$ (We do not need to worry about explicitly restricting $x$ in the set on the set in the middle and on the right, because the interval over which $4(3x-x^2)$ is positive coincides with the domain of the original problem, so we have not introduced/lost any solutions.)  
$4x^2-12x+(y^2-3)^2=0$ has real solutions $\iff$ the discriminant is $\ge 0$. 
$(-12)^2-4(4)(y^2-3)^2 \ge 0$
$144-16(y-3)^2 \ge 0$
$9 \ge (y^2-3)^2$
Because we already established $y \ge \sqrt 3$, we only need to worry about the positive root
$3 \ge y^2-3$
$6 \ge y^2$
Again, since we established $y \ge \sqrt 3$, we can take
$\sqrt 6 \ge y$ 
$\sqrt 6$ is only a "boundary" for $y$; we still need to establish that $y=\sqrt 6$ is part of at least one ordered pair in the solution set:
$(\sqrt 6)^2-3 = 2 \sqrt {x(3-x)}$
$ 3 =   2\sqrt {x(3-x)}$
This gives the solution $\dfrac 32$, implying that $y=\sqrt 6$ is contained in the range.
We established a minimum value of $\sqrt 3$ and a maximum value of $\sqrt 6$, both on the interval $[0, 3]$/ Because the original function is continuous on $[0, 3]$, we can safely conclude that the range is $[\sqrt 3, \sqrt 6].$
 A: (too long for a comment)

I will have to look at the continuity later; but on the issue of checking that $\sqrt 3$ and $\sqrt 6$; is it necessary? 

I think checking that both $\sqrt 3$ and $\sqrt 6$ are in the range is necessary in your "solution" because you use continuity.

My reasoning is that these values are in a sense "necessary but not sufficient", meaning that we know the range must be somewhere between $\sqrt 3$ and $\sqrt 6$, but we don't know it's precisely $[\sqrt 3,\sqrt 6]$

I understand what you mean, and I found nothing wrong in your "solution". 
By the way, what I was trying to say in the linked solution is not the same as your understanding, and I think it's worth showing with details my proof without using continuity (hence no need to check "the boundary") : 
First of all, to say that the range of $y=\sqrt x+\sqrt{3-x}$ is $\sqrt 3\le y\le\sqrt 6$, we have to prove the following two things :
(1) (necessity) If there exists a pair of real numbers $(x,y)$ such that $y=\sqrt x+\sqrt{3-x}$, then $\sqrt 3\le y\le \sqrt 6$.
(2) (sufficiency) For every $y$ satisfying $\sqrt 3\le y\le\sqrt 6$, there exists at least one real $x$ such that $y=\sqrt x+\sqrt{3-x}$.
To make it easy to understand the proof, let us prove $(1)$ and $(2)$ separately.
Proof for (1) (necessity) : 
We consider $$y=\sqrt x+\sqrt{3-x}\tag3$$
First of all, we have to have $$0\le x\le 3\tag4$$
If $y\lt 0$, then the LHS of $(3)$ is negative, and the RHS of $(3)$ is non-negative, so there is no $(x,y)$ satisfying $(3)$. So, we have to have
$$y\ge 0\tag5$$
Under $(4)(5)$, we know that $(3)$ is equivalent to 
$$y^2=(\sqrt x+\sqrt{3-x})^2,$$
i.e.
$$y^2-3=2\sqrt{x(3-x)}\tag6$$
Here, if $y^2-3\lt 0$, then the LHS of $(6)$ is negative, and the RHS of $(6)$ is non-negative, so there is no $(x,y)$ satisfying $(6)$. So, we have to have
$$y^2-3\ge 0\tag7$$
Under $(4)(7)$, we know that $(6)$ is equivalent to
$$(y^2-3)^2=\left(2\sqrt{x(3-x)}\right)^2,$$
i.e.
$$4x^2-12x+y^4-6y^2+9=0\tag8$$
We want $(8)$ to have at least one real root $x$, so we have to have that the discriminant is non-negative :
$$(-12)^2-4\cdot 4(y^4-6y^2+9)\ge 0,$$
i.e.
$$-\sqrt 6\le y\le \sqrt 6\tag9$$ 
As a result, if there exists a pair of $(x,y)$ such that $y=\sqrt x+\sqrt{3-x}$, we have to have $(5)(7)(9)$, i.e. $\sqrt 3\le y\le\sqrt 6.$ $\blacksquare$
Proof for (2) (sufficiency) :
As we've seen in the proof for (1), if $\sqrt 3\le y\le \sqrt 6$, then $(3)$ is equivalent to $(8)$. 
So, it is sufficient to prove that if $\sqrt 3\le y\le\sqrt 6$, then there exists at least one real $x$ where $0\le x\le 3$ such that $(8)$.
By the way, if $\sqrt 3\le y\le \sqrt 6$, then we have $(-12)^2-4\cdot 4(y^4-6y^2+9)\ge 0$, and so
$$\begin{align}&0\le \frac{12-\sqrt{(-12)^2-4\cdot 4(y^4-6y^2+9)}}{8}\le 3\tag{10}\\\\&\iff 0\le\frac{3-\sqrt{-y^4+6y^2}}{2}\le 3\\\\&\iff 0\le 3-\sqrt{-y^4+6y^2}\le 6\\\\&\iff -3\le -\sqrt{-y^4+6y^2}\le 3\\\\&\iff -3\le\sqrt{-y^4+6y^2}\le 3\\\\&\iff -y^4+6y^2\le 3^2\\\\&\iff (y^2-3)^2\ge 0\end{align}$$
This holds for every $y$ satisfying $\sqrt 3\le y\le\sqrt 6$. So, $(10)$ holds for every $y$ satisfying $\sqrt 3\le y\le\sqrt 6$. $\blacksquare$
