Find the range of $y = \sqrt{x} + \sqrt{3 -x}$ I have the function $y = \sqrt{x} + \sqrt{3 -x}$. The range in wolfram is $y \in\mathbb R:  \sqrt{3} \leq y \leq  \sqrt{6}$
(solution after correction of @mathlove)
$\sqrt{x} + \sqrt{3 -x} = y$
$$
\begin{cases}
x \geq 0\\
x \leq 3
\end{cases}
$$
then 
$(\sqrt{x} + \sqrt{3 -x})^2 = y^2$
$x+3-x+2\sqrt{x(3-x)}=y^2$
$2\sqrt{x(3-x)}=y^2-3$
irrational equation, therefore:
$$
\begin{cases}
y^2-3 \geq 0\\
4(3x-x^2)=(y^2-3)^2
\end{cases}
$$
The $y^2 \geq 3$ is verified when $y \leq -\sqrt{3}$ or $y \geq \sqrt{3}$
regarding the second element of the system $4(3x-x^2)=(y^2-3)^2$
$4(3x-x^2)=(y^2-3)^2$
$12x-4x^2=y^4+9-6y^2$
$12x-4x^2-y^4-9+6y^2 = 0$
$12x-4x^2-y^4-9+6y^2 = 0$
$4x^2-12x+y^4-6y^2+9 = 0$
the quadratic equation is verified when the discriminant is $\geq 0$, then
$b^2 - 4ac = (-12)^2-16(y^4-6y^2+9) \geq 0$
$144-16y^4+96y^2-144 \geq 0$
$16y^4-96y^2 \leq 0$
change $t=y^2$ and $t^2=y^4$
$16t^2-96t \leq 0$
the inequality is verified when $t_1 \leq t \leq t_2$ because the discriminant in t is $\geq 0$
$t(16t^2-96) = 0$
then
$t_1 = 0$ and $t_2 = 6$ but $t=y^2$ and 
$\sqrt(0) \leq y \leq \sqrt{6}$
$-\sqrt{6} \leq y \leq \sqrt{6}$
Finally, the solutions in $y$ in the system are:
$y\leq -\sqrt{3}$ or $y\geq \sqrt{3}$ 
and
$-\sqrt{6} \leq y \leq \sqrt{6}$
the solutions in $y$ (=range) of the system is:
$\sqrt{3} \leq y \leq \sqrt{6}$ because the function in the domain is satisfy only for $0 \leq x \leq 3$ 

Note that $y$ is positive, and $$y^2=3+2\sqrt{x(3-x)}$$ It's clear from this that a minimum value for $y$ is $\sqrt{3}$. 
Meanwhile, the largest that $x(3-x)$ can get is $\frac32\left(3-\frac32\right)$, since the expression $x(3-x)$ is quadratic with negative leading coefficient, and $x=\frac32$ is the axis of symmetry. So the maximum that $y^2$ can be is $3+2\sqrt{\frac32\left(3-\frac32\right)}=6$. Again since $y$ is positive, $y$ can maximally be $\sqrt{6}$.
Lastly since the function is continuous, the range is $\left[\sqrt{3},\sqrt{6}\right]$, with $\sqrt{3}$ achieved at $x=0$ and $x=3$, and $\sqrt{6}$ achieved at $x=\frac{3}{2}$.
