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 $0 \leq x \leq 3$ 
 A: 
$4x^2-12x+y^4-6y^2+9 = 0$
the quadratic equation is verified when the discriminant is $>= 0$, then
$b^2 - 4ac = (-12)^2-4(y^4-6y^2+9) >= 0$

This is the part where you have an error :
$$b^2-4ac=(-12)^2-4\cdot 4(y^4-6y^2+9)\ge 0$$
This is equivalent to $-\sqrt 6\le y\le \sqrt 6$.
With $y^2-3\ge 0$ and $y\ge 0$, the range is $\sqrt 3\le y\le\sqrt 6$.
A: 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 $x=\frac{3}{2}$.
A: $f(x) = \sqrt{x} + \sqrt{3-x}$.  From this you have correctly deduced that the domain is $[0,3]$.
\begin{align*}
  f'(x) &= \frac{1}{2\sqrt x} - \frac{1}{2\sqrt{3-x}}\\[0.3cm]
    &= \frac{\sqrt{3-x}}{2\sqrt x \sqrt{3-x}} - \frac{\sqrt{x}}{2\sqrt x\sqrt{3-x}}
\end{align*}
$f'(x) = 0$ when $\sqrt{3-x} = \sqrt{x}$, which is when $x = 3/2$.  Also, $f'(x)$ is undefined when $x = 0$ or when $x=3$.  Luckily these coincide with the endpoints of the domain (which we'd also need to check separately if they didn't coincide).
So we have $f(0) = f(3) = \sqrt{3}$, and
$$ f(3/2) = \sqrt\frac{3}{2} + \sqrt\frac{3}{2} = \frac{2\sqrt{3}}{\sqrt{2}} = \sqrt{6}.$$
So the absolute min is $\sqrt{3}$ and the absolute max is $\sqrt{6}$.  Also, since the function is continuous on $[0,3]$ then by the Intermediate Value Theorem, the function attains every value between $\sqrt{3}$ and $\sqrt{6}$ for input values coming from $[0,3]$.  Therefore the range is $[\sqrt{3}, \sqrt{6}]$.
