To find the range of the function $f(x) = \sqrt{3x - x^2 - 2}$, we need to ensure that the expression inside the square root is non-negative, as the square root of a negative number is not real.

Step 1: Find the domain of the function

The expression inside the square root must satisfy:

$3x - x^2 - 2 \geq 0$

This inequality can be rewritten as:

$-x^2 + 3x - 2 \geq 0$

Multiplying through by $-1$ (which reverses the inequality):

$x^2 - 3x + 2 \leq 0$

Step 2: Solve the quadratic inequality

First, solve the corresponding quadratic equation:

$x^2 - 3x + 2 = 0$

Factor the quadratic equation:

$(x - 1)(x - 2) = 0$

The solutions are:

$x = 1 \quad \text{and} \quad x = 2$

Now, test intervals determined by these roots:

1. For $x < 1$, pick $x = 0$: $(0 - 1)(0 - 2) > 0$ so it's outside the range.
2. For $1 \leq x \leq 2$, pick $x = 1.5$: the quadratic expression is negative within this interval.
3. For $x > 2$: leads to positive outcomes, out of range.

Thus, the function's domain is the interval $[1, 2]$.

Step 3: Determine the range

Evaluate the function at the endpoints of the domain.

1. $f(1) = \sqrt{3(1) - 1^2 - 2} = \sqrt{3 - 1 - 2} = \sqrt{0} = 0$
2. $f(2) = \sqrt{3(2) - 2^2 - 2} = \sqrt{6 - 4 - 2} = \sqrt{0} = 0$

The maximum value occurs midrange to maximized, the function’s highest is max value remains 0.

Thus, the range of the function is $[0,0]$.

Any questions or want any more examples here is help:

1. In step verifying methodic?
2. Can inequalities solved thus?
3. Possible max methods locating so.
4. What-other solve roots?
5. Can quadratic-factor checked?

Tip: If quadratics are hard remember make-solving outputs easier!