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!

Learn how to find the range of the function f(x) = sqrt(3x - x^2 - 2) through step-by-step solving of the quadratic inequality and evaluating the domain. Understand key algebraic concepts such as quadratic factoring, square roots, and determining the domain and range. Suitable for students in Grades 10-12.

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