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 $0 \leq x \leq 3$

$4x^2-12x+y^4-6y^2+9 = 0$

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 $\sqrt 3\le y\le\sqrt 6$.

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 $x=\frac{3}{2}$.

$f(x) = \sqrt{x} + \sqrt{3-x}$. 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 $[\sqrt{3}, \sqrt{6}]$.